Course Hero has millions of student submitted documents similar to the one

below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.

Course Hero has millions of student submitted documents similar to the one below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.

Algebra
Jim Linear Hefferon
1 3
2 1
1 3
2 1
x1 1 3
2 1
x1 1 2 x1 3 1
6 8
2 1
6 8
2 1
Notation R N C {. . . . . .} ... V, W, U v, w 0, 0V B, D En = e1 , . . . , en , RepB (v) Pn Mnm [S] M N V W = h, g H, G t, s T, S RepB,D (h) hi,j |T | R(h), N (h) R (h), N (h) real numbers natural numbers: {0, 1, 2, . . .} complex numbers set of . . . such that . . . sequence; like a set but order matters vector spaces vectors zero vector, zero vector of V bases standard basis for Rn basis vectors matrix representing the vector set of n-th degree polynomials set of nm matrices span of the set S direct sum of subspaces isomorphic spaces homomorphisms, linear maps matrices transformations; maps from a space to itself square matrices matrix representing the map h matrix entry from row i, column j determinant of the matrix T rangespace and nullspace of the map h generalized rangespace and nullspace
Lower case Greek alphabet name alpha beta gamma delta epsilon zeta eta theta character name iota kappa lambda mu nu xi omicron pi character o name rho sigma tau upsilon phi chi psi omega character
Cover. This is Cramers Rule for the system x1 + 2x2 = 6, 3x1 + x2 = 8. The size of the rst box is the determinant shown (the absolute value of the size is the area). The size of the second box is x1 times that, and equals the size of the nal box. Hence, x1 is the nal determinant divided by the rst determinant.
Contents
Chapter One: Linear Systems I Solving Linear Systems . . . . . . . . . . 1 Gauss Method . . . . . . . . . . . . . 2 Describing the Solution Set . . . . . . 3 General = Particular + Homogeneous II Linear Geometry of n-Space . . . . . . . 1 Vectors in Space . . . . . . . . . . . . 2 Length and Angle Measures . . . . . III Reduced Echelon Form . . . . . . . . . . 1 Gauss-Jordan Reduction . . . . . . . . 2 Row Equivalence . . . . . . . . . . . . Topic: Computer Algebra Systems . . . . . Topic: Input-Output Analysis . . . . . . . . Topic: Accuracy of Computations . . . . . . Topic: Analyzing Networks . . . . . . . . . . Chapter Two: Vector Spaces I Denition of Vector Space . . . . . . 1 Denition and Examples . . . . . . 2 Subspaces and Spanning Sets . . . II Linear Independence . . . . . . . . . 1 Denition and Examples . . . . . . III Basis and Dimension . . . . . . . . . 1 Basis . . . . . . . . . . . . . . . . . 2 Dimension . . . . . . . . . . . . . . 3 Vector Spaces and Linear Systems 4 Combining Subspaces . . . . . . . Topic: Fields . . . . . . . . . . . . . . . . Topic: Crystals . . . . . . . . . . . . . . Topic: Dimensional Analysis . . . . . . . vii 1 1 2 11 20 32 32 38 46 46 52 62 64 68 72 79 80 80 91 102 102 113 113 119 124 131 141 143 147
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . .
. . . . . . . . . . . . .
. . . . . . . . . . . . .
. . . . . . . . . . . . .
. . . . . . . . . . . . .
. . . . . . . . . . . . .
. . . . . . . . . . . . .
. . . . . . . . . . . . .
. . . . . . . . . . . . .
. . . . . . . . . . . . .
. . . . . . . . . . . . .
. . . . . . . . . . . . .
. . . . . . . . . . . . .
. . . . . . . . . . . . .
. . . . . . . . . . . . .
. . . . . . . . . . . . .
Chapter Three: Maps Between Spaces I Isomorphisms . . . . . . . . . . . . . . . . 1 Definition and Examples . . . . . . . . . 2 Dimension Characterizes Isomorphism . II Homomorphisms . . . . . . . . . . . . . . 1 Denition . . . . . . . . . . . . . . . . . 2 Rangespace and Nullspace . . . . . . . . III Computing Linear Maps . . . . . . . . . . 1 Representing Linear Maps with Matrices 2 Any Matrix Represents a Linear Map . IV Matrix Operations . . . . . . . . . . . . . 1 Sums and Scalar Products . . . . . . . . 2 Matrix Multiplication . . . . . . . . . . 3 Mechanics of Matrix Multiplication . . . 4 Inverses . . . . . . . . . . . . . . . . . . V Change of Basis . . . . . . . . . . . . . . . 1 Changing Representations of Vectors . . 2 Changing Map Representations . . . . . VI Projection . . . . . . . . . . . . . . . . . . 1 Orthogonal Projection Into a Line . . . 2 Gram-Schmidt Orthogonalization . . . 3 Projection Into a Subspace . . . . . . . Topic: Line of Best Fit . . . . . . . . . . . . . Topic: Geometry of Linear Maps . . . . . . . Topic: Markov Chains . . . . . . . . . . . . . Topic: Orthonormal Matrices . . . . . . . . . Chapter Four: Determinants I Definition . . . . . . . . . . . . . . . . 1 Exploration . . . . . . . . . . . . . 2 Properties of Determinants . . . . . 3 The Permutation Expansion . . . . . 4 Determinants Exist . . . . . . . . . II Geometry of Determinants . . . . . . . 1 Determinants as Size Functions . . . III Other Formulas . . . . . . . . . . . . . 1 Laplaces Expansion . . . . . . . . . Topic: Cramers Rule . . . . . . . . . . . . Topic: Speed of Calculating Determinants Topic: Projective Geometry . . . . . . . . Chapter Five: Similarity I Complex Vector Spaces . . . . . . . 1 Factoring and Complex Numbers; 2 Complex Representations . . . . II Similarity . . . . . . . . . . . . . . viii . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
155 155 155 164 172 172 179 191 191 201 208 208 210 218 227 234 234 238 246 246 250 256 265 270 277 283 289 290 290 295 299 308 315 315 322 322 327 330 333 345 345 346 347 349
. . . . . . . A Review . . . . . . . . . . . . . .
1 Denition and Examples . . . . . . 2 Diagonalizability . . . . . . . . . . 3 Eigenvalues and Eigenvectors . . . III Nilpotence . . . . . . . . . . . . . . . 1 Self-Composition . . . . . . . . . 2 Strings . . . . . . . . . . . . . . . IV Jordan Form . . . . . . . . . . . . . . 1 Polynomials of Maps and Matrices 2 Jordan Canonical Form . . . . . . Topic: Method of Powers . . . . . . . . . Topic: Stable Populations . . . . . . . . Topic: Linear Recurrences . . . . . . . . Appendix Propositions . . . . . . . . . . Quantiers . . . . . . . . . . Techniques of Proof . . . . . Sets, Functions, and Relations
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
349 351 355 363 363 366 377 377 384 397 401 403 A-1 A-1 A-3 A-5 A-7
. . . .
. . . .
. . . .
. . . .
. . . .
Note: starred subsections are optional.
ix
Chapter One
Linear Systems
I Solving Linear Systems
Systems of linear equations are common in science and mathematics. These two examples from high school science [Onan] give a sense of how they arise. The rst example is from Physics. Suppose that we are given three objects, one with a mass known to be 2 kg, and are asked to nd the unknown masses. Suppose further that experimentation with a meter stick produces these two balances.
40 50 25 50
h
c
15
2
c
25
2
h
Since the sum of moments on the left of each balance equals the sum of moments on the right (the moment of an object is its mass times its distance from the balance point), the two balances give this system of two equations. 40h + 15c = 100 25c = 50 + 50h The second example of a linear system is from Chemistry. We can mix, under controlled conditions, toluene C7 H8 and nitric acid HNO3 to produce trinitrotoluene C7 H5 O6 N3 along with the byproduct water (conditions have to be controlled very well, indeed trinitrotoluene is better known as TNT). In what proportion should those components be mixed? The number of atoms of each element present before the reaction x C7 H8 + y HNO3 z C7 H5 O6 N3 + w H2 O
must equal the number present afterward. Applying that principle to the ele1
2 ments C, H, N, and O in turn gives this system. 7x = 7z 8x + 1y = 5z + 2w 1y = 3z 3y = 6z + 1w
Chapter One. Linear Systems
To nish each of these examples requires solving a system of equations. In each, the equations involve only the rst power of the variables. This chapter shows how to solve any such system.
I.1 Gauss Method
1.1 Denition A linear equation in variables x1 , x2 , . . . , xn has the form a1 x1 + a2 x2 + a3 x3 + + an xn = d where the numbers a1 , . . . , an R are the equations coecients and d R is the constant. An n-tuple (s1 , s2 , . . . , sn ) Rn is a solution of, or satises, that equation if substituting the numbers s1 , . . . , sn for the variables gives a true statement: a1 s1 + a2 s2 + . . . + an sn = d. A system of linear equations a1,1 x1 + a1,2 x2 + + a1,n xn = d1 a2,1 x1 + a2,2 x2 + + a2,n xn = d2 . . . am,1 x1 + am,2 x2 + + am,n xn = dm has the solution (s1 , s2 , . . . , sn ) if that n-tuple is a solution of all of the equations in the system. 1.2 Example The ordered pair (1, 5) is a solution of this system. 3x1 + 2x2 = 7 x1 + x2 = 6 In contrast, (5, 1) is not a solution. Finding the set of all solutions is solving the system. No guesswork or good fortune is needed to solve a linear system. There is an algorithm that always works. The next example introduces that algorithm, called Gauss method. It transforms the system, step by step, into one with a form that is easily solved.
Section I. Solving Linear Systems 1.3 Example To solve this system 3x3 = 9 x1 + 5x2 2x3 = 2 1 =3 3 x1 + 2x2 we repeatedly transform it until it is in a form that is easy to solve.
swap row 1 with row 3 1 3 x1 x1
3
+ 2x2 =3 + 5x2 2x3 = 2 3x3 = 9
multiply row 1 by 3
x1 + 6x2 =9 x1 + 5x2 2x3 = 2 3x3 = 9 x1 + 6x2 = 9 x2 2x3 = 7 3x3 = 9
add 1 times row 1 to row 2
The third step is the only nontrivial one. Weve mentally multiplied both sides of the rst row by 1, mentally added that to the old second row, and written the result in as the new second row. Now we can nd the value of each variable. The bottom equation shows that x3 = 3. Substituting 3 for x3 in the middle equation shows that x2 = 1. Substituting those two into the top equation gives that x1 = 3 and so the system has a unique solution: the solution set is { (3, 1, 3) }. Most of this subsection and the next one consists of examples of solving linear systems by Gauss method. We will use it throughout this book. It is fast and easy. But, before we get to those examples, we will rst show that this method is also safe in that it never loses solutions or picks up extraneous solutions. 1.4 Theorem (Gauss method) If a linear system is changed to another by one of these operations (1) an equation is swapped with another (2) an equation has both sides multiplied by a nonzero constant (3) an equation is replaced by the sum of itself and a multiple of another then the two systems have the same set of solutions. Each of those three operations has a restriction. Multiplying a row by 0 is not allowed because obviously that can change the solution set of the system. Similarly, adding a multiple of a row to itself is not allowed because adding 1 times the row to itself has the eect of multiplying the row by 0. Finally, swapping a row with itself is disallowed to make some results in the fourth chapter easier to state and remember (and besides, self-swapping doesnt accomplish anything).
4
Chapter One. Linear Systems
Proof. We will cover the equation swap operation here and save the other two
cases for Exercise 29. Consider this swap of row i with row j. a1,1 x1 + a1,2 x2 + a1,n xn = d1 a1,1 x1 + a1,2 x2 + . . . aj,1 x1 + aj,2 x2 + ai,1 x1 + ai,2 x2 + ai,n xn = di . . . ai,1 x1 + ai,2 x2 + aj,1 x1 + aj,2 x2 + aj,n xn = dj . . . am,1 x1 + am,2 x2 + am,n xn = dm am,1 x1 + am,2 x2 + a1,n xn = d1 . . . aj,n xn = dj . . . ai,n xn = di . . . am,n xn = dm
The n-tuple (s1 , . . . , sn ) satises the system before the swap if and only if substituting the values, the ss, for the variables, the xs, gives true statements: a1,1 s1 +a1,2 s2 + +a1,n sn = d1 and . . . ai,1 s1 +ai,2 s2 + +ai,n sn = di and . . . aj,1 s1 + aj,2 s2 + + aj,n sn = dj and . . . am,1 s1 + am,2 s2 + + am,n sn = dm . In a requirement consisting of statements and-ed together we can rearrange the order of the statements, so that this requirement is met if and only if a1,1 s1 + a1,2 s2 + + a1,n sn = d1 and . . . aj,1 s1 + aj,2 s2 + + aj,n sn = dj and . . . ai,1 s1 + ai,2 s2 + + ai,n sn = di and . . . am,1 s1 + am,2 s2 + + am,n sn = dm . This is exactly the requirement that (s1 , . . . , sn ) solves the system after the row swap. QED 1.5 Denition The three operations from Theorem 1.4 are the elementary reduction operations, or row operations, or Gaussian operations. They are swapping, multiplying by a scalar or rescaling, and pivoting. When writing out the calculations, we will abbreviate row i by i . For instance, we will denote a pivot operation by ki + j , with the row that is changed written second. We will also, to save writing, often list pivot steps together when they use the same i . 1.6 Example A typical use of Gauss method is to solve this system. x+ y =0 2x y + 3z = 3 x 2y z = 3 The rst transformation of the system involves using the rst row to eliminate the x in the second row and the x in the third. To get rid of the second rows 2x, we multiply the entire rst row by 2, add that to the second row, and write the result in as the new second row. To get rid of the third rows x, we multiply the rst row by 1, add that to the third row, and write the result in as the new third row.
21 +2 1 +3
x+
y =0 3y + 3z = 3 3y z = 3
Section I. Solving Linear Systems
5
(Note that the two 1 steps 21 + 2 and 1 + 3 are written as one operation.) In this second system, the last two equations involve only two unknowns. To nish we transform the second system into a third system, where the last equation involves only one unknown. This transformation uses the second row to eliminate y from the third row.
2 +3
x+
y 3y +
=0 3z = 3 4z = 0
Now we are set up for the solution. The third row shows that z = 0. Substitute that back into the second row to get y = 1, and then substitute back into the rst row to get x = 1. 1.7 Example For the Physics problem from the start of this chapter, Gauss method gives this. 40h + 15c = 100 50h + 25c = 50
5/41 +2
40h +
15c = 100 (175/4)c = 175
So c = 4, and back-substitution gives that h = 1. (The Chemistry problem is solved later.) 1.8 Example The reduction x+ y+ z=9 2x + 4y 3z = 1 3x + 6y 5z = 0
21 +2 31 +3
x+ y+ z= 9 2y 5z = 17 3y 8z = 27 x+ y+ 2y z= 9 5z = 17 (1/2)z = (3/2)
(3/2)2 +3
shows that z = 3, y = 1, and x = 7. As these examples illustrate, Gauss method uses the elementary reduction operations to set up back-substitution. 1.9 Denition In each row, the rst variable with a nonzero coecient is the rows leading variable. A system is in echelon form if each leading variable is to the right of the leading variable in the row above it (except for the leading variable in the rst row). 1.10 Example The only operation needed in the examples above is pivoting. Here is a linear system that requires the operation of swapping equations. After the rst pivot x y =0 2x 2y + z + 2w = 4 y + w=0 2z + w = 5 xy
21 +2
=0 z + 2w = 4 y + w=0 2z + w = 5
6
Chapter One. Linear Systems
the second equation has no leading y. To get one, we look lower down in the system for a row that has a leading y and swap it in.
2 3
xy y
=0 + w=0 z + 2w = 4 2z + w = 5
(Had there been more than one row below the second with a leading y then we could have swapped in any one.) The rest of Gauss method goes as before.
23 +4
xy y
+ z+
= 0 w= 0 2w = 4 3w = 3
Back-substitution gives w = 1, z = 2 , y = 1, and x = 1. Strictly speaking, the operation of rescaling rows is not needed to solve linear systems. We have included it because we will use it later in this chapter as part of a variation on Gauss method, the Gauss-Jordan method. All of the systems seen so far have the same number of equations as unknowns. All of them have a solution, and for all of them there is only one solution. We nish this subsection by seeing for contrast some other things that can happen. 1.11 Example Linear systems need not have the same number of equations as unknowns. This system x + 3y = 1 2x + y = 3 2x + 2y = 2 has more equations than variables. Gauss method helps us understand this system also, since this
21 +2 21 +3
x + 3y = 1 5y = 5 4y = 4
shows that one of the equations is redundant. Echelon form
(4/5)2 +3
x + 3y = 1 5y = 5 0= 0
gives y = 1 and x = 2. The 0 = 0 is derived from the redundancy. That examples system has more equations than variables. Gauss method is also useful on systems with more variables than equations. Many examples are in the next subsection.
Section I. Solving Linear Systems
7
Another way that linear systems can dier from the examples shown earlier is that some linear systems do not have a unique solution. This can happen in two ways. The rst is that it can fail to have any solution at all. 1.12 Example Contrast the system in the last example with this one. x + 3y = 1 2x + y = 3 2x + 2y = 0
21 +2 21 +3
x+
3y = 1 5y = 5 4y = 2
Here the system is inconsistent: no pair of numbers satises all of the equations simultaneously. Echelon form makes this inconsistency obvious.
(4/5)2 +3
x + 3y = 1 5y = 5 0= 2
The solution set is empty. 1.13 Example The prior system has more equations than unknowns, but that is not what causes the inconsistency Example 1.11 has more equations than unknowns and yet is consistent. Nor is having more equations than unknowns necessary for inconsistency, as is illustrated by this inconsistent system with the same number of equations as unknowns. x + 2y = 8 2x + 4y = 8
21 +2
x + 2y = 8 0 = 8
The other way that a linear system can fail to have a unique solution is to have many solutions. 1.14 Example In this system x+ y=4 2x + 2y = 8 any pair of numbers satisfying the rst equation automatically satises the second. The solution set {(x, y) x + y = 4} is innite; some of its members are (0, 4), (1, 5), and (2.5, 1.5). The result of applying Gauss method here contrasts with the prior example because we do not get a contradictory equation.
21 +2
x+y=4 0=0
Dont be fooled by the 0 = 0 equation in that example. It is not the signal that a system has many solutions.
8
Chapter One. Linear Systems
1.15 Example The absence of a 0 = 0 does not keep a system from having many dierent solutions. This system is in echelon form x+y+z=0 y+z=0 has no 0 = 0, and yet has innitely many solutions. (For instance, each of these is a solution: (0, 1, 1), (0, 1/2, 1/2), (0, 0, 0), and (0, , ). There are innitely many solutions because any triple whose rst component is 0 and whose second component is the negative of the third is a solution.) Nor does the presence of a 0 = 0 mean that the system must have many solutions. Example 1.11 shows that. So does this system, which does not have many solutions in fact it has none despite that when it is brought to echelon form it has a 0 = 0 row. 2z = 6 y+ z=1 2x + y z = 7 3y + 3z = 0 2x 2x
1 +3
2z = 6 y+ z=1 y+ z=1 3y + 3z = 0 2z = 6 y+ z= 1 0= 0 0 = 3
2x
2 +3 32 +4
We will nish this subsection with a summary of what weve seen so far about Gauss method. Gauss method uses the three row operations to set a system up for back substitution. If any step shows a contradictory equation then we can stop with the conclusion that the system has no solutions. If we reach echelon form without a contradictory equation, and each variable is a leading variable in its row, then the system has a unique solution and we nd it by back substitution. Finally, if we reach echelon form without a contradictory equation, and there is not a unique solution (at least one variable is not a leading variable) then the system has many solutions. The next subsection deals with the third case we will see how to describe the solution set of a system with many solutions. Exercises
1.16 Use Gauss method to nd the unique solution for each system. x z=0 2x + 3y = 13 =1 (a) (b) 3x + y x y = 1 x + y + z = 4 1.17 Use Gauss method to solve each system or conclude many solutions or no solutions.
Section I. Solving Linear Systems
9
(a) 2x + 2y = 5 (b) x + y = 1 (c) x 3y + z = 1 x 4y = 0 x+y=2 x + y + 2z = 14 (d) x y = 1 (e) 4y + z = 20 (f ) 2x + z+w= 5 3x 3y = 2 2x 2y + z = 0 y w = 1 x +z= 5 3x zw= 0 x + y z = 10 4x + y + 2z + w = 9 1.18 There are methods for solving linear systems other than Gauss method. One often taught in high school is to solve one of the equations for a variable, then substitute the resulting expression into other equations. That step is repeated until there is an equation with only one variable. From that, the rst number in the solution is derived, and then back-substitution can be done. This method takes longer than Gauss method, since it involves more arithmetic operations, and is also more likely to lead to errors. To illustrate how it can lead to wrong conclusions, we will use the system x + 3y = 1 2x + y = 3 2x + 2y = 0 from Example 1.12. (a) Solve the rst equation for x and substitute that expression into the second equation. Find the resulting y. (b) Again solve the rst equation for x, but this time substitute that expression into the third equation. Find this y. What extra step must a user of this method take to avoid erroneously concluding a system has a solution? 1.19 For which values of k are there no solutions, many solutions, or a unique solution to this system? x y=1 3x 3y = k 1.20 This system is not linear, in some sense, 2 sin cos + 3 tan = 3 4 sin + 2 cos 2 tan = 10 6 sin 3 cos + tan = 9 and yet we can nonetheless apply Gauss method. Do so. Does the system have a solution? 1.21 What conditions must the constants, the bs, satisfy so that each of these systems has a solution? Hint. Apply Gauss method and see what happens to the right side. [Anton] (a) x 3y = b1 (b) x1 + 2x2 + 3x3 = b1 3x + y = b2 2x1 + 5x2 + 3x3 = b2 x + 7y = b3 x1 + 8x3 = b3 2x + 4y = b4 1.22 True or false: a system with more unknowns than equations has at least one solution. (As always, to say true you must prove it, while to say false you must produce a counterexample.) 1.23 Must any Chemistry problem like the one that starts this subsection a balance the reaction problem have innitely many solutions? 1.24 Find the coecients a, b, and c so that the graph of f (x) = ax2 + bx + c passes through the points (1, 2), (1, 6), and (2, 3).
10
Chapter One. Linear Systems
1.25 Gauss method works by combining the equations in a system to make new equations. (a) Can the equation 3x2y = 5 be derived, by a sequence of Gaussian reduction steps, from the equations in this system? x+y=1 4x y = 6 (b) Can the equation 5x3y = 2 be derived, by a sequence of Gaussian reduction steps, from the equations in this system? 2x + 2y = 5 3x + y = 4 (c) Can the equation 6x 9y + 5z = 2 be derived, by a sequence of Gaussian reduction steps, from the equations in the system? 2x + y z = 4 6x 3y + z = 5 1.26 Prove that, where a, b, . . . , e are real numbers and a = 0, if ax + by = c has the same solution set as ax + dy = e then they are the same equation. What if a = 0? 1.27 Show that if ad bc = 0 then ax + by = j cx + dy = k has a unique solution. 1.28 In the system ax + by = c dx + ey = f each of the equations describes a line in the xy-plane. By geometrical reasoning, show that there are three possibilities: there is a unique solution, there is no solution, and there are innitely many solutions. 1.29 Finish the proof of Theorem 1.4. 1.30 Is there a two-unknowns linear system whose solution set is all of R2 ? 1.31 Are any of the operations used in Gauss method redundant? That is, can any of the operations be synthesized from the others? 1.32 Prove that each operation of Gauss method is reversible. That is, show that if two systems are related by a row operation S1 S2 then there is a row operation to go back S2 S1 .
? 1.33 A box holding pennies, nickels and dimes contains thirteen coins with a total value of 83 cents. How many coins of each type are in the box? [Anton] ? 1.34 Four positive integers are given. Select any three of the integers, nd their arithmetic average, and add this result to the fourth integer. Thus the numbers 29, 23, 21, and 17 are obtained. One of the original integers is:
Section I. Solving Linear Systems
11
(a) 19 (b) 21 (c) 23 (d) 29 (e) 17 [Con. Prob. 1955] ? 1.35 Laugh at this: AHAHA + TEHE = TEHAW. It resulted from substituting a code letter for each digit of a simple example in addition, and it is required to identify the letters and prove the solution unique. [Am. Math. Mon., Jan. 1935] ? 1.36 The Wohascum County Board of Commissioners, which has 20 members, recently had to elect a President. There were three candidates (A, B, and C); on each ballot the three candidates were to be listed in order of preference, with no abstentions. It was found that 11 members, a majority, preferred A over B (thus the other 9 preferred B over A). Similarly, it was found that 12 members preferred C over A. Given these results, it was suggested that B should withdraw, to enable a runo election between A and C. However, B protested, and it was then found that 14 members preferred B over C! The Board has not yet recovered from the resulting confusion. Given that every possible order of A, B, C appeared on at least one ballot, how many members voted for B as their rst choice? [Wohascum no. 2] ? 1.37 This system of n linear equations with n unknowns, said the Great Mathematician, has a curious property. Good heavens! said the Poor Nut, What is it? Note, said the Great Mathematician, that the constants are in arithmetic progression. Its all so clear when you explain it! said the Poor Nut. Do you mean like 6x + 9y = 12 and 15x + 18y = 21? Quite so, said the Great Mathematician, pulling out his bassoon. Indeed, the system has a unique solution. Can you nd it? Good heavens! cried the Poor Nut, I am baed. Are you? [Am. Math. Mon., Jan. 1963]
I.2 Describing the Solution Set
A linear system with a unique solution has a solution set with one element. A linear system with no solution has a solution set that is empty. In these cases the solution set is easy to describe. Solution sets are a challenge to describe only when they contain many elements. 2.1 Example This system has many solutions because in echelon form 2x +z=3 xyz=1 3x y =4
(1/2)1 +2 (3/2)1 +3
2x
+ z= 3 y (3/2)z = 1/2 y (3/2)z = 1/2 + z= 3 y (3/2)z = 1/2 0= 0
2 +3
2x
not all of the variables are leading variables. The Gauss method theorem showed that a triple satises the rst system if and only if it satises the third. Thus, the solution set {(x, y, z) 2x + z = 3 and x y z = 1 and 3x y = 4}
12
Chapter One. Linear Systems
can also be described as {(x, y, z) 2x + z = 3 and y 3z/2 = 1/2}. However, this second description is not much of an improvement. It has two equations instead of three, but it still involves some hard-to-understand interaction among the variables. To get a description that is free of any such interaction, we take the variable that does not lead any equation, z, and use it to describe the variables that do lead, x and y. The second equation gives y = (1/2) (3/2)z and the rst equation gives x = (3/2) (1/2)z. Thus, the solution set can be described as {(x, y, z) = ((3/2) (1/2)z, (1/2) (3/2)z, z) z R}. For instance, (1/2, 5/2, 2) is a solution because taking z = 2 gives a rst component of 1/2 and a second component of 5/2. The advantage of this description over the ones above is that the only variable appearing, z, is unrestricted it can be any real number. 2.2 Denition The non-leading variables in an echelon-form linear system are free variables. In the echelon form system derived in the above example, x and y are leading variables and z is free. 2.3 Example A linear system can end with more than one variable free. This row reduction x+ y+ z w= 1 y z + w = 1 3x + 6z 6w = 6 y + z w = 1 x+
31 +3
y+ z w= 1 y z + w = 1 3y + 3z 3w = 3 y + z w = 1
32 +3 2 +4
x+y+zw= 1 y z + w = 1 0= 0 0= 0
ends with x and y leading, and with both z and w free. To get the description that we prefer we will start at the bottom. We rst express y in terms of the free variables z and w with y = 1 + z w. Next, moving up to the top equation, substituting for y in the rst equation x + (1 + z w) + z w = 1 and solving for x yields x = 2 2z + 2w. Thus, the solution set is {2 2z + 2w, 1 + z w, z, w) z, w R}. We prefer this description because the only variables that appear, z and w, are unrestricted. This makes the job of deciding which four-tuples are system solutions into an easy one. For instance, taking z = 1 and w = 2 gives the solution (4, 2, 1, 2). In contrast, (3, 2, 1, 2) is not a solution, since the rst component of any solution must be 2 minus twice the third component plus twice the fourth.
Section I. Solving Linear Systems 2.4 Example After this reduction =0 z + 3w = 2 3x 3y =0 x y + 2z + 6w = 4 2x 2y 2x 2y
(3/2)1 +3 (1/2)1 +4
13
=0 z + 3w = 2 0=0 2z + 6w = 4 =0 z + 3w = 2 0=0 0=0
2x 2y
22 +4
x and z lead, y and w are free. The solution set is {(y, y, 2 3w, w) y, w R}. For instance, (1, 1, 2, 0) satises the system take y = 1 and w = 0. The fourtuple (1, 0, 5, 4) is not a solution since its rst coordinate does not equal its second. We refer to a variable used to describe a family of solutions as a parameter and we say that the set above is paramatrized with y and w. (The terms parameter and free variable do not mean the same thing. Above, y and w are free because in the echelon form system they do not lead any row. They are parameters because they are used in the solution set description. We could have instead paramatrized with y and z by rewriting the second equation as w = 2/3 (1/3)z. In that case, the free variables are still y and w, but the parameters are y and z. Notice that we could not have paramatrized with x and y, so there is sometimes a restriction on the choice of parameters. The terms parameter and free are related because, as we shall show later in this chapter, the solution set of a system can always be paramatrized with the free variables. Consequenlty, we shall paramatrize all of our descriptions in this way.) 2.5 Example This is another system with innitely many solutions. x + 2y =1 2x +z =2 3x + 2y + z w = 4
21 +2 31 +3
x+
2y =1 4y + z =0 4y + z w = 1 2y 4y + z =1 =0 w = 1
2 +3
x+
The leading variables are x, y, and w. The variable z is free. (Notice here that, although there are innitely many solutions, the value of one of the variables is xed w = 1.) Write w in terms of z with w = 1 + 0z. Then y = (1/4)z. To express x in terms of z, substitute for y into the rst equation to get x = 1 (1/2)z. The solution set is {(1 (1/2)z, (1/4)z, z, 1) z R}. We nish this subsection by developing the notation for linear systems and their solution sets that we shall use in the rest of this book. 2.6 Denition An mn matrix is a rectangular array of numbers with m rows and n columns. Each number in the matrix is an entry,
14
Chapter One. Linear Systems
Matrices are usually named by upper case roman letters, e.g. A. Each entry is denoted by the corresponding lower-case letter, e.g. ai,j is the number in row i and column j of the array. For instance, A= 1 2.2 3 4 5 7
has two rows and three columns, and so is a 2 3 matrix. (Read that twoby-three; the number of rows is always stated rst.) The entry in the second row and rst column is a2,1 = 3. Note that the order of the subscripts matters: a1,2 = a2,1 since a1,2 = 2.2. (The parentheses around the array are a typographic device so that when two matrices are side by side we can tell where one ends and the other starts.) 2.7 Example We can abbreviate this linear system x1 + 2x2 =4 x2 x3 = 0 x1 + 2x3 = 4 with this matrix. 1 2 0 1 1 0 0 1 2 4 0 4
The vertical bar just reminds a reader of the dierence between the coecients on the systemss left hand side and the constants on the right. When a bar is used to divide a matrix into parts, we call it an augmented matrix. In this notation, Gauss method goes this way. 1 2 0 4 1 2 0 4 1 2 0 4 0 1 1 0 1 +3 0 1 1 0 22 +3 0 1 1 0 1 0 2 4 0 2 2 0 0 0 0 0 The second row stands for y z = 0 and the rst row stands for x + 2y = 4 so the solution set is {(4 2z, z, z) z R}. One advantage of the new notation is that the clerical load of Gauss method the copying of variables, the writing of +s and =s, etc. is lighter. We will also use the array notation to clarify the descriptions of solution sets. A description like {(2 2z + 2w, 1 + z w, z, w) z, w R} from Example 2.3 is hard to read. We will rewrite it to group all the constants together, all the coecients of z together, and all the coecients of w together. We will write them vertically, in one-column wide matrices. 2 2 2 1 1 1 { + z + w z, w R} 0 1 0 0 0 1
Section I. Solving Linear Systems
15
For instance, the top line says that x = 2 2z + 2w. The next section gives a geometric interpretation that will help us picture the solution sets when they are written in this way. 2.8 Denition A vector (or column vector ) is a matrix with a single column. A matrix with a single row is a row vector . The entries of a vector are its components. Vectors are an exception to the convention of representing matrices with capital roman letters. We use lower-case roman or greek letters overlined with an arrow: a, b, . . . or , , . . . (boldface is also common: a or ). For instance, this is a column vector with a third component of 7. 1 v = 3 7 2.9 Denition The linear equation a1 x1 + a2 x2 + + an xn = d with unknowns x1 , . . . , xn is satised by s1 . s= . . sn if a1 s1 + a2 s2 + + an sn = d. A vector satises a linear system if it satises each equation in the system. The style of description of solution sets that we use involves adding the vectors, and also multiplying them by real numbers, such as the z and w. We need to dene these operations. 2.10 Denition The vector sum of u and v is this. u1 v1 u1 + v1 . . . . u+v = . + . = . . . un vn u n + vn In general, two matrices with the same number of rows and the same number of columns add in this way, entry-by-entry. 2.11 Denition The scalar multiplication of the real number r and the vector v is this. v1 rv1 . . rv =r . = . . . vn rvn In general, any matrix is multiplied by a real number in this entry-by-entry way.
16
Chapter One. Linear Systems
Scalar multiplication can be written in either order: r v or v r, or without the symbol: rv. (Do not refer to scalar multiplication as scalar product because that name is used for a dierent operation.) 2.12 Example 2 3 2+3 5 3 + 1 = 3 1 = 2 1 4 1+4 5 1 7 4 28 7 = 1 7 3 21
Notice that the denitions of vector addition and scalar multiplication agree where they overlap, for instance, v + v = 2v. With the notation dened, we can now solve systems in the way that we will use throughout this book. 2.13 Example This system 2x + y w =4 y + w+u=4 x z + 2w =0 reduces in 2 1 0 1 1 0 this way. 0 0 1 1 0 1 1 2 0 4 4 0
(1/2)1 +3
(1/2)2 +3
2 1 0 1 0 1/2 2 1 0 0 1 0 0 0 1
0 1 0 0 1 1 1 5/2 0 1 0 1 1 3 1/2
4 4 0
4 4 2
The solution set is {(w + (1/2)u, 4 w u, 3w + (1/2)u, w, u) w, u R}. We write that in vector form. x 0 1 1/2 y 4 1 1 z = 0 + 3 w + 1/2 u w, u R} { w 0 1 0 u 0 0 1 Note again how well vector notation sets o the coecients of each parameter. For instance, the third row of the vector form shows plainly that if u is held xed then z increases three times as fast as w. That format also shows plainly that there are innitely many solutions. For example, we can x u as 0, let w range over the real numbers, and consider the rst component x. We get innitely many rst components and hence innitely many solutions.
Section I. Solving Linear Systems
17
Another thing shown plainly is that setting both w and u to zero gives that this x 0 y 4 z = 0 w 0 u 0 is a particular solution of the linear system. 2.14 Example In the same way, this system x y+ z=1 3x + z=3 5x 2y + 3z = 5 reduces 1 1 1 3 0 1 5 2 3 1 1 1 31 +2 3 0 3 51 +3 5 0 3 1 1 1 2 +3 0 0 3 0 0 0 1 0 0
1 2 2
1 2 0
to a one-parameter solution set. 1 1/3 {0 + 2/3 z z R} 0 1 Before the exercises, we pause to point out some things that we have yet to do. The rst two subsections have been on the mechanics of Gauss method. Except for one result, Theorem 1.4 without which developing the method doesnt make sense since it says that the method gives the right answers we have not stopped to consider any of the interesting questions that arise. For example, can we always describe solution sets as above, with a particular solution vector added to an unrestricted linear combination of some other vectors? The solution sets we described with unrestricted parameters were easily seen to have innitely many solutions so an answer to this question could tell us something about the size of solution sets. An answer to that question could also help us picture the solution sets, in R2 , or in R3 , etc. Many questions arise from the observation that Gauss method can be done in more than one way (for instance, when swapping rows, we may have a choice of which row to swap with). Theorem 1.4 says that we must get the same solution set no matter how we proceed, but if we do Gauss method in two dierent ways must we get the same number of free variables both times, so that any two solution set descriptions have the same number of parameters? Must those be the same variables (e.g., is it impossible to solve a problem one way and get y and w free or solve it another way and get y and z free)?
18
Chapter One. Linear Systems
In the rest of this chapter we answer these questions. The answer to each is yes. The rst question is answered in the last subsection of this section. In the second section we give a geometric description of solution sets. In the nal section of this chapter we tackle the last set of questions. Consequently, by the end of the rst chapter we will not only have a solid grounding in the practice of Gauss method, we will also have a solid grounding in the theory. We will be sure of what can and cannot happen in a reduction. Exercises
2.15 Find the indicated entry of the matrix, if it is dened. A= 1 2 3 1 1 4
(a) a2,1 (b) a1,2 (c) a2,2 (d) a3,1 2.16 Give the size of each matrix. 1 1 1 0 4 5 10 1 (a) (b) 1 (c) 2 1 5 10 5 3 1 2.17 Do the indicated vector operation, if it is dened. 3 3 2 1 3 4 2 +9 (a) 1 + 0 (b) 5 (c) 5 1 (d) 7 5 1 1 4 1 1 1 1 1 2 3 1 + 2 (e) (f ) 6 1 4 0 + 2 1 2 5 3 3 1 2.18 Solve each system using matrix notation. Express the solution using vectors. (a) 3x + 6y = 18 (b) x + y = 1 (c) x1 + x3 = 4 x + 2y = 6 x y = 1 x1 x2 + 2x3 = 5 4x1 x2 + 5x3 = 17 (d) 2a + b c = 2 (e) x + 2y z =3 (f ) x +z+w=4 2a +c=3 2x + y +w=4 2x + y w=2 ab =0 x y+z+w=1 3x + y + z =7 2.19 Solve each system using matrix notation. Give each solution set in vector notation. (a) 2x + y z = 1 (b) x z =1 (c) x y + z =0 4x y =3 y + 2z w = 3 y +w=0 x + 2y + 3z w = 7 3x 2y + 3z + w = 0 y w=0 (d) a + 2b + 3c + d e = 1 3a b + c + d + e = 3 2.20 The vector is in the set. What value of the parameters produces that vector? 5 1 (a) ,{ k k R} 5 1 1 2 3 2 , { 1 i + 0 j i, j R} (b) 1 0 1
Section I. Solving Linear Systems
0 1 2 4 , { 1 m + 0 n m, n R} 2 0 1 2.21 Decide if the vector is in the set. 3 6 (a) ,{ k k R} 1 2 (c)
19
5 5 ,{ j j R} 4 4 2 0 1 1 ,{ 3 (c) + 1 r r R} 1 7 3 1 2 3 (d) 0 , { 0 j + 1 k j, k R} 1 1 1 2.22 Paramatrize the solution set of this one-equation system. x1 + x2 + + xn = 0 2.23 (a) Apply Gauss method to the left-hand side to solve x + 2y w=a 2x +z =b x+ y + 2w = c for x, y, z, and w, in terms of the constants a, b, and c. (b) Use your answer from the prior part to solve this. x + 2y w= 3 2x +z = 1 x+ y + 2w = 2 2.24 Why is the comma needed in the notation ai,j for matrix entries? 2.25 Give the 44 matrix whose i, j-th entry is (a) i + j; (b) 1 to the i + j power. 2.26 For any matrix A, the transpose of A, written Atrans , is the matrix whose columns are the rows of A. Find the transpose of each of these. 1 1 2 3 2 3 5 10 (a) (b) (c) (d) 1 4 5 6 1 1 10 5 0 2 2.27 (a) Describe all functions f (x) = ax + bx + c such that f (1) = 2 and f (1) = 6. (b) Describe all functions f (x) = ax2 + bx + c such that f (1) = 2. 2.28 Show that any set of ve points from the plane R2 lie on a common conic section, that is, they all satisfy some equation of the form ax2 + by 2 + cxy + dx + ey + f = 0 where some of a, . . . , f are nonzero. 2.29 Make up a four equations/four unknowns system having (a) a one-parameter solution set; (b) a two-parameter solution set; (c) a three-parameter solution set. ? 2.30 (a) Solve the system of equations. ax + y = a2 x + ay = 1 For what values of a does the system fail to have solutions, and for what values of a are there innitely many solutions? (b)
20
(b) Answer the above question for the system. ax + y = a3 x + ay = 1
Chapter One. Linear Systems
[USSR Olympiad no. 174] ? 2.31 In air a gold-surfaced sphere weighs 7588 grams. It is known that it may contain one or more of the metals aluminum, copper, silver, or lead. When weighed successively under standard conditions in water, benzene, alcohol, and glycerine its respective weights are 6588, 6688, 6778, and 6328 grams. How much, if any, of the forenamed metals does it contain if the specic gravities of the designated substances are taken to be as follows? Aluminum 2.7 Alcohol 0.81 Copper 8.9 Benzene 0.90 Gold 19.3 Glycerine 1.26 Lead 11.3 Water 1.00 Silver 10.8 [Math. Mag., Sept. 1952]
I.3 General = Particular + Homogeneous
The prior subsection has many descriptions of solution sets. They all t a pattern. They have a vector that is a particular solution of the system added to an unrestricted combination of some other vectors. The solution set from Example 2.13 illustrates. 0 1 1/2 4 1 1 0 + w 3 + u 1/2 w, u R} { 0 1 0 0 0 1
particular solution unrestricted combination
The combination is unrestricted in that w and u can be any real numbers there is no condition like such that 2w u = 0 that would restrict which pairs w, u can be used to form combinations. That example shows an innite solution set conforming to the pattern. We can think of the other two kinds of solution sets as also tting the same pattern. A one-element solution set ts in that it has a particular solution, and the unrestricted combination part is a trivial sum (that is, instead of being a combination of two vectors, as above, or a combination of one vector, it is a combination of no vectors). A zero-element solution set ts the pattern since there is no particular solution, and so the set of sums of that form is empty. We will show that the examples from the prior subsection are representative, in that the description pattern discussed above holds for every solution set.
Section I. Solving Linear Systems
21
3.1 Theorem For any linear system there are vectors 1 , . . . , k such that the solution set can be described as {p + c1 1 + + ck k c1 , . . . , ck R} where p is any particular solution, and where the system has k free variables. This description has two parts, the particular solution p and also the unrestricted linear combination of the s. We shall prove the theorem in two corresponding parts, with two lemmas. We will focus rst on the unrestricted combination part. To do that, we consider systems that have the vector of zeroes as one of the particular solutions, so that p + c1 1 + + ck k can be shortened to c1 1 + + ck k . 3.2 Denition A linear equation is homogeneous if it has a constant of zero, that is, if it can be put in the form a1 x1 + a2 x2 + + an xn = 0. (These are homogeneous because all of the terms involve the same power of their variable the rst power including a 0x0 that we can imagine is on the right side.) 3.3 Example With any linear system like 3x + 4y = 3 2x y = 1 we associate a system of homogeneous equations by setting the right side to zeros. 3x + 4y = 0 2x y = 0 Our interest in the homogeneous system associated with a linear system can be understood by comparing the reduction of the system 3x + 4y = 3 2x y = 1
(2/3)1 +2
3x +
4y = 3 (11/3)y = 1
with the reduction of the associated homogeneous system. 3x + 4y = 0 2x y = 0
(2/3)1 +2
3x +
4y = 0 (11/3)y = 0
Obviously the two reductions go in the same way. We can study how linear systems are reduced by instead studying how the associated homogeneous systems are reduced. Studying the associated homogeneous system has a great advantage over studying the original system. Nonhomogeneous systems can be inconsistent. But a homogeneous system must be consistent since there is always at least one solution, the vector of zeros.
22
Chapter One. Linear Systems
3.4 Denition A column or row vector of all zeros is a zero vector , denoted 0. There are many dierent zero vectors, e.g., the one-tall zero vector, the two-tall zero vector, etc. Nonetheless, people often refer to the zero vector, expecting that the size of the one being discussed will be clear from the context. 3.5 Example Some homogeneous systems have the zero vector as their only solution. 3x + 2y + z = 0 6x + 4y =0 y+z=0
21 +2
3x + 2y +
z=0 3x + 2y + z=0 2 3 2z = 0 y+ z=0 y+ z=0 2z = 0
3.6 Example Some homogeneous systems have many solutions. One example is the Chemistry problem from the rst page of this book. 7x 7z =0 8x + y 5z 2k = 0 y 3z =0 3y 6z k = 0 7x
(8/7)1 +2
7z =0 y + 3z 2w = 0 y 3z =0 3y 6z w = 0 y+ 7z =0 3z 2w = 0 6z + 2w = 0 15z + 5w = 0
7x
2 +3 32 +4
7x
(5/2)3 +4
7z =0 y + 3z 2w = 0 6z + 2w = 0 0=0
The solution set:
1/3 1 { w w R} 1/3 1
has many vectors besides the zero vector (if we interpret w as a number of molecules then solutions make sense only when w is a nonnegative multiple of 3). We now have the terminology to prove the two parts of Theorem 3.1. The rst lemma deals with unrestricted combinations. 3.7 Lemma For any homogeneous linear system there exist vectors 1 , . . . , k such that the solution set of the system is {c1 1 + + ck k c1 , . . . , ck R} where k is the number of free variables in an echelon form version of the system.
Section I. Solving Linear Systems
23
Before the proof, we will recall the back substitution calculations that were done in the prior subsection. Imagine that we have brought a system to this echelon form. x + 2y z + 2w = 0 3y + z =0 w = 0 We next perform back-substitution to express each variable in terms of the free variable z. Working from the bottom up, we get rst that w is 0 z, next that y is (1/3) z, and then substituting those two into the top equation x + 2((1/3)z) z + 2(0) = 0 gives x = (1/3) z. So, back substitution gives a paramatrization of the solution set by starting at the bottom equation and using the free variables as the parameters to work row-by-row to the top. The proof below follows this pattern. Comment: That is, this proof just does a verication of the bookkeeping in back substitution to show that we havent overlooked any obscure cases where this procedure fails, say, by leading to a division by zero. So this argument, while quite detailed, doesnt give us any new insights. Nevertheless, we have written it out for two reasons. The rst reason is that we need the result the computational procedure that we employ must be veried to work as promised. The second reason is that the row-by-row nature of back substitution leads to a proof that uses the technique of mathematical induction. This is an important, and non-obvious, proof technique that we shall use a number of times in this book. Doing an induction argument here gives us a chance to see one in a setting where the proof material is easy to follow, and so the technique can be studied. Readers who are unfamiliar with induction arguments should be sure to master this one and the ones later in this chapter before going on to the second chapter.
Proof. First use Gauss method to reduce the homogeneous system to echelon
form. We will show that each leading variable can be expressed in terms of free variables. That will nish the argument because then we can use those free variables as the parameters. That is, the s are the vectors of coecients of the free variables (as in Example 3.6, where the solution is x = (1/3)w, y = w, z = (1/3)w, and w = w). We will proceed by mathematical induction, which has two steps. The base step of the argument will be to focus on the bottom-most non-0 = 0 equation and write its leading variable in terms of the free variables. The inductive step of the argument will be to argue that if we can express the leading variables from the bottom t rows in terms of free variables, then we can express the leading variable of the next row up the t + 1-th row up from the bottom in terms of free variables. With those two steps, the theorem will be proved because by the base step it is true for the bottom equation, and by the inductive step the fact that it is true for the bottom equation shows that it is true for the next one up, and then another application of the inductive step implies it is true for third equation up, etc.
More information on mathematical induction is in the appendix.
24
Chapter One. Linear Systems
For the base step, consider the bottom-most non-0 = 0 equation (the case where all the equations are 0 = 0 is trivial). We call that the m-th row: am,
m
x
m
+ am,
m +1
x
m +1
+ + am,n xn = 0
where am, m = 0. (The notation here has stand for leading, so am, m means the coecient, from the row m of the variable leading row m.) Either there are variables in this equation other than the leading one x m or else there are not. If there are other variables x m +1 , etc., then they must be free variables because this is the bottom non-0 = 0 row. Move them to the right and divide by am, m x
m
= (am,
m +1
/am,
m
)x
m +1
+ + (am,n /am,
m
)xn
to expresses this leading variable in terms of free variables. If there are no free variables in this equation then x m = 0 (see the tricky point noted following this proof). For the inductive step, we assume that for the m-th equation, and for the (m 1)-th equation, . . . , and for the (m t)-th equation, we can express the leading variable in terms of free variables (where 0 t < m). To prove that the same is true for the next equation up, the (m (t + 1))-th equation, we take each variable that leads in a lower-down equation x m , . . . , x mt and substitute its expression in terms of free variables. The result has the form am(t+1),
m(t+1)
x
m(t+1)
+ sums of multiples of free variables = 0
where am(t+1), m(t+1) = 0. We move the free variables to the right-hand side and divide by am(t+1), m(t+1) , to end with x m(t+1) expressed in terms of free variables. Because we have shown both the base step and the inductive step, by the principle of mathematical induction the proposition is true. QED We say that the set {c1 1 + + ck k c1 , . . . , ck R} is generated by or spanned by the set of vectors {1 , . . . , k }. There is a tricky point to this denition. If a homogeneous system has a unique solution, the zero vector, then we say the solution set is generated by the empty set of vectors. This ts with the pattern of the other solution sets: in the proof above the solution set is derived by taking the cs to be the free variables and if there is a unique solution then there are no free variables. This proof incidentally shows, as discussed after Example 2.4, that solution sets can always be paramatrized using the free variables. The next lemma nishes the proof of Theorem 3.1 by considering the particular solution part of the solution sets description. 3.8 Lemma For a linear system, where p is any particular solution, the solution set equals this set. {p + h h satises the associated homogeneous system}
Section I. Solving Linear Systems
25
Proof. We will show mutual set inclusion, that any solution to the system is
in the above set and that anything in the set is a solution to the system. For set inclusion the rst way, that if a vector solves the system then it is in the set described above, assume that s solves the system. Then s p solves the associated homogeneous system since for each equation index i between 1 and n, ai,1 (s1 p1 ) + + ai,n (sn pn ) = (ai,1 s1 + + ai,n sn ) (ai,1 p1 + + ai,n pn ) = di di =0 where pj and sj are the j-th components of p and s. We can write s p as h, where h solves the associated homogeneous system, to express s in the required p + h form. For set inclusion the other way, take a vector of the form p + h, where p solves the system and h solves the associated homogeneous system, and note that it solves the given system: for any equation index i, ai,1 (p1 + h1 ) + + ai,n (pn + hn ) = (ai,1 p1 + + ai,n pn ) + (ai,1 h1 + + ai,n hn ) = di + 0 = di where hj is the j-th component of h.
QED
The two lemmas above together establish Theorem 3.1. We remember that theorem with the slogan General = Particular + Homogeneous. 3.9 Example This system illustrates Theorem 3.1. x + 2y z = 1 2x + 4y =2 y 3z = 0 Gauss method
21 +2
x + 2y z = 1 x + 2y z = 1 2 3 2z = 0 y 3z = 0 y 3z = 0 2z = 0
shows that the general solution is a singleton set. 1 {0} 0
More information on equality of sets is in the appendix.
26
Chapter One. Linear Systems
That single vector is, of course, a particular solution. The associated homogeneous system reduces via the same row operations x + 2y z = 0 2x + 4y =0 y 3z = 0 to also give a singleton set.
21 +2 2 3
x + 2y z = 0 y 3z = 0 2z = 0
0 {0} 0
As the theorem states, and as discussed at the start of this subsection, in this single-solution case the general solution results from taking the particular solution and adding to it the unique solution of the associated homogeneous system. 3.10 Example Also discussed there is that the case where the general solution set is empty ts the General = Particular+Homogeneous pattern. This system illustrates. Gauss method x + z + w = 1 2x y + w= 3 x + y + 3z + 2w = 1
21 +2 1 +3
x
+ z + w = 1 y 2z w = 5 y + 2z + w = 2
shows that it has no solutions. The associated homogeneous system, of course, has a solution. x + z+ w=0 2x y + w=0 x + y + 3z + 2w = 0
21 +2 2 +3 1 +3
x
+ z+w=0 y 2z w = 0 0=0
In fact, the solution set of the homogeneous system is innite. 1 1 2 z + 1 w z, w R} { 0 1 0 1 However, because no particular solution of the original system exists, the general solution set is empty there are no vectors of the form p + h because there are no p s. 3.11 Corollary Solution sets of linear systems are either empty, have one element, or have innitely many elements.
Proof. Weve seen examples of all three happening so we need only prove that
those are the only possibilities. First, notice a homogeneous system with at least one non-0 solution v has innitely many solutions because the set of multiples sv is innite if s = 1 then sv v = (s 1)v is easily seen to be non-0, and so sv = v.
Section I. Solving Linear Systems Now, apply Lemma 3.8 to conclude that a solution set {p + h h solves the associated homogeneous system}
27
is either empty (if there is no particular solution p), or has one element (if there is a p and the homogeneous system has the unique solution 0), or is innite (if there is a p and the homogeneous system has a non-0 solution, and thus by the prior paragraph has innitely many solutions). QED This table summarizes the factors aecting the size of a general solution. number of solutions of the associated homogeneous system one unique solution no solutions innitely many innitely many solutions no solutions
particular solution exists?
yes no
The factor on the top of the table is the simpler one. When we perform Gauss method on a linear system, ignoring the constants on the right side and so paying attention only to the coecients on the left-hand side, we either end with every variable leading some row or else we nd that some variable does not lead a row, that is, that some variable is free. (Of course, ignoring the constants on the right is formalized by considering the associated homogeneous system. We are simply putting aside for the moment the possibility of a contradictory equation.) A nice insight into the factor on the top of this table at work comes from considering the case of a system having the same number of equations as variables. This system will have a solution, and the solution will be unique, if and only if it reduces to an echelon form system where every variable leads its row, which will happen if and only if the associated homogeneous system has a unique solution. Thus, the question of uniqueness of solution is especially interesting when the system has the same number of equations as variables. 3.12 Denition A square matrix is nonsingular if it is the matrix of coecients of a homogeneous system with a unique solution. It is singular otherwise, that is, if it is the matrix of coecients of a homogeneous system with innitely many solutions. 3.13 Example The systems from Example 3.3, Example 3.5, and Example 3.9 each have an associated homogeneous system with a unique solution. Thus these matrices are nonsingular. 3 2 1 1 2 1 3 4 6 4 0 2 4 0 2 1 0 1 1 0 1 3
28
Chapter One. Linear Systems
The Chemistry problem from Example 3.6 is a homogeneous system with more than one solution so its matrix is singular. 7 0 7 0 8 1 5 2 0 1 3 0 0 3 6 1 3.14 Example The rst of these matrices is nonsingular while the second is singular 1 2 1 2 3 4 3 6 because the rst of these homogeneous systems has a unique solution while the second has innitely many solutions. x + 2y = 0 3x + 4y = 0 x + 2y = 0 3x + 6y = 0
We have made the distinction in the denition because a system (with the same number of equations as variables) behaves in one of two ways, depending on whether its matrix of coecients is nonsingular or singular. A system where the matrix of coecients is nonsingular has a unique solution for any constants on the right side: for instance, Gauss method shows that this system x + 2y = a 3x + 4y = b has the unique solution x = b 2a and y = (3a b)/2. On the other hand, a system where the matrix of coecients is singular never has a unique solutions it has either no solutions or else has innitely many, as with these. x + 2y = 1 3x + 6y = 2 x + 2y = 1 3x + 6y = 3
Thus, singular can be thought of as connoting troublesome, or at least not ideal. The above table has two factors. We have already considered the factor along the top: we can tell which column a given linear system goes in solely by considering the systems left-hand side the constants on the right-hand side play no role in this factor. The tables other factor, determining whether a particular solution exists, is tougher. Consider these two 3x + 2y = 5 3x + 2y = 5 3x + 2y = 5 3x + 2y = 4
with the same left sides but dierent right sides. Obviously, the rst has a solution while the second does not, so here the constants on the right side
Section I. Solving Linear Systems
29
decide if the system has a solution. We could conjecture that the left side of a linear system determines the number of solutions while the right side determines if solutions exist, but that guess is not correct. Compare these two systems 3x + 2y = 5 4x + 2y = 4 3x + 2y = 5 3x + 2y = 4
with the same right sides but dierent left sides. The rst has a solution but the second does not. Thus the constants on the right side of a system dont decide alone whether a solution exists; rather, it depends on some interaction between the left and right sides. For some intuition about that interaction, consider this system with one of the coecients left as the parameter c. x + 2y + 3z = 1 x+ y+ z=1 cx + 3y + 4z = 0 If c = 2 this system has no solution because the left-hand side has the third row as a sum of the rst two, while the right-hand does not. If c = 2 this system has a unique solution (try it with c = 1). For a system to have a solution, if one row of the matrix of coecients on the left is a linear combination of other rows, then on the right the constant from that row must be the same combination of constants from the same rows. More intuition about the interaction comes from studying linear combinations. That will be our focus in the second chapter, after we nish the study of Gauss method itself in the rest of this chapter. Exercises
3.15 Solve each system. Express the solution set using vectors. Identify the particular solution and the solution set of the homogeneous system. (a) 3x + 6y = 18 (b) x + y = 1 (c) x1 + x3 = 4 x + 2y = 6 x y = 1 x1 x2 + 2x3 = 5 4x1 x2 + 5x3 = 17 (d) 2a + b c = 2 (e) x + 2y z =3 (f ) x +z+w=4 2a +c=3 2x + y +w=4 2x + y w=2 ab =0 x y+z+w=1 3x + y + z =7 3.16 Solve each system, giving the solution set in vector notation. Identify the particular solution and the solution of the homogeneous system. (a) 2x + y z = 1 (b) x z =1 (c) x y + z =0 4x y =3 y + 2z w = 3 y +w=0 x + 2y + 3z w = 7 3x 2y + 3z + w = 0 y w=0 (d) a + 2b + 3c + d e = 1 3a b + c + d + e = 3 3.17 For the system 2x y w= 3 y + z + 2w = 2 x 2y z = 1
30
Chapter One. Linear Systems
which of these can be used as the particular solution part of some general solution? 0 2 1 3 1 4 (a) (b) (c) 5 1 8 0 0 1 3.18 Lemma 3.8 says that any particular solution may be used for p. Find, if possible, a general solution to this system x y +w=4 2x + 3y z =0 y+z+w=4 that uses the given vector its particular solution. as 0 5 2 0 1 1 (a) (b) (c) 0 7 1 4 10 1 3.19 One of these is nonsingular while the other is singular. Which is which? 1 3 1 3 (b) 4 12 4 12 3.20 Singular or nonsingular? 1 2 1 2 1 2 1 (a) (b) (c) (Careful!) 1 3 3 6 1 3 1 1 2 1 2 2 1 1 0 5 (d) 1 1 3 (e) 3 4 7 1 1 4 3.21 Is the given vector in the set generated by the given set? 2 1 1 (a) ,{ , } 3 4 5 1 2 1 0 ,{ 1 , 0 } (b) 1 0 1 4 3 2 1 1 (c) 3 , { 0 , 1 , 3 , 2 } 1 4 5 0 0 1 2 3 0 1 0 (d) , { , } 1 0 0 1 1 2 3.22 Prove that any linear system with a nonsingular matrix of coecients has a solution, and that the solution is unique. 3.23 To tell the whole truth, there is another tricky point to the proof of Lemma 3.7. What happens if there are no non-0 = 0 equations? (There arent any more tricky points after this one.) 3.24 Prove that if s and t satisfy a homogeneous system then so do these vectors. (a) s + t (b) 3s (c) ks + mt for k, m R Whats wrong with: These three show that if a homogeneous system has one solution then it has many solutions any multiple of a solution is another solution, (a)
Section I. Solving Linear Systems
31
and any sum of solutions is a solution also so there are no homogeneous systems with exactly one solution.? 3.25 Prove that if a system with only rational coecients and constants has a solution then it has at least one all-rational solution. Must it have innitely many?
32
Chapter One. Linear Systems
II
Linear Geometry of n-Space
For readers who have seen the elements of vectors before, in calculus or physics, this section is an optional review. However, later work will refer to this material so it is not optional if it is not a review. In the rst section, we had to do a bit of work to show that there are only three types of solution sets singleton, empty, and innite. But in the special case of systems with two equations and two unknowns this is easy to see. Draw each two-unknowns equation as a line in the plane and then the two lines could have a unique intersection, be parallel, or be the same line.
Unique solution No solutions Innitely many solutions
3x + 2y = 7 x y = 1
3x + 2y = 7 3x + 2y = 4
3x + 2y = 7 6x + 4y = 14
These pictures dont prove the results from the prior section, which apply to any number of linear equations and any number of unknowns, but nonetheless they do help us to understand those results. This section develops the ideas that we need to express our results from the prior section, and from some future sections, geometrically. In particular, while the two-dimensional case is familiar, to extend to systems with more than two unknowns we shall need some higherdimensional geometry.
II.1 Vectors in Space
Higher-dimensional geometry sounds exotic. It is exotic interesting and eye-opening. But it isnt distant or unreachable. We begin by dening one-dimensional space to be the set R1 . To see that denition is reasonable, draw a one-dimensional space
and make the usual correspondence with R: pick a point to label 0 and another to label 1.
0 1
Now, with a scale and a direction, nding the point corresponding to, say +2.17, is easy start at 0 and head in the direction of 1 (i.e., the positive direction), but dont stop there, go 2.17 times as far.
Section II. Linear Geometry of n-Space
33
The basic idea here, combining magnitude with direction, is the key to extending to higher dimensions. An object comprised of a magnitude and a direction is a vector (we will use the same word as in the previous section because we shall show below how to describe such an object with a column vector). We can draw a vector as having some length, and pointing somewhere.
There is a subtlety here these vectors
are equal, even though they start in dierent places, because they have equal lengths and equal directions. Again: those vectors are not just alike, they are equal. How can things that are in dierent places be equal? Think of a vector as representing a displacement (vector is Latin for carrier or traveler). These squares undergo the same displacement, despite that those displacements start in dierent places.
Sometimes, to emphasize this property vectors have of not being anchored, they are referred to as free vectors. Thus, these free vectors are equal as each is a displacement of one over and two up.
More generally, vectors in the plane are the same if and only if they have the same change in rst components and the same change in second components: the vector extending from (a1 , a2 ) to (b1 , b2 ) equals the vector from (c1 , c2 ) to (d1 , d2 ) if and only if b1 a1 = d1 c1 and b2 a2 = d2 c2 . An expression like the vector that, were it to start at (a1 , a2 ), would extend to (b1 , b2 ) is awkward. We instead describe such a vector as b1 a1 b2 a2 so that, for instance, the one over and two up arrows shown above picture this vector. 1 2
34
Chapter One. Linear Systems
We often draw the arrow as starting at the origin, and we then say it is in the canonical position (or natural position). When the vector b1 a1 b2 a2 is in its canonical position then it extends to the endpoint (b1 a1 , b2 a2 ). We typically just refer to the point 1 2 rather than the endpoint of the canonical position of that vector. Thus, we will call both of these sets Rn . {(x1 , x2 ) x1 , x2 R} { x1 x2 x1 , x2 R}
In the prior section we dened vectors and vector operations with an algebraic motivation; r v1 v2 = rv1 rv2 v1 v2 + w1 w2 = v1 + w1 v2 + w2
we can now interpret those operations geometrically. For instance, if v represents a displacement then 3v represents a displacement in the same direction but three times as far, and 1v represents a displacement of the same distance as v but in the opposite direction.
v 3v v
And, where v and w represent displacements, v + w represents those displacements combined.
v+w w
v
The long arrow is the combined displacement in this sense: if, in one minute, a ships motion gives it the displacement relative to the earth of v and a passengers motion gives a displacement relative to the ships deck of w, then v + w is the displacement of the passenger relative to the earth. Another way to understand the vector sum is with the parallelogram rule. Draw the parallelogram formed by the vectors v1 , v2 and then the sum v1 + v2 extends along the diagonal to the far corner.
Section II. Linear Geometry of n-Space
v+w w
35
v
The above drawings show how vectors and vector operations behave in R2 . We can extend to R3 , or to even higher-dimensional spaces where we have no pictures, with the obvious generalization: the free vector that, if it starts at (a1 , . . . , an ), ends at (b1 , . . . , bn ), is represented by this column b1 a1 . . . bn an (vectors are equal if they have the same representation), we arent too careful to distinguish between a point and the vector whose canonical representation ends at that point, v1 . n R = { . v1 , . . . , vn R} . vn and addition and scalar multiplication are component-wise. Having considered points, we now turn to the lines. In R2 , the line through (1, 2) and (3, 1) is comprised of (the endpoints of) the vectors in this set { 2 1 +t 1 2 t R}
That description expresses this picture.
2 1 = 3 1 1 2
The vector associated with the parameter t has its whole body in the line it is a direction vector for the line. Note that points on the line to the left of x = 1 are described using negative values of t. In R3 , the line through (1, 2, 1) and (2, 3, 2) is the set of (endpoints of) vectors of this form
{ 1 2 1 +t 1 1 1 t R}
36
Chapter One. Linear Systems
and lines in even higher-dimensional spaces work in the same way. If a line uses one parameter, so that there is freedom to move back and forth in one dimension, then a plane must involve two. For example, the plane through the points (1, 0, 5), (2, 1, 3), and (2, 4, 0.5) consists of (endpoints of) the vectors in 1 1 3 {0 + t 1 + s 4 t, s R} 5 8 4.5 (the column vectors associated with the parameters 2 1 3 1 2 1 4 = 4 0 1 = 1 0 4.5 0.5 5 5 8 3 are two vectors whose whole bodies lie in the plane). As with the line, note that some points in this plane are described with negative ts or negative ss or both. A description of planes that is often encountered in algebra and calculus uses a single equation as the condition that describes the relationship among the rst, second, and third coordinates of points in a plane.
P ={
x y z
2x + y + z = 4}
The translation from such a description to the vector description that we favor in this book is to think of the condition as a one-equation linear system and paramatrize x = (1/2)(4 y z).
P ={
2 0 0
+
0.5 1 0
y+
0.5 0 1
z
y, z R}
Generalizing from lines and planes, we dene a k-dimensional linear surface (or k-at) in Rn to be {p + t1 v1 + t2 v2 + + tk vk t1 , . . . , tk R} where v1 , . . . , vk Rn . For example, in R4 , 2 1 0 { 3 + t 0 t R} 0.5 0
Section II. Linear Geometry of n-Space is a line, 0 1 2 0 1 0 { + t + s t, s R} 0 0 1 0 1 0 3 0 1 2 1 0 0 0 { + r + s + t r, s, t R} 2 0 1 1 0.5 1 0 0
37
is a plane, and
is a three-dimensional linear surface. Again, the intuition is that a line permits motion in one direction, a plane permits motion in combinations of two directions, etc. A linear surface description can be misleading about the dimension this 1 1 2 0 1 2 L = { + t + s t, s R} 1 0 0 2 1 2 is a degenerate plane because it is actually a line. 1 1 0 1 L = { + r r R} 1 0 2 1 We shall see in the Linear Independence section of Chapter Two what relationships among vectors causes the linear surface they generate to be degenerate. We nish this subsection by restating our conclusions from the rst section in geometric terms. First, the solution set of a linear system with n unknowns is a linear surface in Rn . Specically, it is a k-dimensional linear surface, where k is the number of free variables in an echelon form version of the system. Second, the solution set of a homogeneous linear system is a linear surface passing through the origin. Finally, we can view the general solution set of any linear system as being the solution set of its associated homogeneous system oset from the origin by a vector, namely by any particular solution. Exercises
1.1 Find the canonical name for each vector. (a) the vector from (2, 1) to (4, 2) in R2 (b) the vector from (3, 3) to (2, 5) in R2 (c) the vector from (1, 0, 6) to (5, 0, 3) in R3 (d) the vector from (6, 8, 8) to (6, 8, 8) in R3 1.2 Decide if the two vectors are equal. (a) the vector from (5, 3) to (6, 2) and the vector from (1, 2) to (1, 1)
38
Chapter One. Linear Systems
(b) the vector from (2, 1, 1) to (3, 0, 4) and the vector from (5, 1, 4) to (6, 0, 7) 1.3 Does (1, 0, 2, 1) lie on the line through (2, 1, 1, 0) and (5, 10, 1, 4)? 1.4 (a) Describe the plane through (1, 1, 5, 1), (2, 2, 2, 0), and (3, 1, 0, 4). (b) Is the origin in that plane? 1.5 Describe the plane that contains this point and line. 2 0 3 1.6 Intersect these planes. 1 { 1 1 t+ 0 1 3 s t, s R} 1 { 1 0 + 0 3 0 k+ 2 0 4 m k, m R} { 1 0 4 + 1 1 2 t t R}
1.7 Intersect each pair, if possible. 1 0 1 0 (a) { 1 + t 1 t R}, { 3 +s 1 s R} 2 1 2 2 0 1 0 2 s, w R} t R}, {s 1 + w 4 (b) { 0 + t 1 1 1 2 1 1.8 When a plane does not pass through the origin, performing operations on vectors whose bodies lie in it is more complicated than when the plane passes through the origin. Consider the picture in this subsection of the plane 2 { 0 0 + 0.5 1 0 y+ 0.5 0 1 z y, z R}
and the three vectors it shows, with endpoints (2, 0, 0), (1.5, 1, 0), and (1.5, 0, 1). (a) Redraw the picture, including the vector in the plane that is twice as long as the one with endpoint (1.5, 1, 0). The endpoint of your vector is not (3, 2, 0); what is it? (b) Redraw the picture, including the parallelogram in the plane that shows the sum of the vectors ending at (1.5, 0, 1) and (1.5, 1, 0). The endpoint of the sum, on the diagonal, is not (3, 1, 1); what is it? 1.9 Show that the line segments (a1 , a2 )(b1 , b2 ) and (c1 , c2 )(d1 , d2 ) have the same lengths and slopes if b1 a1 = d1 c1 and b2 a2 = d2 c2 . Is that only if? 1.10 How should R0 be dened? ? 1.11 A person traveling eastward at a rate of 3 miles per hour nds that the wind appears to blow directly from the north. On doubling his speed it appears to come from the north east. What was the winds velocity? [Math. Mag., Jan. 1957] 1.12 Euclid describes a plane as a surface which lies evenly with the straight lines on itself. Commentators (e.g., Heron) have interpreted this to mean (A plane surface is) such that, if a straight line pass through two points on it, the line coincides wholly with it at every spot, all ways. (Translations from [Heath], pp. 171-172.) Do planes, as described in this section, have that property? Does this description adequately dene planes?
Section II. Linear Geometry of n-Space
39
II.2 Length and Angle Measures
Weve translated the rst sections results about solution sets into geometric terms for insight into how those sets look. But we must watch out not to be mislead by our own terms; labeling subsets of Rk of the forms {p + tv t R} and {p + tv + sw t, s R} as lines and planes doesnt make them act like the lines and planes of our prior experience. Rather, we must ensure that the names suit the sets. While we cant prove that the sets satisfy our intuition we cant prove anything about intuition in this subsection well observe that a result familiar from R2 and R3 , when generalized to arbitrary Rk , supports the idea that a line is straight and a plane is at. Specically, well see how to do Euclidean geometry in a plane by giving a denition of the angle between two Rn vectors in the plane that they generate. 2.1 Denition The length of a vector v Rn is this. v =
2 2 v1 + + vn
2.2 Remark This is a natural generalization of the Pythagorean Theorem. A classic discussion is in [Polya]. We can use that denition to derive a formula for the angle between two vectors. For a model of what to do, consider two vectors in R3 .
v
u
Put them in canonical position and, in the plane that they determine, consider the triangle formed by u, v, and u v.
Apply the Law of Cosines, u v 2 = u 2 + v 2 2 u is the angle between the vectors. Expand both sides (u1 v1 )2 + (u2 v2 )2 + (u3 v3 )2
v cos , where
2 2 2 (u2 = + u2 + u2 ) + (v1 + v2 + v3 ) 2 u 1 2 3
v cos
and simplify. = arccos(
u1 v1 + u2 v2 + u3 v3 ) u v
40
Chapter One. Linear Systems
In higher dimensions no picture suces but we can make the same argument analytically. First, the form of the numerator is clear it comes from the middle terms of the squares (u1 v1 )2 , (u2 v2 )2 , etc. 2.3 Denition The dot product (or inner product, or scalar product) of two n-component real vectors is the linear combination of their components. u v = u1 v1 + u2 v2 + + un vn Note that the dot product of two vectors is a real number, not a vector, and that the dot product of a vector from Rn with a vector from Rm is dened only when n equals m. Note also this relationship between dot product and length: dotting a vector with itself gives its length squared u u = u1 u1 + + un un = u 2 . 2.4 Remark The wording in that denition allows one or both of the two to be a row vector instead of a column vector. Some books require that the rst vector be a row vector and that the second vector be a column vector. We shall not be that strict. Still reasoning with letters, but guided by the pictures, we use the next theorem to argue that the triangle formed by u, v, and u v in Rn lies in the planar subset of Rn generated by u and v. 2.5 Theorem (Triangle Inequality) For any u, v Rn , u+v u + v with equality if and only if one of the vectors is a nonnegative scalar multiple of the other one. This inequality is the source of the familiar saying, The shortest distance between two points is in a straight line.
nish u+v v
start
u
Proof. (Well use some algebraic properties of dot product that we have not
yet checked, for instance that u (a + b) = u a + u b and that u v = v u. See Exercise 17.) The desired inequality holds if and only if its square holds. u+v
2
( u + v )2 v + v 2 v +v v
(u + v) (u + v) u 2 + 2 u u u+u v+v u+v v u u+2 u 2u v 2 u v
Section II. Linear Geometry of n-Space
41
That, in turn, holds if and only if the relationship obtained by multiplying both sides by the nonnegative numbers u and v 2( v u) ( u v) 2 u and rewriting 0 u is true. But factoring 0 ( u v v u) ( u v v u) shows that this certainly is true since it only says that the square of the length of the vector u v v u is not negative. As for equality, it holds when, and only when, u v v u is 0. The check that u v = v u if and only if one vector is a nonnegative real scalar multiple of the other is easy. QED This result supports the intuition that even in higher-dimensional spaces, lines are straight and planes are at. For any two points in a linear surface, the line segment connecting them is contained in that surface (this is easily checked from the denition). But if the surface has a bend then that would allow for a shortcut (shown here grayed, while the segment from P to Q that is contained in the surface is solid).
P Q
2
v
2
2
v
2
2( v u) ( u v) + u
2
v
2
Because the Triangle Inequality says that in any Rn , the shortest cut between two endpoints is simply the line segment connecting them, linear surfaces have no such bends. Back to the denition of angle measure. The heart of the Triangle Inequalitys proof is the u v u v line. At rst glance, a reader might wonder if some pairs of vectors satisfy the inequality in this way: while u v is a large number, with absolute value bigger than the right-hand side, it is a negative large number. The next result says that no such pair of vectors exists. 2.6 Corollary (Cauchy-Schwartz Inequality) For any u, v Rn , |u v| u v
with equality if and only if one vector is a scalar multiple of the other.
Proof. The Triangle Inequalitys proof shows that u v u
v so if u v is positive or zero then we are done. If u v is negative then this holds. | u v | = ( u v ) = (u ) v The equality condition is Exercise 18. u v = u v
QED
42
Chapter One. Linear Systems
The Cauchy-Schwartz inequality assures us that the next denition makes sense because the fraction has absolute value less than or equal to one. 2.7 Denition The angle between two nonzero vectors u, v Rn is = arccos( u v ) u v
(the angle between the zero vector and any other vector is dened to be a right angle). Thus vectors from Rn are orthogonal if and only if their dot product is zero. 2.8 Example These vectors are orthogonal. 1 1 1 1
=0
The arrows are shown away from canonical position but nevertheless the vectors are orthogonal. 2.9 Example The R3 angle formula given at the start of this subsection is a special case of the denition. Between these two
0 3 2
1 1 0
the angle is arccos( (1)(0) + (1)(3) + (0)(2) 3 ) = arccos( ) 2 + 12 + 02 02 + 32 + 22 2 13 1
approximately 0.94 radians. Notice that these vectors are not orthogonal. Although the yz-plane may appear to be perpendicular to the xy-plane, in fact the two planes are that way only in the weak sense that there are vectors in each orthogonal to all vectors in the other. Not every vector in each is orthogonal to all vectors in the other. Exercises
2.10 Find the length of each vector.
Section II. Linear Geometry of n-Space
1 1 (e) 1 0
43
(a)
3 1
(b)
1 2
(c)
4 1 1
(d)
0 0 0
2.11 Find the angle between each two, if it is dened. 1 0 1 1 1 1 , , 4 (a) (b) 2 , 4 (c) 2 4 2 0 1 1 2.12 During maneuvers preceding the Battle of Jutland, the British battle cruiser Lion moved as follows (in nautical miles): 1.2 miles north, 6.1 miles 38 degrees east of south, 4.0 miles at 89 degrees east of north, and 6.5 miles at 31 degrees east of north. Find the distance between starting and ending positions. [Ohanian] 2.13 Find k so that these two vectors are perpendicular. k 1 4 3
2.14 Describe the set of vectors in R3 orthogonal to this one. 1 3 1 2.15 (a) Find the angle between the diagonal of the unit square in R2 and one of the axes. (b) Find the angle between the diagonal of the unit cube in R3 and one of the axes. (c) Find the angle between the diagonal of the unit cube in Rn and one of the axes. (d) What is the limit, as n goes to , of the angle between the diagonal of the unit cube in Rn and one of the axes? 2.16 Is any vector perpendicular to itself? 2.17 Describe the algebraic properties of dot product. (a) Is it right-distributive over addition: (u + v) w = u w + v w? (b) Is is left-distributive (over addition)? (c) Does it commute? (d) Associate? (e) How does it interact with scalar multiplication? As always, any assertion must be backed by either a proof or an example. 2.18 Verify the equality condition in Corollary 2.6, the Cauchy-Schwartz Inequality. (a) Show that if u is a negative scalar multiple of v then u v and v u are less than or equal to zero. (b) Show that |u v| = u v if and only if one vector is a scalar multiple of the other. 2.19 Suppose that u v = u w and u = 0. Must v = w? 2.20 Does any vector have length zero except a zero vector? (If yes, produce an example. If no, prove it.) 2.21 Find the midpoint of the line segment connecting (x1 , y1 ) with (x2 , y2 ) in R2 . Generalize to Rn . 2.22 Show that if v = 0 then v/ v has length one. What if v = 0? 2.23 Show that if r 0 then rv is r times as long as v. What if r < 0?
44
Chapter One. Linear Systems
2.24 A vector v Rn of length one is a unit vector. Show that the dot product of two unit vectors has absolute value less than or equal to one. Can less than happen? Can equal to ? 2.25 Prove that u + v 2 + u v 2 = 2 u 2 + 2 v 2 . 2.26 Show that if x y = 0 for every y then x = 0. 2.27 Is u1 + + un u1 + + un ? If it is true then it would generalize the Triangle Inequality. 2.28 What is the ratio between the sides in the Cauchy-Schwartz inequality? 2.29 Why is the zero vector dened to be perpendicular to every vector? 2.30 Describe the angle between two vectors in R1 . 2.31 Give a simple necessary and sucient condition to determine whether the angle between two vectors is acute, right, or obtuse. 2.32 Generalize to Rn the converse of the Pythagorean Theorem, that if u and v are perpendicular then u + v 2 = u 2 + v 2 . 2.33 Show that u = v if and only if u + v and u v are perpendicular. Give an example in R2 . 2.34 Show that if a vector is perpendicular to each of two others then it is perpendicular to each vector in the plane they generate. (Remark. They could generate a degenerate plane a line or a point but the statement remains true.) 2.35 Prove that, where u, v Rn are nonzero vectors, the vector u v + u v bisects the angle between them. Illustrate in R2 . 2.36 Verify that the denition of angle is dimensionally correct: (1) if k > 0 then the cosine of the angle between ku and v equals the cosine of the angle between u and v, and (2) if k < 0 then the cosine of the angle between ku and v is the negative of the cosine of the angle between u and v. 2.37 Show that the inner product operation is linear : for u, v, w Rn and k, m R, u (kv + mw) = k(u v) + m(u w). 2.38 The geometric mean of two positive reals x, y is xy. It is analogous to the arithmetic mean (x + y)/2. Use the Cauchy-Schwartz inequality to show that the geometric mean of any x, y R is less than or equal to the arithmetic mean. ? 2.39 A ship is sailing with speed and direction v1 ; the wind blows apparently (judging by the vane on the mast) in the direction of a vector a; on changing the direction and speed of the ship from v1 to v2 the apparent wind is in the direction of a vector b. Find the vector velocity of the wind. [Am. Math. Mon., Feb. 1933] 2.40 Verify the Cauchy-Schwartz inequality by rst proving Lagranges identity:
2
aj b j
1jn
=
1jn
a2 j
1jn
b2 j
(ak bj aj bk )2
1k<jn
and then noting that the nal term is positive. (Recall the meaning aj b j = a1 b 1 + a2 b 2 + + an b n
1jn
and
1jn
aj 2 = a1 2 + a2 2 + + an 2
Section II. Linear Geometry of n-Space
45
of the notation.) This result is an improvement over Cauchy-Schwartz because it gives a formula for the dierence between the two sides. Interpret that dierence in R2 .
46
Chapter One. Linear Systems
III
Reduced Echelon Form
After developing the mechanics of Gauss method, we observed that it can be done in more than one way. One example is that we sometimes have to swap rows and there can be more than one row to choose from. Another example is that from this matrix 2 2 4 3 Gauss method could derive any of these echelon form matrices. 2 0 2 1 1 0 1 1 2 0 0 1
The rst results from 21 + 2 . The second comes from following (1/2)1 with 41 + 2 . The third comes from 21 + 2 followed by 22 + 1 (after the rst pivot the matrix is already in echelon form so the second one is extra work but it is nonetheless a legal row operation). The fact that the echelon form outcome of Gauss method is not unique leaves us with some questions. Will any two echelon form versions of a system have the same number of free variables? Will they in fact have exactly the same variables free? In this section we will answer both questions yes. We will do more than answer the questions. We will give a way to decide if one linear system can be derived from another by row operations. The answers to the two questions will follow from this larger result.
III.1 Gauss-Jordan Reduction
Gaussian elimination coupled with back-substitution solves linear systems, but its not the only method possible. Here is an extension of Gauss method that has some advantages. 1.1 Example To solve x + y 2z = 2 y + 3z = 7 x z = 1 we can start by going to 1 1 +3 0 0 echelon form as usual. 1 2 2 1 1 2 +3 7 0 1 1 3 1 1 0 0 1 2 3 4 2 7 8
We can keep going to a second stage 1 (1/4)3 0 0
by making the leading entries into ones 1 2 2 1 3 7 2 0 1
Section III. Reduced Echelon Form and then to a third stage that uses the leading entries to other entries in each column by pivoting upwards. 1 1 0 2 1 0 0 33 +2 2 +1 0 1 0 1 0 1 0 23 +1 0 0 1 2 0 0 1 The answer is x = 1, y = 1, and z = 2.
47 eliminate all of the 1 1 2
Note that the pivot operations in the rst stage proceed from column one to column three while the pivot operations in the third stage proceed from column three to column one. 1.2 Example We often combine the operations of the middle stage into a single step, even though they are operations on dierent rows. 2 1 4 2 7 6
21 +2
2 0 1 0
1 4 1/2 1
7 8 7/2 2 5/2 2
(1/2)1 (1/4)2 (1/2)2 +1
1 0 0 1
The answer is x = 5/2 and y = 2. This extension of Gauss method is Gauss-Jordan reduction. It goes past echelon form to a more rened, more specialized, matrix form. 1.3 Denition A matrix is in reduced echelon form if, in addition to being in echelon form, each leading entry is a one and is the only nonzero entry in its column. The disadvantage of using Gauss-Jordan reduction to solve a system is that the additional row operations mean additional arithmetic. The advantage is that the solution set can just be read o. In any echelon form, plain or reduced, we can read o when a system has an empty solution set because there is a contradictory equation, we can read o when a system has a one-element solution set because there is no contradiction and every variable is the leading variable in some row, and we can read o when a system has an innite solution set because there is no contradiction and at least one variable is free. In reduced echelon form we can read o not just what kind of solution set the system has, but also its description. Whether or not the echelon form is reduced, we have no trouble describing the solution set when it is empty, of course. The two examples above show that when the system has a single solution then the solution can be read o from the right-hand column. In the case when the solution set is innite, its parametrization can also be read o
48
Chapter One. Linear Systems that is shown
of the reduced echelon form. Consider, for example, this system brought to echelon form and then to reduced echelon form. 2 6 1 2 5 2 6 1 2 5 2 +3 0 3 1 4 1 0 3 1 4 1 0 3 1 2 5 0 0 0 2 4 1 0 1/2 (1/2)1 (4/3)3 +2 32 +1 0 1 1/3 3 +1 (1/3)2 0 0 0 (1/2)3
0 0 1
9/2 3 2
Starting with the middle matrix, the echelon form version, back substitution produces 2x4 = 4 so that x4 = 2, then another back substitution gives 3x2 + x3 + 4(2) = 1 implying that x2 = 3 (1/3)x3 , and then the nal back substitution gives 2x1 + 6(3 (1/3)x3 ) + x3 + 2(2) = 5 implying that x1 = (9/2) + (1/2)x3 . Thus the solution set is this. x1 9/2 1/2 x 3 1/3 S = { 2 = x3 0 + 1 x3 x3 R} x4 2 0 Now, considering the nal matrix, the reduced echelon form version, note that adjusting the parametrization by moving the x3 terms to the other side does indeed give the description of this innite solution set. Part of the reason that this works is straightforward. While a set can have many parametrizations that describe it, e.g., both of these also describe the above set S (take t to be x3 /6 and s to be x3 1) 9/2 3 4 1/2 3 2 8/3 1/3 { { 0 + 6 t t R} 1 + 1 s s R} 2 0 2 0 nonetheless we have in this book stuck to a convention of parametrizing using the unmodied free variables (that is, x3 = x3 instead of x3 = 6t). We can easily see that a reduced echelon form version of a system is equivalent to a parametrization in terms of unmodied free variables. For instance, 1 0 2 4 x1 = 4 2x3 0 1 1 3 x2 = 3 x3 0 0 0 0 (to move from left to right we also need to know how many equations are in the system). So, the convention of parametrizing with the free variables by solving each equation for its leading variable and then eliminating that leading variable from every other equation is exactly equivalent to the reduced echelon form conditions that each leading entry must be a one and must be the only nonzero entry in its column.
Section III. Reduced Echelon Form
49
Not as straightforward is the other part of the reason that the reduced echelon form version allows us to read o the parametrization that we would have gotten had we stopped at echelon form and then done back substitution. The prior paragraph shows that reduced echelon form corresponds to some parametrization, but why the same parametrization? A solution set can be parametrized in many ways, and Gauss method or the Gauss-Jordan method can be done in many ways, so a rst guess might be that we could derive many dierent reduced echelon form versions of the same starting system and many dierent parametrizations. But we never do. Experience shows that starting with the same system and proceeding with row operations in many dierent ways always yields the same reduced echelon form and the same parametrization (using the unmodied free variables). In the rest of this section we will show that the reduced echelon form version of a matrix is unique. It follows that the parametrization of a linear system in terms of its unmodied free variables is unique because two dierent ones would give two dierent reduced echelon forms. We shall use this result, and the ones that lead up to it, in the rest of the book but perhaps a restatement in a way that makes it seem more immediately useful may be encouraging. Imagine that we solve a linear system, parametrize, and check in the back of the book for the answer. But the parametrization there appears dierent. Have we made a mistake, or could these be dierent-looking descriptions of the same set, as with the three descriptions above of S? The prior paragraph notes that we will show here that dierent-looking parametrizations (using the unmodied free variables) describe genuinely dierent sets. Here is an informal argument that the reduced echelon form version of a matrix is unique. Consider again the example that started this section of a matrix that reduces to three dierent echelon form matrices. The rst matrix of the three is the natural echelon form version. The second matrix is the same as the rst except that a row has been halved. The third matrix, too, is just a cosmetic variant of the rst. The denition of reduced echelon form outlaws this kind of fooling around. In reduced echelon form, halving a row is not possible because that would change the rows leading entry away from one, and neither is combining rows possible, because then a leading entry would no longer be alone in its column. This informal justication is not a proof; we have argued that no two dierent reduced echelon form matrices are related by a single row operation step, but we have not ruled out the possibility that multiple steps might do. Before we go to that proof, we nish this subsection by rephrasing our work in a terminology that will be enlightening. Many dierent matrices yield the same reduced echelon form matrix. The three echelon form matrices from the start of this section, and the matrix they were derived from, all give this reduced echelon form matrix. 1 0 0 1 We think of these matrices as related to each other. The next result speaks to
50 this relationship.
Chapter One. Linear Systems
1.4 Lemma Elementary row operations are reversible.
Proof. For any matrix A, the eect of swapping rows is reversed by swapping
them back, multiplying a row by a nonzero k is undone by multiplying by 1/k, and adding a multiple of row i to row j (with i = j) is undone by subtracting the same multiple of row i from row j. A A
i j j i
A A
ki (1/k)i
A
ki +j ki +j
A
QED
(The i = j conditions is needed. See Exercise 13.)
This lemma suggests that reduces to is misleading where A B, we shouldnt think of B as after A or simpler than A. Instead we should think of them as interreducible or interrelated. Below is a picture of the idea. The matrices from the start of this section and their reduced echelon form version are shown in a cluster. They are all interreducible; these relationships are shown also.
2 0 0 1 1 0 1 1
2 4
2 3 1 0 0 1
2 0
2 1
We say that matrices that reduce to each other are equivalent with respect to the relationship of row reducibility. The next result veries this statement using the denition of an equivalence. 1.5 Lemma Between matrices, reduces to is an equivalence relation.
Proof. We must check the conditions (i) reexivity, that any matrix reduces to
itself, (ii) symmetry, that if A reduces to B then B reduces to A, and (iii) transitivity, that if A reduces to B and B reduces to C then A reduces to C. Reexivity is easy; any matrix reduces to itself in zero row operations. That the relationship is symmetric is Lemma 1.4 if A reduces to B by some row operations then also B reduces to A by reversing those operations. For transitivity, suppose that A reduces to B and that B reduces to C. Linking the reduction steps from A B with those from B C gives a reduction from A to C. QED 1.6 Denition Two matrices that are interreducible by the elementary row operations are row equivalent.
More information on equivalence relations is in the appendix.
Section III. Reduced Echelon Form
51
The diagram below shows the collection of all matrices as a box. Inside that box, each matrix lies in some class. Matrices are in the same class if and only if they are interreducible. The classes are disjoint no matrix is in two distinct classes. The collection of matrices has been partitioned into row equivalence classes.
A B
...
One of the classes in this partition is the cluster of matrices shown above, expanded to include all of the nonsingular 22 matrices. The next subsection proves that the reduced echelon form of a matrix is unique; that every matrix reduces to one and only one reduced echelon form matrix. Rephrased in terms of the row-equivalence relationship, we shall prove that every matrix is row equivalent to one and only one reduced echelon form matrix. In terms of the partition what we shall prove is: every equivalence class contains one and only one reduced echelon form matrix. So each reduced echelon form matrix serves as a representative of its class. After that proof we shall, as mentioned in the introduction to this section, have a way to decide if one matrix can be derived from another by row reduction. We just apply the Gauss-Jordan procedure to both and see whether or not they come to the same reduced echelon form. Exercises
1.7 Use Gauss-Jordan reduction to solve each system. (a) x + y = 2 (b) x z=4 (c) 3x 2y = 1 xy=0 2x + 2y =1 6x + y = 1/2 (d) 2x y = 1 x + 3y z = 5 y + 2z = 5 1.8 Find the reduced echelon form of each matrix. 1 3 1 1 0 3 1 2 2 1 2 0 4 (a) (b) (c) 1 4 2 1 5 1 3 1 3 3 3 4 8 1 2 0 1 3 2 (d) 0 0 5 6 1 5 1 5 1.9 Find each solution set by using Gauss-Jordan reduction, then reading o the parametrization. (a) 2x + y z = 1 (b) x z =1 (c) x y + z =0 4x y =3 y + 2z w = 3 y +w=0 x + 2y + 3z w = 7 3x 2y + 3z + w = 0 y w=0 (d) a + 2b + 3c + d e = 1 3a b + c + d + e = 3
More information on partitions and class representatives is in the appendix.
52
1.10 Give two distinct echelon form versions 2 1 1 6 4 1 1 5 1
Chapter One. Linear Systems
of this matrix. 3 2 5
1.11 List the reduced echelon forms possible for each size. (a) 22 (b) 23 (c) 32 (d) 33 1.12 What results from applying Gauss-Jordan reduction to a nonsingular matrix? 1.13 The proof of Lemma 1.4 contains a reference to the i = j condition on the row pivoting operation. (a) The denition of row operations has an i = j condition on the swap operation i j . Show that in A
i j i j
A this condition is not needed.
(b) Write down a 22 matrix with nonzero entries, and show that the 11 +1 operation is not reversed by 1 1 + 1 . (c) Expand the proof of that lemma to make explicit exactly where the i = j condition on pivoting is used.
III.2 Row Equivalence
We will close this section and this chapter by proving that every matrix is row equivalent to one and only one reduced echelon form matrix. The ideas that appear here will reappear, and be further developed, in the next chapter. The underlying theme here is that one way to understand a mathematical situation is by being able to classify the cases that can happen. We have met this theme several times already. We have classied solution sets of linear systems into the no-elements, one-element, and innitely-many elements cases. We have also classied linear systems with the same number of equations as unknowns into the nonsingular and singular cases. We adopted these classications because they give us a way to understand the situations that we were investigating. Here, where we are investigating row equivalence, we know that the set of all matrices breaks into the row equivalence classes. When we nish the proof here, we will have a way to understand each of those classes its matrices can be thought of as derived by row operations from the unique reduced echelon form matrix in that class. To understand how row operations act to transform one matrix into another, we consider the eect that they have on the parts of a matrix. The crucial observation is that row operations combine the rows linearly. 2.1 Denition A linear combination of x1 , . . . , xm is an expression of the form c1 x1 + c2 x2 + + cm xm where the cs are scalars. (We have already used the phrase linear combination in this book. The meaning is unchanged, but the next results statement makes a more formal denition in order.)
Section III. Reduced Echelon Form
53
2.2 Lemma (Linear Combination Lemma) A linear combination of linear combinations is a linear combination.
Proof. Given the linear combinations c1,1 x1 + + c1,n xn through cm,1 x1 +
+ cm,n xn , consider a combination of those d1 (c1,1 x1 + + c1,n xn ) + + dm (cm,1 x1 + + cm,n xn ) where the ds are scalars along with the cs. Distributing those ds and regrouping gives = d1 c1,1 x1 + + d1 c1,n xn + d2 c2,1 x1 + + dm c1,1 x1 + + dm c1,n xn = (d1 c1,1 + + dm cm,1 )x1 + + (d1 c1,n + + dm cm,n )xn which is indeed a linear combination of the xs.
QED
In this subsection we will use the convention that, where a matrix is named with an upper case roman letter, the matching lower-case greek letter names the rows. 1 1 2 2 A= B= . . . . . . m m 2.3 Corollary Where one matrix row reduces to another, each row of the second is a linear combination of the rows of the rst. The proof below uses induction on the number of row operations used to reduce one matrix to the other. Before we proceed, here is an outline of the argument (readers unfamiliar with induction may want to compare this argument with the one used in the General = Particular + Homogeneous proof). First, for the base step of the argument, we will verify that the proposition is true when reduction can be done in zero row operations. Second, for the inductive step, we will argue that if being able to reduce the rst matrix to the second in some number t 0 of operations implies that each row of the second is a linear combination of the rows of the rst, then being able to reduce the rst to the second in t + 1 operations implies the same thing. Together, this base step and induction step prove this result because by the base step the proposition is true in the zero operations case, and by the inductive step the fact that it is true in the zero operations case implies that it is true in the one operation case, and the inductive step applied again gives that it is therefore true in the two operations case, etc.
Proof. We proceed by induction on the minimum number of row operations
that take a rst matrix A to a second one B.
More information on mathematical induction is in the appendix.
54
Chapter One. Linear Systems
In the base step, that zero reduction operations suce, the two matrices are equal and each row of B is obviously a combination of As rows: i = 0 1 + + 1 i + + 0 m . For the inductive step, assume the inductive hypothesis: with t 0, if a matrix can be derived from A in t or fewer operations then its rows are linear combinations of the As rows. Consider a B that takes t+1 operations. Because there are more than zero operations, there must be a next-to-last matrix G so that A G B. This G is only t operations away from A and so the inductive hypothesis applies to it, that is, each row of G is a linear combination of the rows of A. If the last operation, the one from G to B, is a row swap then the rows of B are just the rows of G reordered and thus each row of B is also a linear combination of the rows of A. The other two possibilities for this last operation, that it multiplies a row by a scalar and that it adds a multiple of one row to another, both result in the rows of B being linear combinations of the rows of G. But therefore, by the Linear Combination Lemma, each row of B is a linear combination of the rows of A. With that, we have both the base step and the inductive step, and so the proposition follows. QED 2.4 Example In the reduction 0 2 1 1
1 2
1 0
1 2
(1/2)2
1 0
1 1
2 +1
1 0 0 1
call the matrices A, D, G, and B. The methods of the proof show that there are three sets of linear relationships. 1 = 0 1 + 1 2 2 = 1 1 + 0 2 1 = 0 1 + 1 2 2 = (1/2)1 + 0 2 1 = (1/2)1 + 1 2 2 = (1/2)1 + 0 2
The prior result gives us the insight that Gauss method works by taking linear combinations of the rows. But to what end; why do we go to echelon form as a particularly simple, or basic, version of a linear system? The answer, of course, is that echelon form is suitable for back substitution, because we have isolated the variables. For instance, in this matrix 2 3 7 8 0 0 0 0 1 5 1 1 R= 0 0 0 3 3 0 0 0 0 0 2 1 x1 has been removed from x5 s equation. That is, Gauss method has made x5 s row independent of x1 s row. Independence of a collection of row vectors, or of any kind of vectors, will be precisely dened and explored in the next chapter. But a rst take on it is that we can show that, say, the third row above is not comprised of the other
Section III. Reduced Echelon Form
55
rows, that 3 = c1 1 + c2 2 + c4 4 . For, suppose that there are scalars c1 , c2 , and c4 such that this relationship holds. 0 0 0 3 3 0 = c1 2 3 7 8 1 0 5 0 0 1 2 0 1 1
+ c2 0 0 + c4 0 0
The rst rows leading entry is in the rst column and narrowing our consideration of the above relationship to consideration only of the entries from the rst column 0 = 2c1 +0c2 +0c4 gives that c1 = 0. The second rows leading entry is in the third column and the equation of entries in that column 0 = 7c1 + 1c2 + 0c4 , along with the knowledge that c1 = 0, gives that c2 = 0. Now, to nish, the third rows leading entry is in the fourth column and the equation of entries in that column 3 = 8c1 + 5c2 + 0c4 , along with c1 = 0 and c2 = 0, gives an impossibility. The following result shows that this eect always holds. It shows that what Gauss linear elimination method eliminates is linear relationships among the rows. 2.5 Lemma In an echelon form matrix, no nonzero row is a linear combination of the other rows.
Proof. Let R be in echelon form. Suppose, to obtain a contradiction, that
some nonzero row is a linear combination of the others. i = c1 1 + . . . + ci1 i1 + ci+1 i+1 + . . . + cm m We will rst use induction to show that the coecients c1 , . . . , ci1 associated with rows above i are all zero. The contradiction will come from consideration of i and the rows below it. The base step of the induction argument is to show that the rst coecient c1 is zero. Let the rst rows leading entry be in column number 1 be the column number of the leading entry of the rst row and consider the equation of entries in that column. i,
1
= c1 1, 1 + . . . + ci1 i1, 1 + ci+1 i+1, 1 + . . . + cm m,
1
The matrix is in echelon form so the entries 2, 1 , . . . , m, 1 , including i, 1 , are all zero. 0 = c1 1, 1 + + ci1 0 + ci+1 0 + + cm 0 Because the entry 1, 1 is nonzero as it leads its row, the coecient c1 must be zero. The inductive step is to show that for each row index k between 1 and i 2, if the coecient c1 and the coecients c2 , . . . , ck are all zero then ck+1 is also zero. That argument, and the contradiction that nishes this proof, is saved for QED Exercise 21.
56
Chapter One. Linear Systems
We can now prove that each matrix is row equivalent to one and only one reduced echelon form matrix. We will nd it convenient to break the rst half of the argument o as a preliminary lemma. For one thing, it holds for any echelon form whatever, not just reduced echelon form. 2.6 Lemma If two echelon form matrices are row equivalent then the leading entries in their rst rows lie in the same column. The same is true of all the nonzero rows the leading entries in their second rows lie in the same column, etc. For the proof we rephrase the result in more technical terms. Dene the form of an m n matrix to be the sequence 1 , 2 , . . . , m where i is the column number of the leading entry in row i and i = if there is no leading entry in that column. The lemma says that if two echelon form matrices are row equivalent then their forms are equal sequences.
Proof. Let B and D be echelon form matrices that are row equivalent. Because
they are row equivalent they must be the same size, say nm. Let the column number of the leading entry in row i of B be i and let the column number of the leading entry in row j of D be kj . We will show that 1 = k1 , that 2 = k2 , etc., by induction. This induction argument relies on the fact that the matrices are row equivalent, because the Linear Combination Lemma and its corollary therefore give that each row of B is a linear combination of the rows of D and vice versa: i = si,1 1 + si,2 2 + + si,m m and j = tj,1 1 + tj,2 2 + + tj,m m
where the ss and ts are scalars. The base step of the induction is to verify the lemma for the rst rows of the matrices, that is, to verify that 1 = k1 . If either row is a zero row then the entire matrix is a zero matrix since it is in echelon form, and hterefore both matrices are zero matrices (by Corollary 2.3), and so both 1 and k1 are . For the case where neither 1 nor 1 is a zero row, consider the i = 1 instance of the linear relationship above. 1 = s1,1 1 + s1,2 2 + + s1,m m 0 b1,
1
= s1,1 0 + s1,2 0 . . .
d1,k1 0 0
+ s1,m 0
First, note that 1 < k1 is impossible: in the columns of D to the left of column k1 the entries are are all zeroes (as d1,k1 leads the rst row) and so if 1 < k1 then the equation of entries from column 1 would be b1, 1 = s1,1 0+ +s1,m 0, but b1, 1 isnt zero since it leads its row and so this is an impossibility. Next, a symmetric argument shows that k1 < 1 also is impossible. Thus the 1 = k1 base case holds.
Section III. Reduced Echelon Form
57
The inductive step is to show that if 1 = k1 , and 2 = k2 , . . . , and r = kr , then also r+1 = kr+1 (for r in the interval 1 .. m 1). This argument is saved for Exercise 22. QED That lemma answers two of the questions that we have posed (i) any two echelon form versions of a matrix have the same free variables, and consequently (ii) any two echelon form versions have the same number of free variables. There is no linear system and no combination of row operations such that, say, we could solve the system one way and get y and z free but solve it another way and get y and w free, or solve it one way and get two free variables while solving it another way yields three. We nish now by specializing to the case of reduced echelon form matrices. 2.7 Theorem Each matrix is row equivalent to a unique reduced echelon form matrix.
Proof. Clearly any matrix is row equivalent to at least one reduced echelon
form matrix, via Gauss-Jordan reduction. For the other half, that any matrix is equivalent to at most one reduced echelon form matrix, we will show that if a matrix Gauss-Jordan reduces to each of two others then those two are equal. Suppose that a matrix is row equivalent the two reduced echelon form matrices B and D, which are therefore row equivalent to each other. The Linear Combination Lemma and its corollary allow us to write the rows of one, say B, as a linear combination of the rows of the other i = ci,1 1 + + ci,m m . The preliminary result, Lemma 2.6, says that in the two matrices, the same collection of rows are nonzero. Thus, if 1 through r are the nonzero rows of B then the nonzero rows of D are 1 through r . Zero rows dont contribute to the sum so we can rewrite the relationship to include just the nonzero rows. i = ci,1 1 + + ci,r r ()
The preliminary result also says that for each row j between 1 and r, the leading entries of the j-th row of B and D appear in the same column, denoted j . Rewriting the above relationship to focus on the entries in the j -th column bi,
j
= ci,1 + ci,2 . . . + ci,r
d1,
j j
d2, dr,
j
gives this set of equations for i = 1 up to i = r. b1, bj, br,
j
= c1,1 d1, j + + c1,j dj, j + + c1,r dr, . . . = cj,1 d1, j + + cj,j dj, j + + cj,r dr, . . . = cr,1 d1, j + + cr,j dj, j + + cr,r dr,
j
j
j
j
j
58
Chapter One. Linear Systems
Since D is in reduced echelon form, all of the ds in column j are zero except for dj, j , which is 1. Thus each equation above simplies to bi, j = ci,j dj, j = ci,j 1. But B is also in reduced echelon form and so all of the bs in column j are zero except for bj, j , which is 1. Therefore, each ci,j is zero, except that c1,1 = 1, and c2,2 = 1, . . . , and cr,r = 1. We have shown that the only nonzero coecient in the linear combination labelled () is cj,j , which is 1. Therefore j = j . Because this holds for all nonzero rows, B = D. QED We end with a recap. In Gauss method we start with a matrix and then derive a sequence of other matrices. We dened two matrices to be related if one can be derived from the other. That relation is an equivalence relation, called row equivalence, and so partitions the set of all matrices into row equivalence classes.
13 27 13 01
...
(There are innitely many matrices in the pictured class, but weve only got room to show two.) We have proved there is one and only one reduced echelon form matrix in each row equivalence class. So the reduced echelon form is a canonical form for row equivalence: the reduced echelon form matrices are representatives of the classes.
10 01
...
We can answer questions about the classes by translating them into questions about the representatives. 2.8 Example We can decide if matrices are interreducible by seeing if GaussJordan reduction produces the same reduced echelon form result. Thus, these are not row equivalent 1 2
3 6
1 2
3 5
More information on canonical representatives is in the appendix.
Section III. Reduced Echelon Form because their reduced echelon forms are not equal. 1 0 3 0 1 0 0 1
59
2.9 Example Any nonsingular 33 matrix Gauss-Jordan reduces to this. 1 0 0 1 0 0 0 0 1
2.10 Example We can describe the classes by listing all possible reduced echelon form matrices. Any 22 matrix lies in one of these: the class of matrices row equivalent to this, 0 0 0 0 the innitely many classes of matrices row equivalent to one of this type 1 0 a 0
where a R (including a = 0), the class of matrices row equivalent to this, 0 1 0 0 and the class of matrices row equivalent to this 1 0 0 1 (this the class of nonsingular 22 matrices). Exercises
2.11 Decide if the matrices are row equivalent. 1 0 2 0 1 1 2 , (a) (b) 3 1 1 , 1 2 4 8 5 1 5 2 1 1 1 0 2 1 0 , (c) 1 1 (d) 0 2 10 1 4 3 1 1 1 1 0 1 2 (e) , 0 0 3 1 1 1 1 0 2 1 2 0 2 0 2 10 4 3 2 1 5
1 0 , 2 2
2.12 Describe the matrices in each of the classes represented in Example 2.10. 2.13 Describe all matrices in the row equivalence class of these.
60
Chapter One. Linear Systems
1 0 1 2 1 1 (b) (c) 0 0 2 4 1 3 2.14 How many row equivalence classes are there? 2.15 Can row equivalence classes contain dierent-sized matrices? 2.16 How big are the row equivalence classes? (a) Show that the class of any zero matrix is nite. (b) Do any other classes contain only nitely many members? 2.17 Give two reduced echelon form matrices that have their leading entries in the same columns, but that are not row equivalent. 2.18 Show that any two n n nonsingular matrices are row equivalent. Are any two singular matrices row equivalent? 2.19 Describe all of the row equivalence classes containing these. (a) 2 2 matrices (b) 2 3 matrices (c) 3 2 matrices (d) 33 matrices 2.20 (a) Show that a vector 0 is a linear combination of members of the set {1 , . . . , n } if and only there is a linear relationship 0 = c0 0 + + cn n where c0 is not zero. (Watch out for the 0 = 0 case.) (b) Derive Lemma 2.5. 2.21 Finish the proof of Lemma 2.5. (a) First illustrate the inductive step by showing that 2 = k2 . (b) Do the full inductive step: assume that ck is zero for 1 k < i 1, and deduce that ck+1 is also zero. (c) Find the contradiction. 2.22 Finish the induction argument in Lemma 2.6. (a) State the inductive hypothesis, Also state what must be shown to follow from that hypothesis. (b) Check that the inductive hypothesis implies that in the relationship r+1 = sr+1,1 1 + sr+2,2 2 + + sr+1,m m the coecients sr+1,1 , . . . , sr+1,r are each zero. (c) Finish the inductive step by arguing, as in the base case, that r+1 < kr+1 and kr+1 < r+1 are impossible. 2.23 Why, in the proof of Theorem 2.7, do we bother to restrict to the nonzero rows? Why not just stick to the relationship that we began with, i = ci,1 1 + +ci,m m , with m instead of r, and argue using it that the only nonzero coecient is ci,i , which is 1? 2.24 Three truck drivers went into a roadside cafe. One truck driver purchased four sandwiches, a cup of coee, and ten doughnuts for $8.45. Another driver purchased three sandwiches, a cup of coee, and seven doughnuts for $6.30. What did the third truck driver pay for a sandwich, a cup of coee, and a doughnut? [Trono] 2.25 The fact that Gaussian reduction disallows multiplication of a row by zero is needed for the proof of uniqueness of reduced echelon form, or else every matrix would be row equivalent to a matrix of all zeros. Where is it used? 2.26 The Linear Combination Lemma says which equations can be gotten from Gaussian reduction from a given linear system. (a) (1) Produce an equation not implied by this system. 3x + 4y = 8 2x + y = 3
Section III. Reduced Echelon Form
(2) Can any equation be derived from an inconsistent system?
61
2.27 Extend the denition of row equivalence to linear systems. Under your denition, do equivalent systems have the same solution set? [Homan & Kunze] 2.28 In this matrix 1 2 3 3 0 3 1 4 5 the rst and second columns add to the third. (a) Show that remains true under any row operation. (b) Make a conjecture. (c) Prove that it holds.
62
Chapter One. Linear Systems
Topic: Computer Algebra Systems
The linear systems in this chapter are small enough that their solution by hand is easy. But large systems are easiest, and safest, to do on a computer. There are special purpose programs such as LINPACK for this job. Another popular tool is a general purpose computer algebra system, including both commercial packages such as Maple, Mathematica, or MATLAB, or free packages such as SciLab,, MuPAD, or Octave. For example, in the Topic on Networks, we need to solve this. i0 i1 i2 i1 i2 = 0 i5 = 0 i4 + i5 = 0 i3 + i4 i6 = 0 5i1 + 10i3 = 10 2i2 + 4i4 = 10 5i1 2i2 + 50i5 = 0 i3
It can be done by hand, but it would take a while and be error-prone. Using a computer is better. We illustrate by solving that system under Maple (for another system, a users manual would obviously detail the exact syntax needed). The array of coecients can be entered in this way
> A:=array( [[1,-1,-1,0,0,0,0], [0,1,0,-1,0,-1,0], [0,0,1,0,-1,1,0], [0,0,0,1,1,0,-1], [0,5,0,10,0,0,0], [0,0,2,0,4,0,0], [0,5,-1,0,0,10,0]] );
(putting the rows on separate lines is not necessary, but is done for clarity). The vector of constants is entered similarly.
> u:=array( [0,0,0,0,10,10,0] );
Then the system is solved, like magic.
> linsolve(A,u); 7 2 5 2 5 7 [ -, -, -, -, -, 0, - ] 3 3 3 3 3 3
Systems with innitely many solutions are solved in the same way the computer simply returns a parametrization. Exercises
Answers for this Topic use Maple as the computer algebra system. In particular, all of these were tested on Maple V running under MS-DOS NT version 4.0. (On all of them, the preliminary command to load the linear algebra package along with Maples responses to the Enter key, have been omitted.) Other systems have similar commands.
Topic: Computer Algebra Systems
1 Use the computer to solve the two problems that opened this chapter. (a) This is the Statics problem. 40h + 15c = 100 25c = 50 + 50h (b) This is the Chemistry problem. 7h = 7j 8h + 1i = 5j + 2k 1i = 3j 3i = 6j + 1k
63
2 Use the computer to solve these systems from the rst subsection, or conclude many solutions or no solutions. (a) 2x + 2y = 5 (b) x + y = 1 (c) x 3y + z = 1 x 4y = 0 x+y=2 x + y + 2z = 14 (d) x y = 1 (e) 4y + z = 20 (f ) 2x + z+w= 5 3x 3y = 2 2x 2y + z = 0 y w = 1 x +z= 5 3x zw= 0 x + y z = 10 4x + y + 2z + w = 9 3 Use the computer to solve these systems from the second subsection. (a) 3x + 6y = 18 (b) x + y = 1 (c) x1 + x3 = 4 x + 2y = 6 x y = 1 x1 x2 + 2x3 = 5 4x1 x2 + 5x3 = 17 (d) 2a + b c = 2 (e) x + 2y z =3 (f ) x +z+w=4 2a +c=3 2x + y +w=4 2x + y w=2 ab =0 x y+z+w=1 3x + y + z =7 4 What does the computer give for the solution of the general 22 system? ax + cy = p bx + dy = q
64
Chapter One. Linear Systems
Topic: Input-Output Analysis
An economy is an immensely complicated network of interdependences. Changes in one part can ripple out to aect other parts. Economists have struggled to be able to describe, and to make predictions about, such a complicated object. Mathematical models using systems of linear equations have emerged as a key tool. One is Input-Output Analysis, pioneered by W. Leontief, who won the 1973 Nobel Prize in Economics. Consider an economy with many parts, two of which are the steel industry and the auto industry. As they work to meet the demand for their product from other parts of the economy, that is, from users external to the steel and auto sectors, these two interact tightly. For instance, should the external demand for autos go up, that would lead to an increase in the auto industrys usage of steel. Or, should the external demand for steel fall, then it would lead to a fall in steels purchase of autos. The type of Input-Output model we will consider takes in the external demands and then predicts how the two interact to meet those demands. We start with a listing of production and consumption statistics. (These numbers, giving dollar values in millions, are excerpted from [Leontief 1965], describing the 1958 U.S. economy. Todays statistics would be quite dierent, both because of ination and because of technical changes in the industries.) used by steel value of steel value of auto 5 395 48 used by auto 2 664 9 030 used by others total 25 448 30 346
For instance, the dollar value of steel used by the auto industry in this year is 2, 664 million. Note that industries may consume some of their own output. We can ll in the blanks for the external demand. This years value of the steel used by others this year is 17, 389 and this years value of the auto used by others is 21, 268. With that, we have a complete description of the external demands and of how auto and steel interact, this year, to meet them. Now, imagine that the external demand for steel has recently been going up by 200 per year and so we estimate that next year it will be 17, 589. Imagine also that for similar reasons we estimate that next years external demand for autos will be down 25 to 21, 243. We wish to predict next years total outputs. That prediction isnt as simple as adding 200 to this years steel total and subtracting 25 from this years auto total. For one thing, a rise in steel will cause that industry to have an increased demand for autos, which will mitigate, to some extent, the loss in external demand for autos. On the other hand, the drop in external demand for autos will cause the auto industry to use less steel, and so lessen somewhat the upswing in steels business. In short, these two industries form a system, and we need to predict the totals at which the system as a whole will settle.
Topic: Input-Output Analysis
65
For that prediction, let s be next years total production of steel and let a be next years total output of autos. We form these equations. next years production of steel = next years use of steel by steel + next years use of steel by auto + next years use of steel by others next years production of autos = next years use of autos by steel + next years use of autos by auto + next years use of autos by others On the left side of those equations go the unknowns s and a. At the ends of the right sides go our external demand estimates for next year 17, 589 and 21, 243. For the remaining four terms, we look to the table of this years information about how the industries interact. For instance, for next years use of steel by steel, we note that this year the steel industry used 5395 units of steel input to produce 25, 448 units of steel output. So next year, when the steel industry will produce s units out, we expect that doing so will take s (5395)/(25 448) units of steel input this is simply the assumption that input is proportional to output. (We are assuming that the ratio of input to output remains constant over time; in practice, models may try to take account of trends of change in the ratios.) Next years use of steel by the auto industry is similar. This year, the auto industry uses 2664 units of steel input to produce 30346 units of auto output. So next year, when the auto industrys total output is a, we expect it to consume a (2664)/(30346) units of steel. Filling in the other equation in the same way, we get this system of linear equation. 5 395 2 664 s+ a + 17 589 = s 25 448 30 346 48 9 030 s+ a + 21 243 = a 25 448 30 346 Gauss method on this system. (20 053/25 448)s (2 664/30 346)a = 17 589 (48/25 448)s + (21 316/30 346)a = 21 243 gives s = 25 698 and a = 30 311. Looking back, recall that above we described why the prediction of next years totals isnt as simple as adding 200 to last years steel total and subtracting 25 from last years auto total. In fact, comparing these totals for next year to the ones given at the start for the current year shows that, despite the drop in external demand, the total production of the auto industry is predicted to rise. The increase in internal demand for autos caused by steels sharp rise in business more than makes up for the loss in external demand for autos. One of the advantages of having a mathematical model is that we can ask What if . . . ? questions. For instance, we can ask What if the estimates for
66
Chapter One. Linear Systems
next years external demands are somewhat o? To try to understand how much the models predictions change in reaction to changes in our estimates, we can try revising our estimate of next years external steel demand from 17, 589 down to 17, 489, while keeping the assumption of next years external demand for autos xed at 21, 243. The resulting system (20 053/25 448)s (2 664/30 346)a = 17 489 (48/25 448)s + (21 316/30 346)a = 21 243 when solved gives s = 25 571 and a = 30 311. This kind of exploration of the model is sensitivity analysis. We are seeing how sensitive the predictions of our model are to the accuracy of the assumptions. Obviously, we can consider larger models that detail the interactions among more sectors of an economy. These models are typically solved on a computer, using the techniques of matrix algebra that we will develop in Chapter Three. Some examples are given in the exercises. Obviously also, a single model does not suit every case; expert judgment is needed to see if the assumptions underlying the model are reasonable for a particular case. With those caveats, however, this model has proven in practice to be a useful and accurate tool for economic analysis. For further reading, try [Leontief 1951] and [Leontief 1965]. Exercises
Hint: these systems are easiest to solve on a computer. 1 With the steel-auto system given above, estimate next years total productions in these cases. (a) Next years external demands are: up 200 from this year for steel, and unchanged for autos. (b) Next years external demands are: up 100 for steel, and up 200 for autos. (c) Next years external demands are: up 200 for steel, and up 200 for autos. 2 In the steel-auto system, the ratio for the use of steel by the auto industry is 2 664/30 346, about 0.0878. Imagine that a new process for making autos reduces this ratio to .0500. (a) How will the predictions for next years total productions change compared to the rst example discussed above (i.e., taking next years external demands to be 17, 589 for steel and 21, 243 for autos)? (b) Predict next years totals if, in addition, the external demand for autos rises to be 21, 500 because the new cars are cheaper. 3 This table gives the numbers for the auto-steel system from a dierent year, 1947 (see [Leontief 1951]). The units here are billions of 1947 dollars. used by used by used by steel auto others total value of 6.90 1.28 18.69 steel value of 0 4.40 14.27 autos (a) Solve for total output if next years external demands are: steels demand up 10% and autos demand up 15%. (b) How do the ratios compare to those given above in the discussion for the 1958 economy?
Topic: Input-Output Analysis
67
(c) Solve the 1947 equations with the 1958 external demands (note the dierence in units; a 1947 dollar buys about what $1.30 in 1958 dollars buys). How far o are the predictions for total output? 4 Predict next years total productions of each of the three sectors of the hypothetical economy shown below used by used by used by used by farm rail shipping others total value of 25 50 100 800 farm value of 25 50 50 300 rail value of 15 10 0 500 shipping if next years external demands are as stated. (a) 625 for farm, 200 for rail, 475 for shipping (b) 650 for farm, 150 for rail, 450 for shipping 5 This table gives the interrelationships among three segments of an economy (see [Clark & Coupe]). used by used by used by used by total food wholesale retail others value of food 0 2 318 4 679 11 869 value of wholesale 393 1 089 22 459 122 242 value of retail 3 53 75 116 041 We will do an Input-Output analysis on this system. (a) Fill in the numbers for this years external demands. (b) Set up the linear system, leaving next years external demands blank. (c) Solve the system where next years external demands are calculated by taking this years external demands and inating them 10%. Do all three sectors increase their total business by 10%? Do they all even increase at the same rate? (d) Solve the system where next years external demands are calculated by taking this years external demands and reducing them 7%. (The study from which these numbers are taken concluded that because of the closing of a local military facility, overall personal income in the area would fall 7%, so this might be a rst guess at what would actually happen.)
68
Chapter One. Linear Systems
Topic: Accuracy of Computations
Gauss method lends itself nicely to computerization. The code below illustrates. It operates on an nn matrix a, pivoting with the rst row, then with the second row, etc.
for(pivot_row=1;pivot_row<=n-1;pivot_row++){ for(row_below=pivot_row+1;row_below<=n;row_below++){ multiplier=a[row_below,pivot_row]/a[pivot_row,pivot_row]; for(col=pivot_row;col<=n;col++){ a[row_below,col]-=multiplier*a[pivot_row,col]; } } }
(This code is in the C language. Here is a brief translation. The loop construct for(pivot row=1;pivot row<=n-1;pivot row++){ } sets pivot row to 1 and then iterates while pivot row is less than or equal to n 1, each time through incrementing pivot row by one with the ++ operation. The other non-obvious construct is that the -= in the innermost loop amounts to the a[row below,col] = multiplier a[pivot row,col] + a[row below,col] operation.) While this code provides a quick take on how Gauss method can be mechanized, it is not ready to use. It is naive in many ways. The most glaring way is that it assumes that a nonzero number is always found in the pivot row, pivot row position for use as the pivot entry. To make it practical, one way in which this code needs to be reworked is to cover the case where nding a zero in that location leads to a row swap, or to the conclusion that the matrix is singular. Adding some if statements to cover those cases is not hard, but we will instead consider some more subtle ways in which the code is naive. There are pitfalls arising from the computers reliance on nite-precision oating point arithmetic. For example, we have seen above that we must handle as a separate case a system that is singular. But systems that are nearly singular also require care. Consider this one. x + 2y = 3 1.000 000 01x + 2y = 3.000 000 01 By eye we get the solution x = 1 and y = 1. But a computer has more trouble. A computer that represents real numbers to eight signicant places (as is common, usually called single precision) will represent the second equation internally as 1.000 000 0x + 2y = 3.000 000 0, losing the digits in the ninth place. Instead of reporting the correct solution, this computer will report something that is not even close this computer thinks that the system is singular because the two equations are represented internally as equal. For some intuition about how the computer could come up with something that far o, we can graph the system.
Topic: Accuracy of Computations
69
(1, 1)
At the scale of this graph, the two lines cannot be resolved apart. This system is nearly singular in the sense that the two lines are nearly the same line. Nearsingularity gives this system the property that a small change in the system can cause a large change in its solution; for instance, changing the 3.000 000 01 to 3.000 000 03 changes the intersection point from (1, 1) to (3, 0). This system changes radically depending on a ninth digit, which explains why the eightplace computer has trouble. A problem that is very sensitive to inaccuracy or uncertainties in the input values is ill-conditioned. The above example gives one way in which a system can be dicult to solve on a computer. It has the advantage that the picture of nearly-equal lines gives a memorable insight into one way that numerical diculties can arise. Unfortunately this insight isnt very useful when we wish to solve some large system. We cannot, typically, hope to understand the geometry of an arbitrary large system. In addition, there are ways that a computers results may be unreliable other than that the angle between some of the linear surfaces is quite small. For an example, consider the system below, from [Hamming]. 0.001x + y = 1 xy=0 ()
The second equation gives x = y, so x = y = 1/1.001 and thus both variables have values that are just less than 1. A computer using two digits represents the system internally in this way (we will do this example in two-digit oating point arithmetic, but a similar one with eight digits is easy to invent). (1.0 102 )x + (1.0 100 )y = 1.0 100 (1.0 100 )x (1.0 100 )y = 0.0 100 The computers row reduction step 10001 + 2 produces a second equation 1001y = 999, which the computer rounds to two places as (1.0 103 )y = 1.0 103 . Then the computer decides from the second equation that y = 1 and from the rst equation that x = 0. This y value is fairly good, but the x is quite bad. Thus, another cause of unreliable output is a mixture of oating point arithmetic and a reliance on pivots that are small. An experienced programmer may respond that we should go to double precision where sixteen signicant digits are retained. This will indeed solve many problems. However, there are some diculties with it as a general approach. For one thing, double precision takes longer than single precision (on a 486
70
Chapter One. Linear Systems
chip, multiplication takes eleven ticks in single precision but fourteen in double precision [Programmers Ref.]) and has twice the memory requirements. So attempting to do all calculations in double precision is just not practical. And besides, the above systems can obviously be tweaked to give the same trouble in the seventeenth digit, so double precision wont x all problems. What we need is a strategy to minimize the numerical trouble arising from solving systems on a computer, and some guidance as to how far the reported solutions can be trusted. Mathematicians have made a careful study of how to get the most reliable results. A basic improvement on the naive code above is to not simply take the entry in the pivot row , pivot row position for the pivot, but rather to look at all of the entries in the pivot row column below the pivot row row, and take the one that is most likely to give reliable results (e.g., take one that is not too small). This strategy is partial pivoting. For example, to solve the troublesome system () above, we start by looking at both equations for a best rst pivot, and taking the 1 in the second equation as more likely to give good results. Then, the pivot step of .0012 + 1 gives a rst equation of 1.001y = 1, which the computer will represent as (1.0100 )y = 1.0100 , leading to the conclusion that y = 1 and, after back-substitution, x = 1, both of which are close to right. The code from above can be adapted to this purpose.
for(pivot_row=1;pivot_row<=n-1;pivot_row++){ /* find the largest pivot in this column (in row max) */ max=pivot_row; for(row_below=pivot_row+1;pivot_row<=n;row_below++){ if (abs(a[row_below,pivot_row]) > abs(a[max,row_below])) max=row_below; } /* swap rows to move that pivot entry up */ for(col=pivot_row;col<=n;col++){ temp=a[pivot_row,col]; a[pivot_row,col]=a[max,col]; a[max,col]=temp; } /* proceed as before */ for(row_below=pivot_row+1;row_below<=n;row_below++){ multiplier=a[row_below,pivot_row]/a[pivot_row,pivot_row]; for(col=pivot_row;col<=n;col++){ a[row_below,col]-=multiplier*a[pivot_row,col]; } } }
A full analysis of the best way to implement Gauss method is outside the scope of the book (see [Wilkinson 1965]), but the method recommended by most experts is a variation on the code above that rst nds the best pivot among the candidates, and then scales it to a number that is less likely to give trouble. This is scaled partial pivoting.
Topic: Accuracy of Computations
71
In addition to returning a result that is likely to be reliable, most well-done code will return a number, called the conditioning number that describes the factor by which uncertainties in the input numbers could be magnied to become inaccuracies in the results returned (see [Rice]). The lesson of this discussion is that just because Gauss method always works in theory, and just because computer code correctly implements that method, and just because the answer appears on green-bar paper, doesnt mean that the answer is reliable. In practice, always use a package where experts have worked hard to counter what can go wrong. Exercises
1 Using two decimal places, add 253 and 2/3. 2 This intersect-the-lines problem contrasts with the example discussed above.
(1, 1)
x + 2y = 3 3x 2y = 1
Illustrate that in this system some small change in the numbers will produce only a small change in the solution by changing the constant in the bottom equation to 1.008 and solving. Compare it to the solution of the unchanged system. 3 Solve this system by hand ([Rice]). 0.000 3x + 1.556y = 1.569 0.345 4x 2.346y = 1.018 (a) Solve it accurately, by hand. (b) Solve it by rounding at each step to four signicant digits. 4 Rounding inside the computer often has an eect on the result. Assume that your machine has eight signicant digits. (a) Show that the machine will compute (2/3) + ((2/3) (1/3)) as unequal to ((2/3) + (2/3)) (1/3). Thus, computer arithmetic is not associative. (b) Compare the computers version of (1/3)x + y = 0 and (2/3)x + 2y = 0. Is twice the rst equation the same as the second? 5 Ill-conditioning is not only dependent on the matrix of coecients. This example [Hamming] shows that it can arise from an interaction between the left and right sides of the system. Let be a small real. 3x + 2y + z = 6 2x + 2y + 2z = 2 + 4 x + 2y z = 1 + (a) Solve the system by hand. Notice that the s divide out only because there is an exact cancelation of the integer parts on the right side as well as on the left. (b) Solve the system by hand, rounding to two decimal places, and with = 0.001.
72
Chapter One. Linear Systems
Topic: Analyzing Networks
The diagram below shows some of a cars electrical network. The battery is on the left, drawn as stacked line segments. The wires are drawn as lines, shown straight and with sharp right angles for neatness. Each light is a circle enclosing a loop.
Brake Actuated Switch 12V Light Switch O Dimmer Hi Door Actuated Switch Dome Light
Lo
L
R Brake Lights
L
R Parking Lights
L
R Rear Lights
L
R
L
R
Headlights
The designer of such a network needs to answer questions like: How much electricity ows when both the hi-beam headlights and the brake lights are on? Below, we will use linear systems to analyze simpler versions of electrical networks. For the analysis we need two facts about electricity and two facts about electrical networks. The rst fact about electricity is that a battery is like a pump: it provides a force impelling the electricity to ow through the circuits connecting the batterys ends, if there are any such circuits. We say that the battery provides a potential to ow. Of course, this network accomplishes its function when, as the electricity ows through a circuit, it goes through a light. For instance, when the driver steps on the brake then the switch makes contact and a circuit is formed on the left side of the diagram, and the electrical current owing through that circuit will make the brake lights go on, warning drivers behind. The second electrical fact is that in some kinds of network components the amount of ow is proportional to the force provided by the battery. That is, for each such component there is a number, its resistance, such that the potential is equal to the ow times the resistance. The units of measurement are: potential is described in volts, the rate of ow is in amperes, and resistance to the ow is in ohms. These units are dened so that volts = amperes ohms. Components with this property, that the voltage-amperage response curve is a line through the origin, are called resistors. (Light bulbs such as the ones shown above are not this kind of component, because their ohmage changes as they heat up.) For example, if a resistor measures 2 ohms then wiring it to a 12 volt battery results in a ow of 6 amperes. Conversely, if we have ow of electrical current of 2 amperes through it then there must be a 4 volt potential
Topic: Analyzing Networks
73
dierence between its ends. This is the voltage drop across the resistor. One way to think of a electrical circuits like the one above is that the battery provides a voltage rise while the other components are voltage drops. The two facts that we need about networks are Kirchhos Laws. Current Law. For any point in a network, the ow in equals the ow out. Voltage Law. Around any circuit the total drop equals the total rise. In the above network there is only one voltage rise, at the battery, but some networks have more than one. For a start we can consider the network below. It has a battery that provides the potential to ow and three resistors (resistors are drawn as zig-zags). When components are wired one after another, as here, they are said to be in series.
2 ohm resistance 3 ohm resistance
20 volt potential
5 ohm resistance
By Kirchhos Voltage Law, because the voltage rise is 20 volts, the total voltage drop must also be 20 volts. Since the resistance from start to nish is 10 ohms (the resistance of the wires is negligible), we get that the current is (20/10) = 2 amperes. Now, by Kirchhos Current Law, there are 2 amperes through each resistor. (And therefore the voltage drops are: 4 volts across the 2 oh m resistor, 10 volts across the 5 ohm resistor, and 6 volts across the 3 ohm resistor.) The prior network is so simple that we didnt use a linear system, but the next network is more complicated. In this one, the resistors are in parallel. This network is more like the car lighting diagram shown earlier.
20 volt
12 ohm
8 ohm
We begin by labeling the branches, shown below. Let the current through the left branch of the parallel portion be i1 and that through the right branch be i2 , and also let the current through the battery be i0 . (We are following Kirchos Current Law; for instance, all points in the right branch have the same current, which we call i2 . Note that we dont need to know the actual direction of ow if current ows in the direction opposite to our arrow then we will simply get a negative number in the solution.)
74
Chapter One. Linear Systems
i0
i1
i2
The Current Law, applied to the point in the upper right where the ow i0 meets i1 and i2 , gives that i0 = i1 + i2 . Applied to the lower right it gives i1 + i2 = i0 . In the circuit that loops out of the top of the battery, down the left branch of the parallel portion, and back into the bottom of the battery, the voltage rise is 20 while the voltage drop is i1 12, so the Voltage Law gives that 12i1 = 20. Similarly, the circuit from the battery to the right branch and back to the battery gives that 8i2 = 20. And, in the circuit that simply loops around in the left and right branches of the parallel portion (arbitrarily taken clockwise), there is a voltage rise of 0 and a voltage drop of 8i2 12i1 so the Voltage Law gives that 8i2 12i1 = 0. i0 i0 + i1 i2 = 0 i1 + i2 = 0 12i1 = 20 8i2 = 20 12i1 + 8i2 = 0
The solution is i0 = 25/6, i1 = 5/3, and i2 = 5/2, all in amperes. (Incidentally, this illustrates that redundant equations do indeed arise in practice.) Kirchhos laws can be used to establish the electrical properties of networks of great complexity. The next diagram shows ve resistors, wired in a seriesparallel way.
5 ohm 10 volt 10 ohm 50 ohm 4 ohm 2 ohm
This network is a Wheatstone bridge (see Exercise 4). To analyze it, we can place the arrows in this way.
i1 i0 i3 i5 i4 i2
Topic: Analyzing Networks
75
Kirchos Current Law, applied to the top node, the left node, the right node, and the bottom node gives these. i0 i1 i2 + i5 i3 + i4 = i1 + i2 = i3 + i5 = i4 = i0
Kirchhos Voltage Law, applied to the inside loop (the i0 to i1 to i3 to i0 loop), the outside loop, and the upper loop not involving the battery, gives these. 5i1 + 10i3 = 10 2i2 + 4i4 = 10 5i1 + 50i5 2i2 = 0 Those suce to determine the solution i0 = 7/3, i1 = 2/3, i2 = 5/3, i3 = 2/3, i4 = 5/3, and i5 = 0. Networks of other kinds, not just electrical ones, can also be analyzed in this way. For instance, networks of streets are given in the exercises. Exercises
Many of the systems for these problems are mostly easily solved on a computer. 1 Calculate the amperages in each part of each network. (a) This is a simple network.
3 ohm 9 volt 2 ohm 2 ohm
(b) Compare this one with the parallel case discussed above.
3 ohm 9 volt 2 ohm 2 ohm 2 ohm
(c) This is a reasonably complicated network.
3 ohm 9 volt 3 ohm 2 ohm 2 ohm 2 ohm 3 ohm 4 ohm
76
Chapter One. Linear Systems
2 In the rst network that we analyzed, with the three resistors in series, we just added to get that they acted together like a single resistor of 10 ohms. We can do a similar thing for parallel circuits. In the second circuit analyzed,
20 volt
12 ohm
8 ohm
the electric current through the battery is 25/6 amperes. Thus, the parallel portion is equivalent to a single resistor of 20/(25/6) = 4.8 ohms. (a) What is the equivalent resistance if we change the 12 ohm resistor to 5 ohms? (b) What is the equivalent resistance if the two are each 8 ohms? (c) Find the formula for the equivalent resistance if the two resistors in parallel are r1 ohms and r2 ohms. 3 For the car dashboard example that opens this Topic, solve for these amperages (assume that all resistances are 2 ohms). (a) If the driver is stepping on the brakes, so the brake lights are on, and no other circuit is closed. (b) If the hi-beam headlights and the brake lights are on. 4 Show that, in this Wheatstone Bridge,
r1 rg r2 r4 r3
r2 /r1 equals r4 /r3 if and only if the current owing through rg is zero. (The way that this device is used in practice is that an unknown resistance at r4 is compared to the other three r1 , r2 , and r3 . At rg is placed a meter that shows the current. The three resistances r1 , r2 , and r3 are varied typically they each have a calibrated knob until the current in the middle reads 0, and then the above equation gives the value of r4 .) There are networks other than electrical ones, and we can ask how well Kircho s laws apply to them. The remaining questions consider an extension to networks of streets. 5 Consider this trac circle.
North Avenue Main Street
Pier Boulevard
Topic: Analyzing Networks
77
This is the trac volume, in units of cars per ve minutes. North Pier Main into 100 150 25 out of 75 150 50 We can set up equations to model how the trac ows. (a) Adapt Kirchos Current Law to this circumstance. Is it a reasonable modelling assumption? (b) Label the three between-road arcs in the circle with a variable. Using the (adapted) Current Law, for each of the three in-out intersections state an equation describing the trac ow at that node. (c) Solve that system. (d) Interpret your solution. (e) Restate the Voltage Law for this circumstance. How reasonable is it? 6 This is a network of streets. Shelburne St
Willow west Winooski Ave Jay Ln east
The hourly ow of cars into this networks entrances, and out of its exits can be observed. east Winooski west Winooski Willow Jay Shelburne into 80 50 65 40 out of 30 5 70 55 75 (Note that to reach Jay a car must enter the network via some other road rst, which is why there is no into Jay entry in the table. Note also that over a long period of time, the total in must approximately equal the total out, which is why both rows add to 235 cars.) Once inside the network, the trac may ow in dierent ways, perhaps lling Willow and leaving Jay mostly empty, or perhaps owing in some other way. Kirchhos Laws give the limits on that freedom. (a) Determine the restrictions on the ow inside this network of streets by setting up a variable for each block, establishing the equations, and solving them. Notice that some streets are one-way only. (Hint: this will not yield a unique solution, since trac can ow through this network in various ways; you should get at least one free variable.) (b) Suppose that some construction is proposed for Winooski Avenue East between Willow and Jay, so trac on that block will be reduced. What is the least amount of trac ow that can be allowed on that block without disrupting the hourly ow into and out of the network?
Index
accuracy of Gauss method, 6871 rounding error, 69 angle, 42 augmented matrix, 14 back-substitution, 5 C language, 68 canonical form for row equivalence, 58 Cauchy-Schwartz Inequality, 41 Chemistry problem, 1, 9 chemistry problem, 22 circuits parallel, 73 series, 73 series-parallel, 74 column, 13 vector, 15 component, 15 computer algebra systems, 6263 conditioning number, 71 direction vector, 35 dot product, 40 double precision, 69 echelon form, 5 free variable, 12 leading variable, 5 reduced, 47 elementary reduction operations, 4 pivoting, 4 rescaling, 4 swapping, 4 elementary row operations, 4 entry, 13 equivalence relation row equivalence, 50 at, 36 form, 56 free variable, 12 Gauss method, 2 accuracy, 6871 back-substitution, 5 elementary operations, 4 Gauss-Jordan, 47 Gauss-Jordan, 47 homogeneous equation, 21 ill-conditioned, 69 induction, 23 inner product, 40 Input-Output Analysis, 6467 Kirchhos Laws, 73 leading variable, 5 length, 39 Leontief, W., 64 linear combination, 52 Linear Combination Lemma, 53 linear equation, 2 coecients, 2 constant, 2 homogeneous, 21 satised by a vector, 15 solution of, 2 Gauss method, 3 Gauss-Jordan, 47 system of, 2 linear surface, 36 LINPACK, 62 Maple, 62 Mathematica, 62 mathematical induction, 23 MATLAB, 62
matrix, 13 augmented, 14 column, 13 conditioning number, 71 entry, 13 nonsingular, 27 row, 13 row equivalence, 50 singular, 27 transpose, 19 matrix:form, 56 mean arithmetic, 44 geometric, 44 MuPAD, 62 networks, 7277 Kirchhos Laws, 73 nonsingular matrix, 27 Octave, 62 orthogonal, 42 parallelogram rule, 34 parameter, 13 partial pivoting, 70 partition row equivalence classes, 51 perpendicular, 42 Physics problem, 1 pivoting full, 70 partial scaled, 70 pivoting on rows, 4 potential, 72 proof techniques induction, 23 reduced echelon form, 47 representative for row equivalence classes, 58 rescaling rows, 4 resistance, 72 resistance:equivalent, 76 resistor, 72 row, 13 vector, 15 row equivalence, 50
scalar multiple vector, 15, 34 scalar product, 40 scaled partial pivoting, 70 Schwartz Inequality, 41 SciLab, 62 single precision, 68 singular matrix, 27 Statics problem, 5 sum vector, 15, 34 swapping rows, 4 system of linear equations, 2 Gauss method, 2 solving, 2 transpose, 19 Triangle Inequality, 40 vector, 15, 33 angle, 42 canonical position, 34 column, 15 component, 15 direction, 35 dot product, 40 free, 33 length, 39 orthogonal, 42 row, 15 satises an equation, 15 scalar multiple, 15, 34 sum, 15, 34 unit, 44 zero, 22 voltage drop, 73 Wheatstone bridge, 74 zero vector, 22

**Find millions of documents on Course Hero - Study Guides, Lecture Notes, Reference Materials, Practice Exams and more. Course Hero has millions of course specific materials providing students with the best way to expand their education.**

Below is a small sample set of documents:

Texas El Paso - GEO - 5336

Linear AlgebraJim Hefferon1 32 11 32 1x1 1 32 1x1 1 2 x1 3 16 82 16 82 1Notation R N C {. . . . . .} . V, W, U v, w 0, 0V B, D En = e1 , . . . , en , RepB (v) Pn Mnm [S] M N V W = h, g H, G t, s T, S RepB,D (h) hi,j |T | R(

Texas El Paso - GEO - 3412

From: Subject: Date: To:"Jose Miguel Hurtado, Jr." <jhurtado@utep.edu> transmountain road Tue,September 16, 2008 12:08:30 PM MDT Soto Saul <sasoto3@miners.utep.edu>, Zamora Hector <hzamora2@miners.utep.edu>, "Esparza Nidia J." <njesparza@miners.ute

Texas El Paso - GEO - 3412

Geology 3412: "Geoscience Processes" Friday, October 3rd, 2008Lab 5Dr. HurtadoLab 5: North Campus Mapping Projectdue October 10th, 2008 Goals: 1. 2. 3. 4. 5. 6. To familiarize yourself with the geologic mapping process; To practice making geol

Texas El Paso - GEOL - 1312

GEOL 1312 Midterm Practice Exam1. Which of the following best describes an element? a. a naturally occurring, inorganic solid with a characteristic chemical composition and a crystalline structure b. a small, dense, positively charged subatomic part

Texas El Paso - GEOL - 1312

GEOL1312 - PRINCIPLES OF EARTH SCIENCE 2 - SPRING 2009 http:/www.geo.utep.edu/pub/jasper/geol_1312Instructors: Office: Phone: Email*: Office Hours:Dr. Jasper Konter GEOL 404a 747-5507 jgkonter@utep.eduDr. Bridget Konter GEOL. 305c 747-6118 brko

Texas El Paso - GEOL - 1312

GEOL 1312 - Principles of Earth Science II Study GuideMidterm Exam 1 The midterm exam will be comprehensive and multiple-choice. This review sheet is a guideline only there may be a few questions on the exam not specifically addressed here but cove

Texas El Paso - GEOL - 1312

Class 11: Astronomy 101Very Large Array (VLA) Soccoro, NMClass 11: Astronomy 101Class updates: Reading: 22.1-22.3, 22.5, 24.1 Todays topics: Where are we in the universe? Important people Tools of AstronomyWhere are we in the universe?Pl

Texas El Paso - GEOL - 1312

Class 15: MarsClass 15: MarsClass Updates Homework 3 due today Reading: 23.5Todays topics: Mars Mars Exploration Tectonics & Volcanism Water?Mars: A Global PerspectiveSimilarities Rotation Mars 24 hr, 37 min Earth 24 hrDifferences Si

Texas El Paso - GEOL - 1312

GEOL 1312: Earth Science 2 Spring 2009Please pick up a syllabus at front of the roomGEOL 1312 IntroductionsEarth Science 2Instructors: Ofces: Phone: Email: Dr. Jasper Konter Dr. Bridget Konter GEOL 404a GEOL 305c 747-5507 747-

Texas El Paso - GEOL - 1312

Class 17: Uranus, Neptune, PlutoClass 17: Uranus, Neptune, PlutoClass Updates Reading: 23.8-23.9 Homework #4 posted due Tues., April 7th Midterm Thurs., April 9th Todays topics: Uranus Neptune PlutoUranus NeptunePlutoUranus, Neptune, a

Texas El Paso - GEOL - 1312

Class 18: Icy MoonsClass 18: Icy MoonsClass Updates Reading: 23.6-23.7 Homework 4 due Tuesday, April 7th Midterm Review Tuesday Midterm next Thursday (April 9th) Todays topics: Galilean moons: Io, Europa, Ganymede, Callisto Saturns moon: Tit

Texas El Paso - GEOL - 1312

Class 12: Celestial Motions Class 12: Celestial Motions Class updates: Reading: 18.4, 22.4 Homework 2 returnedTodays topics: Seasons Lunar phases EclipsesRotation & Revolutionrotationrevolution1Reason for Seasons?North Pole Equa

Texas El Paso - GEOL - 1312

2/17/09 Class 7: Stratigraphy & Age Dating Class updates: Reading: Section 4.1-4.4 Homework 1 due today Homework 2 online, due next Tues. Midterm practice online (website) Midterm review next Tues. Midterm 1 on Feb. 19th (next Thurs.) Todays topic

Texas El Paso - GEOL - 1312

2/4/09Class 6: Fossils, Weathering, ErosionClass 6: Fossils, Weathering, ErosionClass updates: Reading: Section 3.3, 10.1-10.5 Homework #1 due next TuesdayTodays topics: Fossils Weathering Soils, erosionFossils1 2/4/09Hard-body Foss

Texas El Paso - GEOL - 1312

2/2/09Class 5: Sedimentary Rocks & StructuresClass updates: Reading: Section 3.3 Homework 1 assignmentTodays topics: Sedimentary rocks Sedimentary structuresFundamental Rock Types1 2/2/09The Rock CycleSedimentary RocksSedimentary

Texas El Paso - GEOL - 1312

Class 4: Igneous & Metamorphic RocksClass 4: Igneous & Metamorphic RocksClass updates: Course website: http:/www.geo.utep.edu/pub/jasper/geol_1312 Labs begin this weekTodays topics: Igneous rocks & processes Metamorphic rocks & processesFu

Texas El Paso - GEOL - 1312

2/17/09Class 8: Geologic History & ExtinctionsClass updates: Reading: Section 4.0, 4.5 HW 2 due Tues. Midterm review on Tues. Midterm practice online Midterm 1 (2/19) next Thurs.Todays topics: Earth History Mass Extinctions Geologic Time Scal

Texas El Paso - GEOL - 1312

Class 16: Jupiter & SaturnJupiters Turbulent AtmosphereClass 16: Jupiter & SaturnClass Updates Reading: 23.6-23.7 Midterm 2: April 9Todays topics: Jupiter & Saturn Jupiter: Internal structure, magnetic eld, atmosphere, moons Saturn: Explora

Texas El Paso - GEOL - 1312

Class 13: The MoonClass 13: The MoonClass updates: Reading: 22.4, 23.4 Homework 2 returned Homework 3 posted todayTodays topics: Moon Exploration Surface features, cratersExploring the MoonApollo 111959: Luna 2, 3 (USSR) 1964: Ranger 7

Texas El Paso - GEOL - 5215

COMPUTER APPLICATIONS-FALL 2008 -FLEDERMAUSLAB 3-D8: VISUALIZATIONIn this lab, you will learn how to import topography and bathymetry data, display it in 3-D, overlay geologic data of interest, and create a flight path (animated movie) of a

Texas El Paso - GEOL - 5215

http:/www.geo.utep.edu/pub/bkonter/geol_5215GEOL 5215COMPUTER APPLICATIONS IN THE GEOSCIENCES- FALL 2008 -INSTRUCTOR: Dr. Bridget Konter OFFICE: GEOL. 305 OFFICE HOURS: Mondays & Tuesdays 3-4, or by appointment PHONE: 747-6118 EMAIL: brkonter@ut

Texas El Paso - GEOL - 5215

COMPUTER APPLICATIONS IN THE GEOSCIENCES - FALL 2008 -P OWER P OI NT P R ESENTA TION T IP SBoth GSA and AGU have standard recommendations regarding structuring a scientific presentation. They suggest the following slide structure: A title A

Texas El Paso - CS - 5314

3mkehhe y oniehhij y o) p k d p g iq$~3mkehhe y o)gveh@|o p k d dj h g mkehhq ie ry$k oneid h hiyji) $ oy g&p d v@yj eh p h q g kri|jedzyo~&{te|ki|iwihwqpg&km nfh g e3mg lk jigqehkyh$g~pfd fed fgU"g g $my eyhk$g {u&yxdvtysu3w or o dj m

Texas El Paso - CS - 2402

Texas El Paso - CS - 2402

n l ~ Wbwb`pb(~tb` k ~ R( ( b ( b R~t ~ s(bgk ~b ~dbbW b b t btr( ~ (db( ( b`dgbb bw b~ 3~ g g r bdg b 3 b~`b(bww~ ~(Y(bkk (g ~bb(d~g3~(bb `t d b3WdbWbb db b b

Texas El Paso - CS - 5303

x | x x |z x x | | x } v | v sl|x$mzHsqYY{zy%ymzH|m%uu x x %hHlxHqfx x v x xv | x | x x | x x v %hHlxSmW mWrY{z$} Y{zWl|ximzwq} x x smYvl|$hHHx E|% |z x x x v | x | | |z x x | | x } v | | YY{zl|'%Hi

Texas El Paso - CS - 2402

Texas El Paso - CS - 2402

b e vvTy T vy l T ` q` V eq vTyy qe TTy` e vvTy T T ` m BvcqwaBaYxw`aUUYeUcHBatxnvapcqwaBaYxnYacTUl ` S T sq wX T ` eT y m S yX d m l ` S T ` VT yevT V e vvTy T `T u y YYet@fc{Yvcsc6DXYveaw`cUekcTUYUe{YaeDfwaxaBtqaBaYxkeaxYa`aX e XT

Texas El Paso - CS - 2402

@8QlGUQddpRodxp%pSSQbp@XQGQWQgUr`dXdet`hQdF vpS@E dpc f8P s V H WP qiF F fF V R WFii E l iF H W E q w ucPFR wiF gFc f E H FR o R FeQUPewt8`tdoGtyPUdFQSIeycweXQSRWIrXgSeSd@yoGWw@USdQpGtq `X`ahr85 g PPq g e i o e Ei Y F u w HF wicoF H w F

Texas El Paso - CS - 2402

E 0 '33 ' C Y X ) ' % ' G G ) 3 G R X dx42Q47#5B48`7(!8aVQIv2 S 3 G R X E 0 '33 ' C 0 ' X $ ) ' QIv6WF427Q4#y841285g&4yV4IG ' X 0 ) ' % 9 X w ' X R ' C X ) ' 0 G ' 0 ' 0 ' i 0 X R ) X uhV51(xqf(@cH#

Texas El Paso - CS - 5303

6 R1 R2 E fse|vt s$js s B$ Vit TB %$B P$Bs 5 h h<p% l jeBse{ s|Tlseje|ssjsp` j sp{Tsse|jslTsesBtje| $fe tes% jsje|l l $ses%{`l q#it T$ i$B

Texas El Paso - CS - 2402

Texas El Paso - CS - 2402

uvek#yTuCui}X1 f1fe$Ee$uRuu1#X$etT1h RektFkefeXefiut#A C4hUT evek16yhe$uTU RVX0AXAFTU Tff#4 uFCVeTzki5 Y 3 veku p p 3 9 D A@X SFff oe#V$b H Y V y 3 2

Texas El Paso - CS - 2402

6 i = 0 i < n i + + j = 0 j < i j+ = 3 4 i = 0 i < n

Texas El Paso - CS - 2402

STx7hfeEnGfkECS(V98CBSV6h5fCA@Tpf65@p85x7p6e7@6VTiEHB6CfVtVaTR@e 5 f P q cP P 5A9 9 b P c 45 4 c WA f 4 Q e9G 4 Pg A9 WA 9 yP S5 Q c 4 c c 5 W 9 W X e e 5 f y c 45 4 c 5 @4cxB9k@CAfAB6xYxxHTpVP6Tpf69heEE RlskllRdjkiT p w p (

Texas El Paso - CS - 2402

Texas El Paso - CS - 2402

CS2402 Data Structures Exercises Fall 2003Trees Ideas of solutionsExercise 1 Suppose you are given a tree structure, actually a set of node and relations between your nodes. The problem is that you dont know which node is supposed to be the root

Texas El Paso - CS - 2402

5' gH 0)( #5 & & %Q 1 ' ' # Cp5Uz$s$C#Fps#sgi33# # $QpQp3R#sFp sp p Q ` "|$3 " s$|Cp3Xe` ses|#sp3 ! e9|3# p

Texas El Paso - CS - 5314

~ ~ wWhWr}6Wq$2{pwu" })@p{qp3`eetb{p}{ewp{W}pt{ pew{3|`WWq}| `p{ph`} pp{&WB{` )gv bwy qt}tf x o t{pj Ptqu t f{ 3txp wy qt`fl dy$}P|ttg tp{ "py$ &d&eet$ Wplh}d`by { h`} tW p{qh hgs&fpe3dW }p t x f xb ~ E P })twp{hsetpWh}W

Texas El Paso - CS - 2402

1 1 1 1 1 1 n extra credit : 2 4

Texas El Paso - CS - 2402

2 2 2 2 4 2

Texas El Paso - CS - 5303

f 1 (f (X) X z Y 4 4 5 x f 1 (f (X) y f (X), f (x) = y x f 1 (f (X) z = f (y) A(def j wlihWi~I l l oWn|nWh l l y oWnYyofe } vr y uy p o9%rkuwlytaU tlq y yiliyWnino oi{yfedi{jaiuydtr$feWu|ndqhU} u u u l l zr y vr l y y l u z

Texas El Paso - CS - 5303

% lionshu klrlxwleu ke pe i w} p iw r k lg }ylpheugow} ouexpkhjq}ylpeugx{zyexwhejtsqrheGqo}yo! } iw | k w u o r ogg i w w y i w $iX{rvsl %qmli h$iy}wxg$%}q|n}hqgxs$qo } o ge i k ogem i} i p km g o o r i | kg i p g ze pe i

Texas El Paso - CS - 2402

Data Structures Spring 2004 Programming assignment on Trees The Knapsack problem and Trees Due by Tuesday 03/23 periodObjective of this assignmentThe objective of this tutorial is to have students get familiar with the notion of tree from a progra

Texas El Paso - CS - 2402

~t e 9#| w ns n) '3n3 ng$3nii3gg xnt$ ~t e | w 9#Tct sf YniTtv# ggg$nfnz#nm ngggUyt3#Gn sn$3 nngAns nQ H33 $34#gnYn5 $Rn gi n5 ~t e | w

Texas El Paso - CS - 2402

x q 9# { crw r) '3r3 rk$f3frtt3dk |ry$e x q { 9#Tiy ewj YrtTyz# kkk$rjr~#rq rkdkUy3#Gr wr$3 rrkArw rQ H33 $34#drYr5 $Rr kt r5 x q {

Texas El Paso - CS - 2402

g e q W T t H E qI q u F q E w v q q q t E u x `fd7Q VPQQ1Q$bQQfQhGfbyPbbgPpbhgfX1b`XVT RQPAG"DB w H t v u t s r q H i Y a e d a c a Y W U S H II H F E C% A# $( @! 984( &( &0 7( 60 54( 3( 20 1( &z u

Texas El Paso - CS - 2402

CS2402 Data Structures Spring 2004Memo on Trees1 General treesGeneral trees are structures consisting of : 1. a set of nodes, among which one is referred to as the root ; 2. relation "parent-child" between nodes ; relations are such that there ex

Texas El Paso - CS - 2402

u g x ob}gz Qp o h$F 33h`#h3u#j#5$`oo b3 3o$o3@ i is ~hs | iu s | g d i d g p#@hbg5goWb}i r }oh `owopo u i is ~hs | iu s | g d i d x i ghs r Qo5$ hhwho3ogo

Texas El Paso - CS - 2402

uk Fu@}C'B } '%k t yg w q g s i q | rvx)Xvxur}5{ uk F}xg)}q3}FB } '%k t yg w q g s i q | rvx)q8Avxur}5{ uk 5l}CB } '%k t yg w q g s i q | rvx)vxur

Texas El Paso - CS - 2402

l s i l ig s s d s i l s i e {e } d sd wd s p l s fyirr )klfkjhz3pe$}4z4$}kl5rzfudz klfGhrfttz hf4ufzxkwf~fzfl d s ftwfhz3yewfh l s d e g l s l w d s } i } s s l i p p d i i ze j fh S)fyiffqfz$}#4 fhg$}~usfyiksfqh)kl

Texas El Paso - CS - 2402

g R I p C A g D y g q B g A s r g g g p A y q v t UF QExFFYF$WFx"FbFxxxd2bWxwuEWWcEfWdcbSYWUSQI GFE%2A s C p r q p i h g C e T V a ` V X V T R P H C D D C B9 @8& %# ) 7! 6 5) ') 41 3 21 'v i y

Texas El Paso - CS - 2402

i Tf55Y3#G$9nYv #t3ginfU i n$kg#v #3g$gn$k99n9) i niv kg# Q#)nsgzt3ginfU yn$kg#f#Y 3g$RgYn$kg#4 n4#

Texas El Paso - CS - 2402

| q h x t s h nl h Sr&gf3gRrq6q kVzywvu$&{qo 6 rog& ~ t ~ xztx tv o x n uk0p&Voh} | x t s nl jh e d {qzywvu$rqpomkhihgf q 6I d a y w u d s r d p i a g e a c a hxvVtWbq$hfWd 8b` zywvu$rqpo g8io6&g3 hkih&3q oh r w

Texas El Paso - CS - 2402

l i l i Ql'kvi '3 t3$ Ppr$pwx$ l i l i 'ko' ` j`T# 3x 3 g j$tpr$pwj t3rB`m `Pw3$|$3w|o b$3$b$1`3 $ FrR 3$ e e l l i 'ico'k 4tgR#o

Texas El Paso - CS - 2402

Texas El Paso - CS - 2402

5 4 4 i = 0 i < n i = + j = 0 j < n j = + = = 3 s Uv } | C}cu r kt o i r y 3 V sqh U bqfihm uipxrh rpshspwhbimr qf}h 3mh qsd htpwih vpvquqt qq gwvsfv(sfUwvxpz p sshh utptpwwqhrr c bf(}huiph m s V x kg nc

Texas El Paso - CS - 5314

uppk m&n&jdhpimfynyjiht n ih&min yi&pn)qopn o y fpn h f qv h f s v qhpYyvY~bu ypv|Yp$Yi lw w ffh &gjhjmhnfdtih t~ d s P (a b) = P (a) + P (b) P (a b) P (true) = 1 P (f alse) = 0 P (a) [0, 1] P (a|b) a P (a|b) = P (a b) P (b) bf$uswhhYy

Texas El Paso - CS - 5314

Automated University TimetablingLillian Torres Osvaldo Palacios Rafael Pacheco Antonio Cortes February 27, 2006 Articial Intelligence (CS5314/4320) Department of Computer Science University of Texas at El Paso 500 W. University, El Paso, TX 79968, U

Texas El Paso - CS - 2402

} ~ ~ d }~ }~ } ~"} } { 3 ~ } ~"` U t ~ f { } C(} t(}` C ~d "} t }{~ |~{} } ~ g } " t ~r(`(}~rC~}d ~ } } 3(C}dg3~~}d ~f { ( } f d{ "} f ~} }( } } { a~a}{|}(}{ ~ }c g W R| ~(}{ ( g} }R g d`

Texas El Paso - CS - 5314

&tiyz~uxfxl}{|xf{t|)x~}&xzX~)v}q~evx){}zvel}v&exv{x"l} | x x x } 3vvf)u&xq t x x{u|i~| t 3$v t {x~&e&|~z"$t } 3vz{zxuts | x g } u d ~{}zvl}v& y$zev {~xx~&) nx} v $~#iz vy ~x$z{#x U~}&x~z)x 6E3 qnrx {3izyg~yr x{vut U#uw|l3uf