Exam 1(GUIDE FOR POSSIBLE QUESTIONS)-Math 251 Fall 2014 (11.111.5)
1. (A) Describe in words and grapgically the surface or the region in the space represented by the
equation or the inequality: .
a. Words
b. Graph
2. (a) Find a unit vector orthogonal to b
Math 311: Topics in Applied Math 1
6: Eigenvalues
6.3: Diagonalization
Examples
336/1b
5
6
2 2
XDX1 where D is diagonal.
Factor the matrix A =
Summary
into a product
k
THEOREM: If j j=1 are distinct eigenvalues of
an n n matrix A, their corresponding eig
Math 311: Topics in Applied Math 1
5: Orthogonality
5.6: The Gram-Schmidt
Orthogonalization Process
Solution
2
5
and v2 =
.
1
10
They are linearly independent and hence form a
basis for the column space of A, although not an
orthonormal one.
The norm of
Math 311: Topics in Applied Math 1
6: Eigenvalues
6.1: Eigenvalues and Eigenvectors
Solution
Determine eigenvalues.
det (A I) = 0
6
4
= 0
3
1
Summary
Let A be an n n matrix. If Ax = x for a scalar
and a nonzero vector x, then is an eigenvalue
associat
Math 311: Topics in Applied Math 1
5: Orthogonality
5.5: Orthonormal Sets
Let S be a subspace of V with orthonormal basis
cfw_u1 , . . . , un and let x V . If p = ci ui where
ci = x, ui , then p x S . Moreover, p is the
element of S that is closest to x
Math 311: Topics in Applied Math 1
5: Orthogonality
5.4: Inner Product Spaces
THEOREM: If V is an inner product space, then
v =
v, v for all v V denes a norm on V .
For vectors x and y in a normed linear space, the
distance between x and y is x y .
Summ
Math 311: Topics in Applied Math 1
5: Orthogonality
5.1: The Scalar Product in Rn
Summary
Let x, y
Solution
Since x and y be linearly independent, neither vector is a
multiple of the other. Hence the angle between them is
not a multiple of . Hence |cos |
Math 311: Topics in Applied Math 1
5: Orthogonality
5.3: Least Squares Problems
Solution
The coefcient matrix and right-hand side vector are
1
1 1
1
1 1
A=
0 1 1 ,
1
0 1
Summary
Recall an overdetermined system: Ax = b where A
is an m n matrix with m
Math 311: Topics in Applied Math 1
4: Linear Transformations
4.3: Similarity
202/1c
Let E = cfw_e1 , e2 be the standard basis for V = R2 and
F = cfw_u1 , u2 be another basis for R2 where
Summary
u1 =
Let E = cfw_v1 , . . . vn and F = cfw_w1 , . . . ,
Math 311: Topics in Applied Math 1
5: Orthogonality
5.2: Orthogonal Subspaces
Examples
233/1c
4 2
1
3
, determine a basis for
Summary
For the matrix A =
2
1
3
4
Subspaces X and Y of Rn are orthogonal if xT y = 0
each of the subspaces R AT , N (A), R
Math 311: Topics in Applied Math 1
3: Vector Spaces
3.6: Row Space and Column Space
166/11
Let A be an m n matrix. Prove rank (A) min (m, n).
Solution
Summary
The rank of A is the dimension of the row space of
A. Thus rank (A) m, the number of rows of A.
Math 311: Topics in Applied Math 1
4: Linear Transformations
4.1: Denition and Examples
1. L : R2 R2 dened by L (x) = 3x is a linear
operator. A given vector is stretched by a factor of 3.
[More generally, for > 0, L (x) = x stretches
or shrinks a vector
Math 311: Topics in Applied Math 1
4: Linear Transformations
4.2: Matrix Representations of
Linear Transformations
Solution
Let A =
Ax =
Summary
1
1 0
. Then for each x R3 ,
0 1 1
x1
1
1 0
x2 x1
x2 =
= L (x) .
0 1 1
x3 x2
x3
If L : Rn Rm is a linear tra
Math 311: Topics in Applied Math 1
3: Vector Spaces
3.5: Change of Basis
A Special Case (deduced from General Case)
Recall E = cfw_e1 , . . . , en , the standard basis for
V = Rn , where ek is an n-element column vector
with a 1 in the kth row and zeros
Math 311: Topics in Applied Math 1
3: Vector Spaces
3.3: Linear Independence
Examples
143/1(d)(e)
Summary
Determine whether the following vectors are linearly
independent in R2 .
Below, V is a vector space over a eld F of scalars.
(d) v1 =
1
2
(e) w1 =
1
Math 311: Topics in Applied Math 1
3: Vector Spaces
3.4: Basis and Dimension
Solution
(a) Let A =
x1 and x2 .
x1 x2
be the matrix whose columns are
Since det (A) = 6 4 = 2 = 0, the columns of A,
x1 and x2 , are linearly independent.
Summary
The vectors
Math 311: Topics in Applied Math 1
3: Vector Spaces
3.2: Subspaces
Solution
A is not a subspace of R3 since x =
but 2x =
2 0 0
T
1 0 0
T
A
A because 2 + 0 = 1.
/
R3 ,
B is a subspace of
geometrically a line in 3-D space
T
through the origin. For let x =
Math 311: Topics in Applied Math 1
3: Vector Spaces
3.1: Denition and Examples
Notes
For us, the eld F is almost always R, the set of real
numbers. We speak of V as a real vector space.
(The eld could be C, the set of complex numbers.
Then V is a complex
Math 311: Topics in Applied Math 1
1: Matrices & Systems of Equations
1.5: Elementary Matrices
The system of n linear equations in n unknowns
Ax = b has a unique solution if and only if A is
nonsingular.
If A is nonsingular, a systematic method of
compu
Math 311: Topics in Applied Math 1
2: Determinants
2.2: Properties of Determinants
Solution
Expanding across the rst row (note the zeros), we have
3 (1)1+3 (0 8) = 24.
Summary
101/3(b),(c),(e)
Here are effects that the three elementary row or
column oper
Math 311: Topics in Applied Math 1
2: Determinants
2.1: The Determinant of a Matrix
Solution
Summary
22 Matrices
Here n = 1, whence det (A) = a11 = .
Throughout this section, let A = [ai j ] be an n n matrix.
a b
c d
Let A =
Let Mi j denote the (n 1) (n
Spring 2004 Math 253/501503
14 Vector Calculus
14.8 Stokes Theorem
c 2004, Art Belmonte
Thu, 15/Apr
931/2 [revisited]
Summary
Solution
Suppose S is an oriented, piecewise-smooth surface bounded by a
simple, closed, piecewise-smooth boundary curve C with p
Spring 2004 Math 253/501503
14 Vector Calculus
14.9 The Divergence Theorem
c 2004, Art Belmonte
Tue, 19/Apr
The respective compositions w (sk (r, ) are
w (s1 (r, )
w (s2 (r, )
Let E be a simple solid region whose closed boundary surface S
has positive (o
Math 311: Topics in Applied Math 1
1: Matrices & Systems of Equations
1.4: Matrix Algebra
A square n n matrix A is invertible or
nonsingular if there is a matrix B such that
AB = BA = I. If such a matrix B exists, it is unique.
We denote this multiplicat
Spring 2004 Math 253/501503
14 Vector Calculus
14.6 Parametric Surfaces & Areas
c 2004, Art Belmonte
Thu, 08/Apr
Solution
Well, x = 4 y 2 2z 2 for (y, z) D = (y, z) : y 2 + 2z 2 4
is one representation. Its graph, however, looks pinched. Were
trying to t
Spring 2004 Math 253/501503
14 Vector Calculus
14.7 Surface Integrals
c 2004, Art Belmonte
Tue, 13/Apr
Table Notes
1. Other rst-order moments are symmetrically dened.
Mxz =
Mxy =
y dS,
z dS
S
S
2. Other second-order moments are symmetrically dened.
Summar
Spring 2004 Math 253/501503
14 Vector Calculus
14.5 Curl and Divergence
c 2004, Art Belmonte
Thu, 08/Apr
Solution
The curl of F is curl F =
i
x
x2y
Summary
k
y
yz 2
z
x 2z
= 2yz, 2x z, x 2 .
F or
, ,
x 2 y, yz 2 , x 2 z = 2x y + z 2 + x 2 .
x y z
90
Math 311: Topics in Applied Math 1
1: Matrices & Systems of Equations
1.1: Systems of Linear Equations
Examples
These problems are exercises from the section in your
textbook, referenced by page and exercise number.
Summary
System of linear equations: A