Solution to selected problems on HW 1.
Problem 6 a), page 23 of Leon) The augmented matrix of the system is
1 1 | 1
4 3 | 3
.
Its RRE form is (you have to show your work for this)
10| 0
0 1 | 1
.
There is a unique solution
x1 = 0
x2 = 1
Problem 6 b), page
Matrix Theory HW 2. (not to turn in)
1. From Leon, on page 56, do problems 7 and 16. On page 66 do problems 3 and 7.
2. Find an elementary matrix E such that EA = B if
A=
3. Let
a.
b.
4. Let
5.
a.
b.
a.
b.
c.
1 2 3
4 5 6
, B=
5 7 9
4 5 6
.
1 3 5
A = 2 1 4
Solution to selected problems on HW 2.
Problem 16, page 56 of Leon.
Suppose that A is n n. Using the hint, we have that
T
AT (A1 )T = (A1 A)T = In = In
and
T
(A1 )T AT = (AA1 )T = In = In .
Thus AT is invertible, with inverse (AT )1 = (A1 )T .
Problem 7,
Solution to selected problems on HW 3.
Page 97, no. 5. Write A in terms of its columns as A = (A1 , A2 , . . . , An ). Since Det(A)
is multilinear in its columns,
Det(A) = Det(A1 , A2 , . . . , An ) = Det(A1 , A2 , . . . , An )
= 2 Det(A1 , A2 , A3 , . .
Solution to selected problems on HW 4.
Page 125, no. 2. (a) Let
V = cfw_(x1 , x2 , x3 ) | x1 + x3 = 1 .
(0, 0, 0) V , since the sum of the rst and third coordinate is 0 + 0 = 0 = 1. Thus V is
not a subspace of R3 .
Page 125, no. 2. (b) Let
V = cfw_(x1 , x
Solutions to problems on HW 5.
1. a) Use Algorithm 5 from Lecture Note 5.
1
A= 2
3
Compute the reduced row echelon form of
12
3 5 ,
25
which is
101
B = 0 1 1 .
000
The leading ones of B are in the rst and second column, so (1, 2, 3)T , (1, 3, 2)T
basis of
Solutions to selected problems on HW 6.
3. Use the algorithm of Lecture Note
1
0
1
Calculate that the RRE form of this
1
0
0
Thus
6, starting with the matrix
11111
1 0 0 1 1 .
00001
matrix is
0000 1
1 0 0 1 1 .
0 1 1 0 1
(v )
00 1
M = 0 1 1 .
1 0 1
00 1
2
Solutions to selected problems on HW 7.
page 174, 4.
First nd x1 , x2 so that
(7, 5)T = x1 (1, 2)T + x2 (1, 1)T ,
by solving (showing your work) the system
x1 + x2 = 7
2x1 x2 = 5.
We get x1 = 4 and x2 = 3. Since L is linear,
L(7, 5)T ) = L(4(1, 2)T + 3(1,
Solutions to selected problems on HW 8.
Solution to Problem 3. Let
1
2
A=
1
3
2
3
.
1
5
Then W = column space(A) so
W = [column space(A)] = N (AT )
by formula (1) of Lecture Note 8.
AT =
1213
2315
has the RRE form
1 0 1 1
01 1 1
The standard form solution
Solutions to selected problems on HW 9.
Solution to Leon, page 294, no. 3.
Suppose that A is an n n matrix, with characteristic polynomial A (t). We have that
= 0 is an eigenvalue of A if and only if A (0) = 0
and
A is singular if and only if Det(A) = 0.
Solutions to selected problems on HW 10.
Solution to 3.
Since each Jordan block contributes one independent eigenvector, A has
1 = dim E (2) Jordan blocks with eigenvalue 2,
and
3 = dim E (5) Jordan blocks with eigenvalue 5.
Since A is 4 4, each of the fo
Dr. Srinivasan
Review of Matrix Theroy - Linear Algebra
1. Computational problems:
1. Gaussian elimination: Determine the values of b and c for which the following
system will have i. no solution, ii. innitely many solutions and iii. an unique solution.
2
Dr. Srinivasan
Review of Matrix Theroy - Linear Algebra
1. Computational problems:
1. Gaussian elimination: Determine the values of b and c for which the following
system will have i. no solution, ii. innitely many solutions and iii. an unique solution.
2
Jordan Form
In these notes we work over the complex numbers C. All vector spaces are complex
vector spaces. The eigenspace E () of a matrix A will denote the complex eigenspace
C
EA () introduced in the previous Lecture Note. All nullspaces are complex nu
Eigenvalues and Eigenvectors
Denition 0.1. Let A Rnn be an n n (real) matrix. A number R is a (real)
eigenvalue of A if there exists a nonzero vector v Rn such that Av = v . The vector v
is called an eigenvector of A belonging to .
Example 0.2. Let
3
2
1
Inner Product Spaces
Denition 0.1. An inner product on a vector space V is an operation on V that assigns
to each pair of vectors x, y V , a real number < x, y > satisfying:
I. < x, y >=< y, x > for all x, y V .
II.
< x + y, z >=< x, z > + < y, z > and
<
EXAM II-sample-key
1.(16)
1
B= 2
3
1
3
5
7
5
3
2
2
2
Find a basis and compute the dimension of
The row space of B, the column space of B and compute
Starting with
1 1 2
B= 2
3 2
3
5 2
and performing the row operations R1 = R1 , R2
1
B = 0
0
the rank of B
Matrix Theory HW 1. Not to turn in
From 8th edition of Leon, on page 23:
Do problems 1, 6 (Gauss Jordan Reduction means to put the coecient matrix in RRE
form, and then write the solution(s) to the system in standard form, as explained in the
rst lecture
Matrices
Rm is the set of all m 1 column vectors with real coecients. If v Rm , then
v1
v2
v= .
.
.
vm
with v1 , . . . , vm R.
Rn is the set of all 1 n row vectors with real coecients. If w Rn , then
w = (w1 , . . . , wn )
with w1 , . . . , wn R.
Sys
EXAM II-sample
SHOW ALL YOUR WORK
1.(16)
1
B= 2
3
1
3
5
2
2
2
7
5
3
Find a basis and compute the dimension of
The row space of B, the column space of B and compute the rank of B.
1
2. (4+8+6) a. Dene: Dimension of a vector space.
b. S is the set of all 3
Some notation about matrices
0m,n is the m n zero matrix. In is the n n identity matrix. The transpose of a
matrix A is written as AT . Rm is the set of all m 1 column vectors with real coecients.
If v Rm , then
v1
v2
v= .
.
.
vm
with v1 , . . . , vm R
Elementary row operations on a matrix A
I. Interchange two rows.
II. Multiply a row by a nonzero real number.
III. Replace a row by its sum with a multiple of another (dierent) row.
Denition 0.1. A matrix A is in row echeleon (RE) form
i. If the rst nonze
Determinants
Suppose that A = (aij ) is an n n matrix. Dene Mi,j to be the (n 1) (n 1)
matrix obtained from A by removing the i-th row and j -th column.
The determinant of A may be dened recursively in terms of determinants of smaller
submatrices.
Denitio
Vector Spaces
1. Vector Spaces
A (real) vector space V is a set which has two operations:
1. An association of x, y V to an element x + y V . This operation is called vector
addition.
2. The association of c R and x V to an element cx V . this operation i
Algorithms to Compute Bases
1. Dimension.
Denition 0.1. The dimension of a vector space V is the number of elements in a basis
of V . This number is independent of choice of basis of V .
Theorem 0.2. Suppose that W is a subspace of a vector space V . Then
Coordinates and Rank
A problem on change of perspective.
In this section we consider the following problem. Suppose that we rotate the coordinate
system cfw_e1 = (1, 0)T , e2 = (0, 1)T in R2 counterclockwise by 45 degrees, to obtain a new
coordinate syst
Linear Maps and Matrices
Linear maps
Suppose that V and W are sets. A map F : V W is a function; that is, to every
v V there is assigned a unique element w = F (v ) in W . Two functions F : V W and
G : V W are equal if F (v ) = G(v ) for all v V .
Denitio
Prof. Srinivasan
Inner Product Spaces and Orthonormal bases.
Let V be a vector space with inner product . Let B = cfw_v1 , v2 , vn be a basis for
V.
Two vectors v1 and v2 are said to be orthogonal to each other if v1 v2 = 0 .
A basis such as B is said to
Prof. Srinivasan
Inner Product Spaces and Orthonormal bases.
Let V be a vector space with inner product . Let B = cfw_v1 , v2 , vn be a basis for
V.
Two vectors v1 and v2 are said to be orthogonal to each other if v1 v2 = 0 .
A basis such as B is said to
Math 331 - Summary of Chapter VI 1. Eigen Values and Eigen Vectors
V is a vector space over R, the real numbers (or C , the complext numers). Let
T : V V be a linear transformation.
A scalar is called an eigenValue of T if there is a non zero vector v in