JIM 201

Linear Algebra

Assignment 2 due 30th April, 4:30pm

Instructions: You are being evaluated on the step by step workings, as well as the correctness of your answers. Try to answer questions in a clear, direct, and efficient way. You may discuss among your friends, but any duplications of students’ answers detected will lower your mark.

1. (a) Given the matrices A and B to be row equivalent:

31 0 0 021 2 1 2 053 0 1 1 3 0 1 0 0

, 20 2 4 3 1 10 0 1 01 1 6 2 0 2

0 0 0 1 2

A B

− − = = − −

’

(i) Find rank(A) and nullity(A)

Answer: Rank(A)=4 Nullity(A)=1

(ii) Find a basis for the row space of A

Answer :{(1,0,0,0,3/2),(0,1,0,0,-5/2),(0,0,1,0,-1/2),(0,0,0,1,2)} (iii) Find a basis for the column space of A

Answer :

1 2 1 23 0 1 1

, , ,0 2 4 31 1 6 2

− −

(b) Explain the following terms:

(i) The set W={w1,w2,…,wr} is linearly dependent

Answer: The set W={w1,w2,…,wr} is linearly dependent- At least one of the vectors in W can be written as a linear combination of the other vectors in W.

(ii) The set W={w1,w2,…,wr} spans the vector space V1

Answer: The set W={w1,w2,…,wr} spans the vector space V1- The space V1 of W consisting of all linear combinations of vectors in W.

(iii) The set W={w1,w2,…,wr}forms a basis for the vector space V2

Answer: The set W={w1,w2,…,wr} forms a basis for the vector space V2- W is linearly independent and spans V2.

(iv) The dimension of a vector space V3

Answer: The dimension of a vector space V3- if a vector space V3 has a basis with n vectors, then the value of n is called the dimension of V3.

(c) Determine whether the given vectors below are linearly dependent in R3 or not. V1=(3,1,4), V2=(2,-3,5), V3=(5,-2,9), and V4=(1,4,-1)

Answer:The reduced matrix has non-trivial solutions. Thus, the vector is linearly dependent.

2. (a) Show whether or not the following are inner products:

(i) <u, v> = 7u1v1 + 1.2u2v2

Answer: Inner product (ii) <u, v> = -7u1v1 - 1.2u2v2

Answer: Not inner product (iii) <u, v> = 7u1v1 - 1.2u2v2

Answer: Not inner product

(b) Let T be the orthogonal projection onto the plane 2x + 2y + z = 0 in R3. Find the matrix B of T with respect to a basis B of R3 of your choice.

Answer: 4 4 1

1 4 4 19

2 2 1

(c) Matrices A, B and C represent linear transformations. For each of them describe the

transformation they represent.

0.6 0.80.8 0.6

A = −

3 00 3

B =

0.36 0.480.48 0.36

C−

= −

Answer: A- reflection across the x-axis B- Shear transformation C- Projection to x-axis

(d) Determine if a 2 × 2 matrix A exist such that:

(i) The kernel of A = span of 23

.

Answer: The matrix A do not exist

(ii) The image of A ≠ the image of A2.

Answer: The matrix A exist

3. (a) Consider the function ( )1 2 2 00 3 3

T M M Mk

= −

from R2×2 to R2×2, where k is an

arbitrary constant. For the standard basis B of R2×2, find the B-matrix, B of T.

Answer: 21

(b) Find the values of the constant l such that the matrix 1 0 2

5 64 0 3

Aλ

λλ

− = − −

is

invertible? Answer: l λ≠

(c) Let 2 121 5

A−

= − . Find a diagonal matrix D and an invertible matrix P such that A =

PDP-1.

Answer:

1 00 2

4 31 1

d

P

=

=

4. (a) Let W be a set of matrices in M2x2 with the operations defined as follows;

++

+=

+

=+

hdgcbfea

hgfe

dcba

BA

=

=

kdkcka

dcba

kkA1

for any scalar k

If NBA ∈, , determine whether the following axiom holds or fails; ( ) AAA βαβα +=+ for any matrix A in W and any scalars α and β .

Answer: axiom (α + β )A = αA + βA holds for any matrix A in W and any scalars α

and β

(b) Determine whether { }22 324,21,2 xxxx +−+− span P2. Answer: Since the system is consistent, then the set of polynomials span P2

(c) Determine whether the set S = { ( a, b, c ); c = a – b } is a subspace of R3. Answer: set S = { ( a, b, c ); c = a – b } is a subspace of R3 5. (a) The transformation 34: RRT → is defined as follows:

+−+−−−++−

=

4321

4321

4321

4

3

2

1

32223

xxxxxxxxxxxx

xxxx

T

(i) Find a basis for the kernel of T. What is the nullity of T?

Answer:

21/ 2214 /119 / 22

1

− − −

, Nullity = 1

(ii) Find a basis for the range of T. What is the rank of T?

Answer: 3 1 11 , 2 , 11 1 3

− − − −

, Rank = 3

(iii) Is T a one-to-one transformation? Give a reason for your answer. Answer: T is not a one-to-one transformation because Nullity ≠ 0

(iv) Is T onto? Give a reason for your answer.

Answer: T is onto because rank(T)=dim(codomain)=3 (b) Determine whether the following transformation is linear

2R→×23M:L such that ( )fedcbafedcba

222, ++++=

L

Answer: The transformation is linear

6. (a) Consider the matrix

2 2 01 3 02 1 1

− = − − −

A

(i) Find the eigenvalues of A. Answer: 1, 4λ = − −

(ii) Find all the eigenvectors of A.

Answer: 0 10 , 11 1

−

(iii) Is A diagonalizable? Give a reason for your answer.

Answer: Not diagonalizable

(iv) Find basis for the eigenspace corresponding to the largest eigenvalue of A.

Answer: 001

(v) Is A invertible? If so, find all the eigenvalues of A–1.

Answer: 1, 1, 1/ 4λ = − −

(b) Let

1 2 1 2 50 0 1 1 10 0 2 2 20 0 3 3 3

B

=

. Find,

(i) a basis for the null space of B,

Answer:

2 1 41 0 0

, ,0 1 10 1 00 0 1

− − − − −

(ii) a basis for row space of B,

Answer: {(1,2,0,1,4), (0,0,1,1,1)} (iii) a basis for column space of B,

Answer:

1 10 1

,0 20 3

(iv) nullity of BT

Answer:3 (v) rank of B.

Answer:2