# Math test help

MATH 270 TEST 3 REVIEW KEY

1. Given subspaces H and K of a vector space V , the sum of H and K, written as H+K,

is the set of all vectors in V that can be written as the sum of two vectors, one in H

and the other in K; that is

H + K = {w : w = u + v}, ∃u ∈ H and ∃v ∈ K. Show that H + K is a subspace of V .

Proof: Given subspaces H and K of a vector space V , 0 of V ∈ H + K, because 0 ∈ H and 0 ∈ K and 0 = 0 + 0.

Now let w1,w2 ∈ H + K, 3 w1 = u1 + v1 and w2 = u2 + v2 where u1,u2 ∈ H and v1,v2 ∈ K. Then w1 + w2 = (u1 + v1) + (u2 + v2) = (u1 + u2) + (v1 + v2).

So u1 + u2 ∈ H and v1 + v2 ∈ K because H and K are subspaces. This shows w1 + w2 ∈ K. Thus H +K is closed under addition.

Finally ∀c ∈ R, cw1 = c(u1 + v1) = cu1 + cv1. ∴ H +K is closed under multiplication by scalars.

2. Based on problem 1, show that H is a subspace of H+K and K is a subspace of H+K.

Proof: H is a subset of H + K because ∀u ∈ H can be written as u + 0, where 0 ∈ K and 0 ∈ H. Since H contains 0 of H +K, and H is closed under vector addition and scalar multi-

plication (because H is a subspace of V ), H is a subspace of H +K.

The same argument applies when H is replaced by K, so K is also a subspace of

H +K.

3. Find an explicit description of Nul A by listing vectors that span the null space for the

following matrix :

A =

1 −2 0 4 00 0 1 −9 0 0 0 0 0 1

Answer :

2

1

0

0

0

, −4 0

9

1

0

4. Let A =

[ −6 12 −3 6

] and w =

[ 2

1

] .

Determine if w ∈ Col A. Is w ∈ Nul A ?

Answer : w ∈ Col A and w ∈ Nul A.

5. Define T : P2 → R2 by T (p) =

[ p(0)

p(1)

] .

For instance, if p(t) = 3 + 5t + 7t2, then T (p) =

[ 3

15

] .

Show that T is a linear transformation. [Hint : For arbitrary polynomials p, q ∈ P2, compute T (p + q) and T (cp) ].

Proof : Let p, q ∈ P2, and c ∈ R. Then

T (p + q) =

[ (p + q)(0)

(p + q)(1)

] =

[ p(0) + q(0)

p(1) + q(1)

] =

[ p(0)

p(1)

] +

[ q(0)

q(1)

]

= T (p) + T (q) and T (cp) =

[ (cp)(0)

(cp)(1)

] = c

[ p(0)

p(1)

]

= cT (p)

∴ T is a linear transformation.

6. Find a basis for the space spanned by the given vectors v1,v2,v3,v4,v5. 1

0

−3 2

,

0

1

2

−3

, −3 −4 1

6

,

1

−3 −8 7

,

2

1

−6 9

Answer :

1

0

−3 2

,

0

1

2

−3

,

1

−3 −8 7

7. Let v1 =

4−3 7

, v2 = 19 −2

, v3 = 711

6

and H = Span{v1,v2,v3}. It can be verified that 4v1 + 5v2 − 3v3 = 0.

Use this information to find a basis for H.

Answer : Since 4v1 + 5v2 − 3v3 = 0, we see that each of the vectors is a linear combination of the others. Thus the sets {v1,v2}, {v1,v3} and {v2,v3} all span H. Since we may confirm that none of the three vectors is a multiple of any of the others,

the sets {v1,v2}, {v1,v3} and {v2,v3} are linearly independent and thus each forms a basis for H.

8. Find the coordinate vector [x]B of x relative to the given basis B = {b1,b2,b3}.

b1 =

1−1 −3

, b2 = −34

9

, b3 = 2−2

4

, x = 8−9

6

Answer : [x]B =

1−1 −3

9. Use an inverse matrix to find [x]B for the given x and B.

B =

{[ 3

−5

] ,

[ −4 6

]} , x =

[ 2

−6

]

Answer : [x]B =

[ 6

4

]

10. The set B = {1 + t2, t+ t2, 1 + 2t+ t2} is a basis for P2. Find the coordinate vector of p(t) = 1 + 4t+ 7t2 relative to B.

Answer : [p]B =

26 −1

11. Use coordinate vectors to test the linear independence of the set of polynomials.

Explain your work.

1 + 2t3, 2 + t− 3t2,−t+ 2t2 − t3

Answer : The coordinate mapping produces the coordinate vectors 1

0

0

2

,

2

1

−3 0

and

0

−1 2

−1

respectively. We test for linear independence of these vectors by writing them as

columns of a matrix and row reducing: 1 2 0

0 1 −1 0 −3 2 2 0 −1

∼ · · · ∼

1 0 0

0 1 0

0 0 1

0 0 0

Since the matrix has a pivot in each column, its columns (and thus the given polyno-

mials) are linearly independent.

12. Find the dimension of Nul A and Col A for the matrix shown below.

A =

1 −6 9 0 −2 0 1 2 −4 5 0 0 0 5 1

0 0 0 0 0

Answer : dim(Col A) = 3 ; dim(Nul A) = 2

13. Assume matrix A is row equivalent to B. Find bases for Col A, Row A and Nul A of

the matrices shown below.

A =

2 −3 6 2 5 −2 3 −3 −3 −4 4 −6 9 5 9 −2 3 3 −4 1

, B =

2 −3 6 2 5 0 0 3 −1 1 0 0 0 1 3

0 0 0 0 0

Answer : Basis for Col A =

2

−2 4

−2

,

6

−3 9

3

,

2

−3 5

−4

,

Basis for Row A = (2,−3, 6, 2, 5), (0, 0, 3,−1, 1), (0, 0, 0, 1, 3) and

Basis for Nul A =

3 2

1

0

0

0

,

9 2

0

−4 3

−3 1

.

14. If a 3× 8 matrix A has rank 3, find dim(Nul A), dim(Row A) and rank(AT ).

Answer : dim(Nul A) = 5 ; dim(Row A) = 3 ; rank(AT ) = 3

15. Let A = {a1, a2, a3} and B = {b1,b2,b3} be bases for a vector space V and suppose a1 = 4b1−b2, a2 = −b1+b2+b3 and a3 = b2−2b3. Find the change-of-coordinate matrix from A to B. Then find [x]B for x = 3a1 + 4a2 + a3.

Answer : PB←A =

4 −1 0−1 1 1 0 1 −2

and [x]B = 82

2

16. Let B = {b1,b2} and C = {c1, c2} be bases for R2. Find the change-of-coordinate matrix from B to C and the change-of-coordinate matrix from C to B.

b1 =

[ 7

5

] , b2 =

[ −3 −1

] , c1 =

[ 1

−5

] , c2 =

[ −2 2

]

Answer : P C←B

=

[ −3 1 −5 2

] and P

B←C =

[ −2 1 −5 3

]

17. In P2, find the change-of-coordinate matrix from the basis B = {1− 2t+ t2, 3− 5t+ 4t2, 2t+ 3t2} to the standard basis C = {1, t, t2}. Then find the B-coordinate vector for −1 + 2t.

Answer : P C←B

=

1 3 0−2 −5 2 1 4 3

and [x]B = 5−2

1

18. Find a basis for the eigenspace corresponding to the eigenvalue.

A =

4 2 3−1 1 3 2 4 9

, λ = 3

Answer :

−21

0

, −30

1

19. Find the characteristic polynomial and the eigenvalues of the following matrix.[ 3 −2 1 −1

]

Answer : λ2 − 2λ− 1 and λ = 1± √

2

20. Find the characteristic polynomial of the following matrix. 6 −2 0−2 9 0 5 8 3

Answer : −λ3 + 18λ2 − 95λ+ 150

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