MATH 270 TEST 3 REVIEW1. Given subspaces H and K of a vect.docx

MATH 270 TEST 3 REVIEW
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 .
2. Based on problem 1, show that H is a subspace of H +K and
K is 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 =
0 0 0 0 1

4. Let A =
[
−6 12
−3 6
]
and w =
[
2
1
]
.
Determine if w ∈ Col A. Is 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) ].
6. Find a basis for the space spanned by the given vectors
1
0
−3
2
0
1
2
−3

−3
−4
1
6
1
−3
−8
7
2
1
−6
9

7. Let v1 =
7
−2
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.
8. Find the coordinate vector [x]B of x relative to the given
basis B = {b1,b2,b3}.
b1 =

−3
9
4
6
9. Use an inverse matrix to find [x]B for the given x and B.
B =
{[
3

−5
]
,
[
−4
6
]}
, x =
[
2
−6
]
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.
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
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
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
2 −3 6 2 5

0 0 3 −1 1
0 0 0 1 3
0 0 0 0 0
14. If a 3 × 8 matrix A has rank 3, find dim(Nul A), dim(Row
A) and rank(AT ).
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.
16. Let B = {b1,b2} and C = {c1,c2} be bases for R2. Find the
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
]
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.
18. Find a basis for the eigenspace corresponding to the
eigenvalue.
A =
3−1 1 3
2 4 9
19. Find the characteristic polynomial and the eigenvalues of

the following matrix.[
3 −2
1 −1
]
5 8 3
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 =
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
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 =
7
−2
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 =
−3
9
4
6
Answer : [x]B =
−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 =

−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
1
0
0
2
2
1
−3
0
0

−1
2
−1
respectively. We test for linear independence of these vectors
by writing them as
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
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
matrix from A to B. Then find [x]B for x = 3a1 + 4a2 + a3.
Answer : PB←A
=
0 1 −2
2
16. Let B = {b1,b2} and C = {c1, c2} be bases for R2. Find the
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 4 3
1

18. Find a basis for the eigenspace corresponding to the
eigenvalue.
A =
2 4 9
Answer :
0
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
2 9 0
5 8 3
Answer : −λ3 + 18λ2 − 95λ+ 150

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