math test
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Math270Test3ReviewKeySpring2020.pdf
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
