Math 2040 Matrix Theory and Linear Algebra II sample midterm ...

Math 2040 Matrix Theory and Linear Algebra II
sample midterm covering material to the end of section 5.1
Instructions: Answer all questions. For each question part, parenthesized key words tell you
the idea being tested. (Students should work very hard on these questions themselves before any in
class solutions are given.)
1. General Theory. (Value = 40%)
(a) (Definitions.) Give the definition of a subspace of a vector space.
Solution: A subspace of a vector space V is a nonempty subset W of V such that W is
closed under the operations of addition and scalar multiplication on V .
(b) (Axiom consequences.) Prove the cancellation law for addition in a vector space; that
u + v = u + w implies v = w for all vectors u, v and w.
Solution:
(1) u + v = u + w by assumption.
(2) ∴ −u + (u + v) = −u + (u + w) upon adding −u to both sides of (1).
(3) ∴ (−u + u) + v = (−u + u) + w because addition of vectors is associative.
(4) ∴ 0 + v = 0 + w because −u + u = 0, −u being the additive inverse of u.
(5) ∴ v = w because 0 is a unit for vector addition.
(c) (Linearity.) Let T 1 : V 1 −→ V 2 and T 2 : V 1 −→ V 2 be two linear transformations between
vector spaces V 1 and V 2 . Let T 3 : V 1 −→ V 2 be defined by T 3 (v) = 2T 1 (v) − T 2 (v) for each
v ∈ V 1 . Prove T 3 is linear.
Solution: Let u, w ∈ V 1 and let α, β ∈ R. We will show T 3 (αu + βw) = αT 3 (u) + βT 3 (w).
(1) T 3 (αu + βw) = 2T 1 (αu + βw) − T 2 (αu + βw) by the definition of T 3 .
(2) ∴ T 3 (αu + βw) = 2αT 1 (u) + 2βT 2 (w) − αT 2 (u) − βT 2 (w) because T 1 and T 2 are linear,
and because 2 is a scalar (so, multiplication by 2 distributes over addition).
(3) ∴ T 3 (αu + βw) = 2αT 1 (u) − αT 2 (u)) + 2βT 2 (w) − βT 2 (w) by a simple rearrangement
of terms using commutativity of addition.
(4) ∴ T 3 (αu+βw) = α(2T 1 (u)−T 2 (u)))+β(2T 2 (w)−T 2 (w)) because scalar multiplications
by α and β distribute over vector additions.
(5) ∴ T 3 (αu + βw) = αT 3 (u) + βT 3 (w) as required.
(d) (Building examples.) Give an example of a linear transformation T : R 3 −→ R 3 such that
its null space is two dimensional. (Hint: Think about the Rank Theorem.)
Solution: There are many possibilities. Any 3 × 3 matrix with rank 1 will do, say, the
matrix [e 1 e 1 e 1 ].
1

[ ]
[ ]
r1
x1
(e) (Abstract vector spaces.) Let V = { | r
r 1 +r 2 = 1 and r 1 , r 2 ∈ R}. For any ∈ V ,
2 x
[ ]
[ ] [ ]
[ 2
]
y1
x1 y1
x1 + y
∈ V , and α ∈ R, suppose + in V is defined to be
1 − 1
and
y 2 x 2 y 2 x 2 + y
[ ]
[ ]
2
x1
αx1 + (1 − α)
α in V is defined to be
. It turns out that V is a vector space with
x 2 αx 2
this addition and scalar multiplication defined on it. Answer the following four questions
about the vector space V .
i. What is 0 ∈ V [
]
1
Solution: 0 = in V .
0
[ ]
4
ii. What is the additive inverse of in V
−3
[ ]
[ ] [ ]
−2 4
1
Solution: , since adding this to results in .
3
−3
0
[ ]
4
iii. What is 5 in V
−3
[ ] [ ]
5(4) + (1 − 5)
16
Solution:
; that is, .
5(−3)
−15
iv. V is a subset of R 2 , there is a 0 ∈ V , and the given addition and multiplication by
scalars are closed operations on V . However, V a not subspace of R 2 with this vector
space structure defined on [ V . Why ]
0
Solution: The 0 ∈ R 2 is , but this is not in V . Also, the addition and scalar
0
multiplication given for V does not correspond to addition and scalar multiplication
inherited from R 2 .
2. Representations. (Value = 40%)
Let B = {b 1 , b 2 , b 3 } be a basis of a vector space V . Let
c 1 = b 1 + b 2 ,
c 2 = b 2 + b 3 , and
c 3 = b 1 + b 3 .
Let C = {c 1 , c 2 , c 3 }.
(a) (Coordinates.) If [x] B =
⎡
⎣ −1 1
−2
⎤
⎦, then what is x
Solution: x = −b 1 + b 2 − 2b 3
2

(b) (Basis and isomorphism.) Prove C is a basis of V by considering {[c 1 ] B , [c 2 ] B , [c 3 ] B } in R 3 .
What about P B allows this consideration
Solution: P B : R 3 → V is an isomorphism of vector spaces, so C is a basis of V ⎡if and ⎤
only if {[c 1 ] B , [c 2 ] B , [c 3 ] B } is a basis of R 3 . [c 1 ] B = [b 1 + b 2 ] B = [b 1 + b 2 + 0b 3 ] B = ⎣ 1 1
0
⎡ ⎤
⎡
⎦.
⎤
Similarly, by inspection of the given equations, we have [c 2 ] B = ⎣ 0 1 ⎦, and [c 3 ] B = ⎣ 1 0 ⎦.
1
1
{[c ⎡ 1 ] B , [c 2 ] B , [c 3 ] B } is a basis of R 3 since an easy row reduction shows the 3 × 3 matrix
⎣ 1 0 1
⎤
1 1 0 ⎦ has 3 pivot columns. Therefore, C is a basis of V as required.
0 1 1
(c) (Change of coordinates.) Calculate the change-of-coordinate matrix P C←B from B represented
vectors to C represented vectors. (Hint: Think about P B←C instead and its relationship
to P C←B .)
Solution. P B←C = [[c 1 ] B [c 2 ] B [c 3 ] B }] =
⎡
⎡
⎣ 1 0 1
1 1 0
0 1 1
⎤
⎤
⎦. P C←B = P −1
B←C
, so it is simply a matter
of computing the inverse of ⎣ 1 0 1
1 1 0 ⎦. I will not waste time doing this as I am sure all
0 1 1
M2040 students can calculate the inverse.
(d) (Linearity.) Let T : V −→ V be a linear transformation and [T ] B ⎡be the ⎤ representation
of T relative to the basis B. Assume, the first column of [T ] B is
T (2b 2 ) = 2b 1 + 4b 2 + 2b 3 and [T ] B
⎡
⎣ 1 1
1
i. What is the second column of [T ] B
ii. What is the third column of [T ] B
⎤
⎦ =
⎡
⎣ 1 0
3
⎤
⎦.
⎣ 1 0
6
⎦. Also, suppose
Solution. By linearity, T (b 2 ) = b 1 + 2b 2 + b 3 . So now⎡the ⎤second column of [T ] B is
[T ] B e 2 = (P −1
B
T P B)e 2 = P −1
B
T (b) 2 = P −1
B (b 1 + 2b 2 + b 3 ) =
⎣ 1 2
1
⎦.
3

Next, [T ] B
⎡
⎣ 1 1
1
⎤
[T ] B (e 3 ). Therefore,
[T ] B (e 3 ) =
⎡
⎣ −1
−2
−4
⎦ = [T ] B (e 1 + e 2 + e 3 ) = [T ] B (e 1 ) + [T ] B (e 2 ) + [T ] B (e 3 ) =
⎤
⎦.
⎡
⎣ 1 0
3
⎤
⎦ =
⎡
⎣ 1 0
6
⎤
⎦ +
⎡
⎣ 1 2
1
⎤
⎡
⎣ 1 0
6
⎤
⎦ +
⎡
⎣ 1 2
1
⎤
⎦ +
⎦ + [T ] B (e 3 ) and this gives us the third column,
(e) (Diagram chasing.) Let T be as in the last part d) above. For 7 test points, write a matrix
equation expressing [T ] C in terms of the symbols P C←B , [T ] B , and P B←C . For 1 test point,
use this equation to find the matrix [T ] C .
Solution. [T ] C = P C←B [T ] B P B←C . I won’t bother with the matrix multiplications which I
am confident all M2040 students can do.
3. Eigenvalues and Eigenvectors. ⎡
(Value = 20%)
Let A be the matrix ⎣ 4 −1 6
⎤
2 1 6 ⎦.
2 −1 8
(a) (Eigenvalues.) Show⎡λ = 2 is an eigenvalue of A (do not use the characteristic polynomial).
Solution. A − 2I = ⎣ 2 −1 6
⎤ ⎡
2 −1 6 ⎦ ∼ ⎣ 2 −1 0
⎤
0 0 1 ⎦, so Nul(A − 2I), the eigenspace of A
2 −3 8 0 0 0
corresponding to λ = 2, is nontrivial. Therefore, λ = 2 is an eigenvalue of A.
(b) (Eigenvalues.) Show λ = 1 is not an eigenvalue of A (again, do not use the characteristic
polynomial).
⎡
Solution. A − I = ⎣ 3 −1 6
⎤ ⎡
⎤
1 0 2
2 0 6 ⎦ ∼ ⎣ 0 1 0 ⎦, so Nul(A − 2I), the eigenspace of A
2 −3 7 0 0 1
corresponding to λ = 2, is trivial. Therefore, λ = 2 is an eigenvalue of A.
⎡ ⎤
(c) (Eigenvectors.) Is ⎣ 1 1 ⎦ an eigenvector of A If so, then what is its corresponding eigenvalue
and, if not, ⎡then⎤
why ⎡not
1
⎤
Solution. Yes, A
⎣ 1 1
1
⎦ = 8
⎣ 1 1
1
⎦, so the corresponding eigenvalue is λ = 8
4

⎡
⎤
(d) (Eigenvectors.) Is ⎣ 1 1 ⎦ an eigenvector of A If so, then what is its corresponding eigenvalue
and, if not, ⎡then ⎤ why ⎡ not ⎤ ⎡
0
⎤
Solution. No, A
⎣ 1 1
0
⎦ =
⎣ 3 3
12
⎦ ≠ λ
⎣ 1 1
0
⎦ for any λ ∈ R 2 .
(e) (Eigenspace.) Find a basis of the ⎡ eigenspace corresponding to λ = 2.
Solution. Nul(A − 2I) = Nul( ⎣ 1 −1/2 0
⎤ ⎡
0 0 1 ⎦) = Span{ ⎣ 1/2 ⎤
1 ⎦}. Therefore, a basis of
0 0 0
0
⎡ ⎤
the eigenspace of λ = 2 is { ⎣ 1 2
0
⎦}.
End Midterm.
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