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Practice Midterm 2
UCLA: Math 115A, Fall 2017
Instructor: Jens Eberhardt Date: 08 October 2017
- This exam has 4 questions, for a total of 23 points.
- Please print your working and answers neatly.
- Write your solutions in the space provided showing working.
- Indicate your final answer clearly.
- You may write on the reverse of a page or on the blank pages found at the back of the booklet however these will not be graded unless very clearly indicated.
- Non programmable and non graphing calculators are allowed.
Name:
ID number:
Question Points Score 1 7
2 5 3 5 4 6
Total: 23
- Let T : V → W be a one-to-one linear transformation between two finite dimensional vector spaces V and W over a field F. (a) (2 points) Let v 1 , v 2 ,... , vr ∈ V be linearly independent. Prove that T (v 1 ), T (v 2 ),... , T (vr ) are linearly independent as well. (b) (2 points) Prove that dim(V ) ≤ dim(W ). Hint: Use the rank-nullity formula. (c) (2 points) Assume that dim(V ) = dim(W ). Show that T is onto. Hint: Use the rank-nullity formula again. (d) (1 point) Give an example of a linear map, which is one-to-one but not onto.
Solution: (a) Let a 1 ,... , ar ∈ F such that
a 1 T (v 1 ) + · · · + a 2 T (v 2 ) = 0.
We have to show a 1 = · · · = ar = 0. Since T is linear, we get
T (a 1 v 1 + · · · + a 2 v 2 ) = 0.
Since T is one-to-one, N(T) = { 0 }, hence
a 1 v 1 + · · · + a 2 v 2 = 0.
Since v 1... , vr ∈ V are linearly independent, we get a 1 = · · · = ar = 0. Hence T (v 1 ),... , T (vr ) is linearly independent.
(b) By the rank-nullity formula
dim(V ) = rank(T ) + nullity(T ).
Since R(T ) is a subspace of W , we know
rank(T ) = dim(R(T )) ≤ dim(W ).
Furthermore, since T is one-to-one
nullity(T ) = dim(N(T )) = 0.
Putting all equations together, we get
dim(V ) = rank(T ) ≤ dim(W ).
(c) As in the last part, we get dim(V ) = rank(T ) ≤ dim(W ). By assumption we have dim(V ) = dim(W ). Hence
dim(R(T )) = rank(T ) = dim(W ).
Since R(T ) is a subspace of W , it follows that R(T ) = W , which means that T is onto.
(d) Consider the map T : R^2 → R^3 , T (x, y) = (x, y, 0). Then T is clearly one-to-one. But (0, 0 , 1) is not in R(T ), hence T is not onto.
- Let T, U : V → W be linear transformations between two finite dimensional vector spaces over a field F with bases β and γ, respectively. (a) (2 points) Prove that T + U : V → W, (T + U )(v) = T (v) + U (v) is a linear transformation. (b) (1 point) Let a ∈ F. Assume that you know A = [T ]γβ and B = [U ]γβ. Express is [aT + U ]γβ in terms of A and B. (c) (2 points) Let v ∈ V. Prove that
Z = {T : V → W | T linear and T (v) = 0}
is a subspace of L(V, W ) = {T : V → W | T linear}.
Solution: (a) Let v, w ∈ V and c ∈ F. Then
(T + U )(cv + w) = T (cv + w) + U (cv + w) = cT (v) + T (w) + cU (v) + U (w) = c(T (v) + U (v)) + (T (w) + T (w)) = c(T + U )(v) + (T + U )(w),
where we used the linearity of T and U in the second equation. Hence T + U is linear.
(b) We know that [−]γβ is linear. Hence
[aT + U ]γβ = a[T ]γβ + [U ]γβ.
(c) Let T 0 : V → W, T 0 (v) = 0. This is the zero element of L(V, W ). Clearly T 0 (v) = 0. Hence T 0 ∈ Z. Now let T, U ∈ Z and a ∈ F , then
(aT + U )(v) = aT (v) + U (v) = a0 + 0 = 0.
Hence also aT + U ∈ Z and Z is a subspace.
- Let V = P 3 (R) and W = M 2 , 2 (R). Let
β = { 1 , x, x^2 , x^3 }
γ =
w 1 =
[
]
, w 2 =
[
]
, w 3 =
[
]
, w 4 =
[
]}
be the standard ordered bases. Consider the linear map T : V → defined by
T (ax^3 + bx^2 + cx^1 + d) =
[
a + b c + d a + c b + c
]
(a) (2 points) Determine A = [T ]γβ. (b) (3 points) Prove that T is an isomorphism. (c) (1 point) Prove that V and W are isomorphic without using T.
Solution: (a) We have to express T (1),... , T (x^3 ) in the basis γ.
T (1) =
[
]
= w 2
T (x) =
[
]
= w 2 + w 3 + w 4
T (x^2 ) =
[
]
= w 1 + w 4
T (x^3 ) =
[
]
= w 1 + w 3
Hence
A = [T ]γβ =
(b) We know that T is an isomorphism if and only of A is invertible. To show that A is invertible, we compute its determinant by evaluating along the first column. So det(A) = − det(B) where
B =
We compute det B = 2 with our favorite technique. Hence det(A) = − 2 6 = 0. Hence A is invertible and T is an isomorphism.
(c) This follows from dim V = 4 = dim W.
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