Practice midterm, Math 113, Fall 2008 – Solutions.

These also illustrate the level of detail you should try to give in your solutions.However, if you run out of time to write detailed proofs, write an outline of theproof.

Problem 1. No justification is needed. In this question, all vector spaces mentioned arevector spaces over R.(a) True/false: any six vectors (v1, . . . , v6) ∈ R7 are linearly independent.

False: take vi = 0 for all i.(b) True/false: If U is a four-dimensional subspace of R7, and W a three-

dimensional subspace of R7, then R7 = U ⊕int W .False, e.g., if W ⊂ U .

(c) If the list (v) containing only one vector is linearly independent, whatdoes this imply (if anything) about v?Implies that v 6= 0.

(d) Give any basis for R4 where none of the basis vectors lie on any of thefour coordinate axes.(1, 2, 0, 0), (2, 1, 0, 0), (0, 0, 1, 2), (0, 0, 2, 1) (for example).

(e) What is the dimension of the space of linear maps from R4 to R3?12. (We proved in class that dimL(V,W ) = dimV dimW . )

Problem 2. Let C[0, 1] be the space of continuous, real-valued functions from [0, 1] to R.Let Q be a subspace of C[0, 1] of dimension 5. Prove that, if x1, . . . , x4 ∈[0, 1], there exists f ∈ Q so that f(xi) = 0 for all i. Is the same conclusiontrue if we considered x1, . . . , x5 ∈ [0, 1]?

Comment. The question was carelessly phrased: f = 0 gives an exam-ple, so I should have written nonzero f ∈ Q.

Solution. Consider the map Q→ R4 given by

T : f 7→ (f(x1), f(x2), f(x3), f(x4)).

It is linear. By the rank-nullity theorem, dim image(T ) + dim null(T ) = 5.Since the image of T is a subspace of R4, it has dimension ≤ 4. So thenull-space of T has dimension ≥ 1. In particular, there exists a nonzerof ∈ Q belonging to the null-space. Then f(x1) = · · · = f(x4) = 0.

This is not so if we considered x1, . . . , x5. The space of quartic realpolynomials a+bx+cx2 +dx3 +ex4 is five-dimensional, and yet no nonzeroelement of it has five zeros.

Problem 3. Let V be an n-dimensional vector space over a field F . Let ` : V → F bea linear functional. Suppose that `′ : V → F is another linear functionalwhich has the same null space as `. Prove that there exists a scalar λ ∈ Fso that `′ = λ`.

Hint. Take a basis for the null-space of ` and extend it to a basis for V .What does `′ do to this basis?

Comment. Yet again, the question was carelessly phrased: it shouldhave specified that ` 6= 0. The statement is false if ` = 0, of course!

Solution. Since ` 6= 0, the dimension of its image is > 0. Since theimage is a subspace of F , it has dimension ≤ 1. So, dim image ` = 1. By

1

2

the rank-nullity theorem, the dimension of null(`) is n − 1. Take a basise1, . . . , en−1 for it. Extend it to a basis e1, . . . , en for V . Note that en doesnot belong to the null-space of `; thus `(en) 6= 0. Let λ = `′(en)

`(en) . Then(`′ − λ`)(en) = `′(en) − λ`(en) = 0. So the null-space of `′ − λ` containsboth en (as we have just shown) and e1, . . . , en−1 (since they belong to boththe null space of ` and `′). These n vectors span all of V . So the nullspaceof `′ − λ` is all of V . So `′ − λ` = 0, i.e., `′ = λ`.

Problem 4. Let V,W be finite-dimensional vector spaces over the field F . Consider thecomposition map:

C : L(V, F )× L(F,W )→ L(V,W ).

(a) Would it be reasonable to call C a linear map? Prove that any T inthe image of C satisfies dim image(T ) ≤ 1.

(b) Prove the reverse: any T with dim image(T ) ≤ 1 belongs to the imageof C.Solution. To be explicit, the map C sends (A,B) ∈ L(V, F )×L(F,W )to the composition BA ∈ L(V,W ).The first part of the question – would it be reasonable to call C a linearmap? – is a little too vague to appear on the actual midterm! Here iswhat you might say, however: To make sense of “linear map”, we firsthave to clarify what the vector space structure of L(V, F ) × L(F,W )is. The only reasonable structure of vector space (i.e., addition andmultiplication) on L(V, F ) × L(F,W ) is the “direct sum.” If we con-sider C as a map from L(V, F )⊕L(F,W ) to L(V,W ), it is not linear.For example, C(2T, 2S) = 4C(T, S).(i) Suppose (A,B) ∈ L(V, F ) × L(F,W ). Then the image of thecomposition BA ∈ L(V,W ) is contained in the image of B, sinceBA(v) = B(A(v)) ∈ image(B). However, the image of B is at mostone-dimensional by the rank-nullity theorem, since dim imageB ≤dimF = 1. So dim image(BA) ≤ 1, as required.(ii) Conversely, suppose T ∈ L(V,W ) satisfies dim image(T ) ≤ 1. IfT = 0, it is clearly in the image of C: it is the composition of two zeromaps. Otherwise, let w ∈ image(T ) be nonzero. Thus, span(w) =image(T ). For every v ∈ V there exists a unique λv ∈ F so thatT (v) = λvw. We shall write T = `2`, where `, `2 are defined thus:• ` : V → F is defined by ` : v 7→ λv. It is linear: for example,

T (v1 + v2) = T (v1) + T (v2) = λv1w + λv2w = (λv1 + λv2)w,

proving λv1 + λv2 = λv1+v2 ; we check scalar multiplication sim-ilarly.

• `2 : F →W is defined by λ 7→ λw. Clearly it is linear.For every v ∈ V , we have T (v) = λvw = `2(λv) = `2(`(v)). So T isthe composite of `2 and `, and so in the image of C.

Problem 5. Let V be a finite-dimensional vector space. Let A ∈ L(V, V ).(a) Prove that the set of B ∈ L(V, V ) so that BA = 0 is a linear subspace

of L(V, V ).(b) Compute the dimension of this subspace in terms of the dimension of

V and the dimension of image(A).Solution.

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(i) Suppose B1, B2 satisfy BiA = 0.Set C = (B1 +B2)A. We want to check C = 0. For any v ∈ V ,

C(v) = B1 +B2(A(v)) = B1(A(v)) +B2(A(v)) = 0,

so indeed C = 0.Similarly, for any scalar λ ∈ F , (λB1)A(v) = (λB1)(A(v)) = λB1(A(v)) =

0.So the set of such Bs is closed under addition of linear operators and

scalar multiplication; so it is a subspace. Note: This painful level of detailis only required when asked for.

(ii) Let X be the vector space of Bs with BA = 0.Take a basis e1, . . . , em for the image of A. Extend it to a basis

e1, . . . , em, em+1 . . . , en

for V ; thus n = dim(V ). Now, BA = 0 if and only if B(A(v)) = 0 for allv ∈ V , which happens if and only if B(w) = 0 for all w ∈ image(A); whichhappens if and only if B(ei) = 0 for 1 ≤ i ≤ m.

We have seen in class that the map

Φ : L(V, V )→ n by n matrices

which sends B to its matrix with respect to the bases (ei)1≤i≤n, (ei)1≤i≤n

is a linear isomorphism. Our remark above shows that Φ(X) is the setof matrices for which the first m columns are identically zero. Therefore,X is isomorphic to the space of n − m × m matrices, which is evidentlyisomorphic to F (n−m)n. Since isomorphic spaces have the same dimension,dim(X) = (n−m)n.