Math 416 Homework 7. Solutions.
1. (Freidberg, Insel, Spence 2.2.2). Let β, γ be the standard bases for Rn, Rm, respectively. In each of the following cases, you are given a linear transformation T : Rn → Rm, and you
are to compute [T ]γβ. (a) T : R2 → R3 given by T (a 1, a2) = (2a1− a2, 3a1+ 4a2, a1). (b) T : R3 → R2 given by T (a 1, a2, a3) = (2a1+ 3a2− a3, a1+ a3). (c) T : R3 → R given by T (a1, a2, a3) = 2a1+ 3a2− a3. (d) T : R3 → R3 given by T (a 1, a2, a3) = (2a1+ a3, −a1+ 4a2+ 5a3, a1+ a3). (e) T : Rn→ Rn given by T (a 1, a2, . . . , an) = (a1, a1, a1, . . . , a1). (f) T : Rn→ Rn given by T (a 1, a2, . . . , an) = (an, an−1, an−2, . . . , a1). (g) T : Rn→ R given by T (a1, a2, . . . , an) = a1+ an. Solution: (a) 2 −1 3 4 1 0 . (b) 2 3 −1 1 0 1 . (c) 2 1 −3 . (d) 0 2 1 −1 4 5 1 0 1 . (e) 1 0 0 . . . 0 1 0 0 . . . 0 .. . 1 0 0 . . . 0 . (f ) 0 0 . . . 0 0 1 0 0 . . . 0 1 0 0 0 . . . 1 0 0 .
(g)
1 0 0 . . . 0 0 1 .
2. (Freidberg, Insel, Spence 2.2.4). Define T : M2×2(R) → P2(R) a b c d 7→ (a + b) + 2dx + bx2. Choose the ordered bases
β = 1 0 0 0 , 0 1 0 0 , 0 0 1 0 , 0 0 0 1 , γ = {1, x, x2}. Compute [T ]γβ.
Hint: You might find it useful to clarify things if you first work out the dimensions of the matrix you’re trying to compute!
Solution: Let us write the elements of β as vi and the elements of γ as wi in the same order
as above. Then we have
T v1= 1, T v2= 1 + x2, T v3 = 0, T v4 = 2x, so we have [T ]γβ = 1 1 0 0 0 0 0 2 0 1 0 0 .
3. Define T : M2×2(R) → M2×2(R) by T (A) = At (i.e. T (A) is the matrix transpose of A).
Choose β as in the previous problem and compute [T ]ββ.
Solution: We see that we have
T v1= v1, T v2 = v3, T v3 = v2, T v4= v4, so [T ]γβ = 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 1 .
4. Define T : M2×2(R) → R where T (A) is the trace of A, i.e. T (A) = a11+ a22. Choose β as in
the previous two problems, and choose γ = {1}. Compute [T ]γβ.
Solution: We have
T v1 = 1, T v2 = 0, T v3 = 0, T v4= 1,
so
[T ]γβ = 1 0 0 1 .
5. Give an example of A, B both n × n matrices where AB 6= BA. What does this, plus the last problem below, say?
6. Why is it true that when you look in a mirror, it flips left-to-right but does not flip up-to-down?
Solution: There’s a short answer and a long answer to this question. The short answer is “neither”. This is because it flips back-to-front.
The long answer would be that the previous short answer is not entirely satisfactory, since we certainly perceive the mirror as flipping left-to-right. For example, if we hold a newspaper in front of us, the letters in the mirror will always look “backwards” and never “upside down”. But notice that we are now talking about perception. There is no natural law at work here, just our perception of it. The main reason we perceive the “backwards” direction is that our brain is trying to invert the transformation, and of course we have a cognitive bias towards spinning in a circle versus flipping upside down.
To see this, first imagine that you are standing facing away from the mirror, holding a news-paper with large type in front of you. This is what your brain is expecting to see, and when you mentally rotate that image to the one in the mirror, you end up making the words “back-wards”.
But of course, you could also imagine craning your head back really far, looking at yourself upside down, then (if one is nimble enough) putting the top of your head on the floor and then doing a handstand. Then you would look upside down compared to your mental image. Of course, it is clear which way our brain prefers to invert the transformation.
7. Recall that we define the transformation Tθ: R2 → R2 that rotates vectors by θ. Let Rx be
the transformation that reflects in the x-axis. Show that
Next, show that there is some angle ψ such that
Rx◦ Tψ = Tθ◦ Rx.
What is the relationship between θ and ψ? Discuss the geometric meaning of this computa-tion.
Hint: You might find it easier to compute the matrices of Tθ, etc. with respect to some basis
in R2, but in factyou can solve this problem using purely geometric arguments.
Solution: We have seen in class, or can compute now directly, that the matrix of rotation wrt the standard basis is
Aθ= [Rθ] = cos θ sin θ − sin θ cos θ .
Now, the matrix of reflection in the x-axis can be computed as follows: Tx(1, 0) = (1, 0), Tx(0, 1) = (0, −1), so we have B = [Tx] = 1 0 0 −1 . Now we compute AθB = cos θ sin θ − sin θ cos θ 1 0 0 −1 = cos θ − sin θ − sin θ − cos θ , BAθ = 1 0 0 −1 cos θ sin θ − sin θ cos θ = cos θ sin θ sin θ − cos θ . These are clearly not equal, so the transformations do not commute.
The next question is: what angle should we choose so that AψB = BAθ?
Looking at the computations above, this would mean that we have cos ψ − sin ψ − sin ψ − cos ψ = cos θ sin θ sin θ − cos θ , or ψ = −θ.
Goemetrically, this makes some sense (recalling the mirror problem above). Think what hap-pens to our reflection when we rotate a certain direction: our mirror image rotates in the opposite direction that we do. If I rotate in such a way as to bring my left shoulder closer to the mirror, my mirror image rotates, but it is my mirror image’s right shoulder that is getting closer, &c.
8. Write down five linear maps T0, T1, T2, T3, T4, where for all k,
Tk: P4(R) → P4(R),
and the rank of Tk is k.
Solution: There are many ways to do this. One solution is similar to a question we saw on the last homework. For example, we know that if we define Tq to be “the qth derivative”, then
it has a q − 1-dimensional nullspace, since dq
dxqx p = 0,
for all p < q. From Rank–Nullity, this means that the rank of Tq is 5 − q, so just choose Tk
to be the 5 − kth derivative.
9. (Freidberg, Insel, Spence 2.3.12). Let V, W, Z be vector spaces, and let T : V → W and U : W → Z be linear.
(a) Prove that if U T is one, then T is one. Must it be true that U is one-to-one??
(b) Prove that if U T is onto, then U is onto. Must it be true that U is onto??
(c) Prove that if both T and U are one-to-one and onto, then so is U T . (Is the converse true?)
Solution:
(a) Yes. We can prove this by contradiction. Assume that T is not one-to-one. Then by definition there exists x 6= y such that T x = T y. But then U T x = U T y with x 6= y and therefore U T is not one-to-one.
(b) Yes. We can again prove this by contradiction. Assume that U is not onto, i.e. there is a z ∈ Z such that z 6∈ R(U ). Then it follows that z 6∈ R(U T ), since if it were, then U (T x) = z but then z would be in R(U ). Therefore U T is not onto.
(c) First we assume that both T, U are one-to-one. Let x 6= y. Then T x 6= T y since T is one-to-one. Then U (T x) 6= U (T y) since U is one-to-one, and therefore x 6= y implies U T (x) 6= U T (y).
Now assume that both T, U are onto. Since U is onto, for every z ∈ Z, there is a w ∈ W such that U w = z. Since T is onto, there is a v ∈ V such that T v = w. Then U T v = U w = z, so for any z ∈ Z, there is a v ∈ V with U T v = z.
However, the converse of this statement is false: U T being one-to-one and onto does not imply that both T, U are. For example, let V = R, W = Rn, Z = R. Define T (x) = (x, 0, 0, 0, . . . , 0) and U (x1, x2, . . . , xn) = x1. Then it is clear that T is not onto and U is
10. (Freidberg, Insel, Spence 2.3.13). Let A, B be n × n matrices. Recall that the trace of A is defined as Trace(A) = Pn
i=1Aii, i.e. the sum of the diagonal elements. Show that
Trace(AB) = Trace(BA).
Solution: Recall that by definition of trace, Trace(AB) =
n
X
i=1
(AB)ii.
By the definition of matrix multiplication, (AB)ii= n X k=1 AikBki, and thus Trace(AB) = n X i=1 n X k=1 AikBki.
