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Midterm 2 Solutions for Linear Algebra...

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MATH 540 MIDTERM 2

Due 11/17/2020 10:30am eastern time in Gradescope This is an open book/open notes assignment. Rules about collaboration and material you can consult and invoke: • You may consult all Math 540 material covered through our 11/12 lecture from the Linear Algebra Done Right textbook, lecture notes/recordings, problem sets, and their solutions (yours or our official ones). If you want to invoke a result from these sources, such as a theorem/proposition/lemma/corollary proved in the textbook or in lecture, or an exercise proved in a problem set, please give an accurate citation. • You may not consult any other sources or collaborate on the exam. Good luck!

MATH 540 MIDTERM 2

Problem 1. (30pts: 10+10+10) Suppose n is a positive integer and x1 , . . . , xn ∈ R are distinct real numbers. Consider the map T : P(R) → Rn given by T p = (p(x1 ), . . . , p(xn )). It is easy to show that T ∈ L(P (R), Rn ), i.e., that T is linear—you need not prove this. (a) Prove that T is not injective. (b) The restriction of T to the subset Pn−1 (R) of its domain P(R) is denoted T |Pn−1 (R) , but for brevity we’ll denote it by S. It is easy to show that S ∈ L(Pn−1 (R), Rn )— you need not prove this. Prove that S is injective. (c) Let y1 , . . . , yn ∈ R. Prove that there exists exactly one p ∈ Pn−1 (R) that satisfies p(xi ) = yi for every i = 1, . . . , n.

Proof of (a). There are many ways to go about this. For example, we can consider p(x) = (x − x1 ) · · · (x − xn ), which is a nonzero (degree n) polynomial that satisfies T p = (p(x1 ), . . . , p(xn )) = (0, . . . , 0). Therefore, T is not injective.



Proof of (b). By LADR Propositions 3.14 and 3.16, it suffices to prove that the only element p ∈ null T is p = 0. Suppose, then, that p ∈ null T . Then, by definition T p = (0, . . . , 0) ⇐⇒ (p(x1 ), . . . , p(xn )) = (0, . . . , 0) ⇐⇒ p(xi ) = 0 for all i = 1, . . . , n. Therefore, p has at least n distinct roots. By LADR Theorem 4.12 and deg p ≤ n − 1, it follows that p = 0. The result follows.  Proof of (c). We’re looking for p ∈ Pn−1 (R) such that Sp = (p(x1 ), . . . , p(xn )) = (y1 , . . . , yn ). To prove that such a p exists, we will show that S is surjective, i.e., that range S = Rn . It follows from the Fundamental Theorem of Linear Maps that dim range S = dim Pn−1 (R) − dim null S = n − 0 = n. On the other hand, range S ⊆ Rn , and dim Rn = n. Therefore, range S = Rn as desired. The uniqueness of p follows from the injectivity of S shown in (b). Common mistakes in Problem 1 solutions:



MATH 540 MIDTERM 2

(1) A common mistake in part (a) was to invoke LADR 3.23 with an infinite dimensional V = P(R). You cannot invoke results for finite dimensional spaces if your space is infinite dimensional. In many cases, they are no longer true. In this case, it just so happens to be true, but that would have to be proven. A nice way to circumvent this, which some of you did, was to apply LADR 3.23 to Pm (R) with m ≥ n. (2) A common subtle mistake in part (b) was the assertion that LADR 4.24 tells us that all polynomials of degree ≤ n − 1 have ≤ n − 1 (distinct) roots. That is of course incorrect; the zero polynomial is in Pn−1 (R) and has infinitely many roots. (3) A common mistake in (c) was to not realize that you needed to also prove the existence of a polynomial p, not just the uniqueness. Problem 2. (30pts: 10+20) (a) Let T ∈ L(F2 ), where F = R a 2 some basis v1 , v2 of F is 0

orC. Suppose that the matrix of T with respect to 1 , where a ∈ F. Prove that T is not diagonalizable. a

(b) Let T ∈ L(F2 ), where  F = R.  Suppose that the matrix of T with respect to some a b basis v1 , v2 of F2 is , where a, b ∈ F and b 6= 0. Prove that T is not diago−b a nalizable. Then, give an example of how this fails if F = C instead; that is, come up with an example of an operator with such a matrix which is diagonalizable if F = C.

Proof of (a). Since that matrix is upper triangular, it follows from Theorem 5.32 that its diagonal entries are the eigenvalues of T . Therefore, the only eigenvalue of T is λ = a. Let’s find E(a, T ). We’re looking for all v ∈ F2 such that T v = av. Since v1 , v2 span F2 , we’re equivalently looking for c1 , c2 ∈ F such that T (c1 v1 + c2 v2 ) = a(c1 v1 + c2 v2 ) ⇐⇒ c1 T v1 + c2 T v2 = a(c1 v1 + c2 v2 ) ⇐⇒ ac1 v1 + c2 v1 + ac2 v2 = ac1 v1 + ac2 v2 ⇐⇒ c2 v1 = 0 ⇐⇒ c2 = 0. Thus, E(a, T ) = span(v1 ). The fact that T is not diagonalizable follows from the fact that its only eigenvalue is λ = a and dim E(a, T ) = dim span(v1 ) = 1 6= dim F2 , so T is not diagonalizable because Theorem 5.41 (e) fails.  Proof of (b). To find the eigenvalues, we’re looking for λ ∈ R for which there exist c1 , c2 ∈ R, not both zero, such that T (c1 v1 + c2 v2 ) = λ(c1 v1 + c2 v2 ) ⇐⇒ c1 T v1 + c2 T v2 = λ(c1 v1 + c2 v2 ) ⇐⇒ (ac1 +bc2 )v1 +(ac2 −bc1 )v2 = λc1 v1 +λc2 v2 ⇐⇒ ((a−λ)c1 +bc2 )v1 +((a−λ)c2 −bc1 )v2 = 0

MATH 540 MIDTERM 2

(1)

⇐⇒



 (a − λ)c1 + bc2 = 0 . (a − λ)c2 − bc1 = 0

Multiply the first equation by c2 , the second by c1 , and add to get: 2(a − λ)c1 c2 = 0. Therefore, either λ = a, or c1 = 0, or c2 = 0. We will prove each of these cases leads to a contradiction. Suppose λ = a. Then (1) becomes ⇐⇒



 bc2 = 0 bc1 = 0

which implies that c1 = c2 = 0 since b = 6 0. But, c1 and c2 cannot both be zero. This is a contradiction. Suppose c1 = 0. The first equation of (1) implies bc2 = 0, and thus c2 = 0 since b = 6 0. But, c1 and c2 cannot both be zero. This is a contradiction. Suppose c2 = 0, because then the second equation of (1) would similarly imply c1 = 0, but c1 and c2 cannot both be zero. This is a contradiction. Therefore, T has no eigenvalues. It therefore cannot be diagonalizable because, e.g., Theorem 5.41 (b) fails. When F = C, we know from LADR Example 5.8 that T ∈ L(C2 ) given by T (x, y) = (−y, x) has ±i as eigenvalues. Therefore, it must be diagonalizable by Corollary 5.44.  Common mistakes in Problem 2 solutions: (1) A very common mistake in part (b) was to carry out the eigenvalue computation in F = R, get all the way to λ = a ± ib, and assert that these aren’t real numbers, so there are no eigenvalues. That is an improper argument. If we are working in F = R, then λ = a ± ib doesn’t make sense as a solution in the first place. I didn’t penalize anyone much for this, because it is salvageable through minor tweaking in how you present your argument. I’m happy to discuss this in office hours. (2) Another very common mistake was dividing out equations by x or by y without first checking that x 6= 0 or y = 6 0, respectively. Problem 3. (30pts: 20+10) Let T ∈ L(F2 ), where F = C. (a) Prove that, either: (i) T is diagonalizable; or,

MATH 540 MIDTERM 2

(ii) there exists a basis of F2 with respect to which the matrix of T is of the form   a 1 , where a ∈ F. 0 a (Side remark : Problem 2 (a) implies that (i) and (ii) cannot both hold.) (b) Prove that (a) fails when F = R by coming up with an operator on R2 that satisfies neither (i) nor (ii).

Proof of (a). By Theorem 5.27, there exists a basis v1 , v2 of C2 with respect to which T has an upper triangular matrix, say:   a b M(T, (v1 , v2 )) = . 0 c If b = 0, then the matrix M(T, (v1 , v2 )) is diagonal, and thus (i) holds. If a 6= c, then since M(T, (v1 , v2 )) is upper triangular, a and c will both be eigenvalues of T by LADR 5.32, and thus T will be diagonalizable by LADR 5.44. So, let us assume that b 6= 0 and a = c from now on, i.e., that:   a b M(T, (v1 , v2 )) = . 0 a In other words: T v1 = av1 and T v2 = bv1 + av2 . Let’s consider: v′1 = v1 and v′2 = b−1 v2 . These two vectors form a basis of C2 because they are linearly independent: if one vector were a multiple of the other, then the same would have been true of v1 , v2 —a contradiction. Then T v′1 = T v1 = av1 = av′1 , and T v2′ = T b−1 v2 = b−1 T v2 = b−1 (bv1 + av2 ) = v1 + ab−1 v2 = v′1 + av′2 , so the matrix of T with respect to v′1 , v2′ is   a 1 ′ , v′ )) = M(T, (v 1 2 0 a which is of the desired form.

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Proof of (b). Consider T ∈ L(R2 ) given by T (x, y) = (−y, x). We know from LADR Example 5.8 that T has no eigenvalues. Therefore, it does not have an upper triangular matrix with respect to any basis since the first basis vector would have to be an eigenvector. Therefore, neither (i) nor (ii) can hold. 

MATH 540 MIDTERM 2

Common mistakes in Problem 3 solutions:   a b (1) A very common mistake was for people to get to a matrix of the form and 0 a then assert,without proof, that we can choose another basis to turn the matrix a 1 into . First of all, this is wrong unless b 6= 0. Second, even if b = 6 0, why 0 a can we do that? We do not have unlimited freedom over what our matrix entries are. (Here’s an example of a limit on our freedom with choosing matrices for linear transformations: if there is only one eigenvalue, we cannot choose a basis that gives an upper triangular matrix with distinct diagonal entries.) Problem 4. (10pts) Suppose we start with the numbers a1 = 0, a2 = 1, and from them we inductively set an+2 = 5an+1 − 6an for every n ≥ 1. Thus: a3 = 5 · 1 − 6 · 0 = 5, a4 = 5 · 5 − 6 · 1 = 19, and so on. Consider the operator T ∈ L(R2 ) given by T (x, y) = (y, 5y − 6x). This T was not chosen arbitrarily. It was chosen because it has the following property with regard to the previously defined sequence of numbers: (an+1 , an+2 ) = T (an , an+1 ) for every n ≥ 1. Applying this recursively clearly yields:

i.e.,

(an+1 , an+2 ) = T (an , an+1 ) = T (T (an−1 , an )) = . . . = T (T (· · · T (a1 , a2 ))) | {z }

(⋆)

(an+1 , an+2 ) = T n (a1 , a2 ) for every n ≥ 1.

n times

Find, with proof, the eigenvalues and eigenspaces of T . Use this information, together with (⋆), to find (again, with proof) an explicit formula for an in terms of n. (An explicit formula for an should not recursive/inductive but, e.g., of the form an = 1+2n . This is not the formula of course.) We first find the eigenvalues and eigenspaces by looking for λ ∈ R and (x, y) ∈ R2 \{(0, 0)} such that T (x, y) = λ(x, y) ⇐⇒ (y, 5y − 6x) = λ(x, y)     y = λx y = λx ⇐⇒ ⇐⇒ 5y − 6x = λy (λ2 − 5λ + 6)x = 0 Note that x 6= 0, or else the first equation would imply x = y = 0, which we’re excluding. Therefore, we can divide the second equation by x and get     y = λx y = λx ⇐⇒ ⇐⇒ λ = 2 or 3 λ2 − 5λ + 6 = 0

MATH 540 MIDTERM 2

Therefore, the eigenvalues are λ = 2 and 3, and the corresponding eigenvectors are given by (x, y) = (x, λx), with x = 6 0. Therefore, the corresponding eigenspaces are: E(2, T ) = {(x, λx) : x ∈ R} = span((1, 2)), E(3, T ) = {(x, λx) : x ∈ R} = span((1, 3)). Next, seeing as to how (a1 , a2 ) = (0, 1) = (1, 3) − (1, 2), it follows from the linearity of T (and thus T n ) (an+1 , an+2 ) = T n (a1 , a2 ) = T n ((1, 3) − (1, 2)) = T n (1, 3) − T n (1, 2), n ≥ 1. Since (1, 3) is an eigenvector with corresponding eigenvalue 3, T n (1, 3) = 3n (1, 3); likewise, T n (1, 2) = 2n (1, 2). Therefore (an+1 , an+2 ) = 3n (1, 3) − 2n (1, 2) = (3n − 2n , 3n+1 − 2n+1 ), n ≥ 1, and thus an+1 = 3n − 2n , n ≥ 1. Shifting indices over by 1 gives an = 3n−1 − 2n−1 , n ≥ 2. Clearly, this formula also holds when n = 1 (since a1 = 0 and 31−1 − 21−1 = 0 too), so we deduce that an = 3n−1 − 2n−1 for all n ≥ 1. Common mistakes in Problem 4 solutions: (1) While solving for λ (or x, or y), many of you forgot to check that x 6= 0 (or y = 6 0) before dividing an equation by x (or y). (2) Some of you decided to treat T as a matrix. Be careful, T is not a matrix, but it does have an associated matrix. (3) Some of you wanted to use the associated matrix of T , which you declared was   2 0 with respect to the standard basis vectors. It isn’t. This is the matrix of 0 3 T with respect to an eigenvector for λ = 2 and λ = 3, taken in this order. The  0 1 matrix of T with respect to the standard basis vectors is . −6 5 (4) Some of you wanted to incorrectly made use of the isomorphism M(v) between vectors and column-vectors, because you preferred to take the matrix exponential of the matrix associated to T . We didn’t cover M(v) in class,   but that’s ok. 0 Nonetheless, you either didn’t realize that M((0, 1)) is not , or you didn’t 1 indicate that you were taking the isomorphism in the first place, and intermingled matrices with linear operators....