Quiz #9: Tue, Nov 8th MATH 110 with Professor Stankova Section # GSI: Benson Au Name: Steve McQueen You have 10 minutes to complete this quiz. (1) Assume throughout that T : V → V is a linear operator on a finite-dimensional vector space over the field F . (a) Suppose that there exist a non-zero vector v ∈ V , a scalar λ ∈ F , and an integer m > 1 such that (T − λI)m(v) = 0. Then λ is an eigenvalue of T . True False (b) A cycle of generalized eigenvectors is linearly independent. True False (c) Any linear operator on a finite-dimensional vector space has a Jordan canonical form. True False (d) There is exactly one cycle of generalized eigenvectors cor- responding to each eigenvalue of a linear operator on a finite-dimensional vector space. True False (e) For any Jordan block J, the operator LJ has Jordan canon- ical form J. True False (f) Suppose that dim(V ) = n. Then for any eigenvalue λ of n T , we have that Kλ = ker((T − λI) ). True False (g) If two linear operators T,U : V → V have the same char- acteristic polynomial, then they are similar. True False (h) Two matrices are similar if and only if they have same Jordan canonical form. True False (i) If a linear operator on a finite-dimensional vector space has a Jordan canonical basis, then this basis is unique. True False

1 (a) True, (T − λI)k(v) is an eigenvector with eigenvalue λ for some k, where 0 ≤ k ≤ m − 1. (b) True, this is shown in the corollary of Theorem 7.6. (c) False, the characteristic polynomial of the operator needs to split (this is also sufficient). (d) False, there could be multiple cycles corresponding to a single eigenvalue (put two Jordan blocks with the same diagonal element together). (e) True, since the Jordan canonical form is unique up to or- dering of the blocks, and there is only one block in this case. (f) As pointed out in section, I intended for you to assume that the characteristic polynomial splits in this problem. m In this case, the answer is true since Kλ = ker((T − λ) ) by Theroem 7.2, where m is the algebraic multiplicity of the eigenvalue λ. (g) False, a Jordan block J has the same characteristic poly- nomial as the diagonal matrix obtained by removing the superdiagonal elements of J. (h) True, this is the content of Theorem 7.11. (i) False, in general, there are many Jordan bases; however, the so-called dot diagram is unique. (2) Find a Jordan canonical form J for the matrix   0 −1 −1     −3 −1 −2 .   7 5 6 (Hint: one of the eigenvalues is 1.) (3) Let T : V → V be a linear operator on a finite-dimensional vector space. Prove that if rank(T m) = rank(T m+1) for some positive integer m, then rank(T m) = rank(T k) for any positive integer k ≥ m. (4) Let T be a linear operator on a vector space V , and let γ be a cycle of generalized eigenvectors that corresponds to the eigen- value λ. Prove that span(γ) is a T -invariant subpsace of V . (2) We compute the characteristic polynomial   −λ −1 −1     p(λ) = det(A − λI) = det −3 −1 − λ −2   7 5 6 − λ Expanding along the first row, we get that p(λ) = −λ((−1 − λ)(6 − λ) + 10) + (−3(6 − λ) + 14) − (−15 + 7(1 + λ)). Simplifying, p(λ) = −λ(λ2 − 5λ + 4) + 4 − 4λ = −λ(λ − 4)(λ − 1) − 4(λ − 1) = (λ − 1)(−λ(λ − 4) − 4)) = (λ − 1)(−λ2 + 4λ − 4) = −(λ − 1)(λ − 2)2.

We compute the dimension of E2 via the rank-nullity theorem. Note that   −2 −1 −1     A − 2I = −3 −3 −2 .   7 5 4 We compute the rank by performing row operations on A − 2I to see that it has the same rank as   −2 −1 −1    3  ,  0 − 2 −1/2   0 0 0

which clearly has rank 2, and so dim(E2) = 1. Thus, a Jordan canonical form for A is   2 1 0     0 2 0 .   0 0 1 (3) Observe that V ⊃ T (V ) ⊃ T 2(V ) ⊃ · · · Thus, the condition rank(T m) = rank(T m+1) is equivalent to T m(V ) = T m+1(V ), and we need to prove that T m(V ) = T k(V ), ∀k ≥ m. By our initial observation, it suffices to prove that T m(V ) ⊂ T k(V ).

But for any v0 ∈ V , m m+1 T (v0) = T (v1) m = T (T (v1)) m+1 = T (T (v2)) 2 m = T (T (v2)) = ··· k−m m = T (T (vm)) k = T (vm) m m+1 for some elements v1, . . . , vm ∈ V since T (V ) = T (V ) (we have applied this condition iteratively), and so we are done. (4) Let γ = {(T − λ)p−1(v),..., (T − λ)v, v}. We define i vi = (T − λ) (v)

for short. Note that T (vi) = vi+1 + λvi and that vp = 0. This shows that T (vi) ∈ span(γ) for each i = 0, . . . , p − 1. It follows that span(γ) is T -invariant.