HWsolutions13

# B let t be a linear operator on a finite dimensional

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(b) Let T be a linear operator on a finite-dimensional vector space V that has a Jordan canonical form J . If β is any basis for V , then the Jordan canonical form of [ T ] β is J . True. One should be careful here. It is NOT asserting that [ T ] β is the Jordan canonical form for T . Rather, that J is the Jordan canonical form for A = [ T ] β . To see this, let γ be a Jordan canonical basis for T . Then J = [ T ] γ = [ I ] γ β [ T ] β [ I ] β γ . So A is similar to J , and hence has J as its Jordan canonical form by Theorem 7.11. (c) Linear operators having the same characteristic polynomial are similar. False. Consider L A and L B where A = 1 0 0 1 and B = 1 1 0 1 . These linear operators have the same characteristic polynomials, but are not similar, by Theorem 7.11, since they have different Jordan canonical forms. (d) Matrices having the same Jordan canonical form are similar. True. Theorem 7.11. (e) Every matrix is similar to its Jordan canonical form. True. Let A be a matrix and J its Jordan canonical form. It is obvious that J is also the Jordan canonical form of itself (see part (a) above). Therefore A and J are similar, by Theorem 7.11.

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HOMEWORK 13 SOLUTIONS 7 (f) Every linear operator with the characteristic polynomial ( - 1) n ( t - λ ) n has the same Jordan canonical form. False. The same example from part (c) above will suffice. (g) Every linear operator on a finite-dimensional vector space has a unique Jordan canonical basis. False. Take T to be the identity operator (say, on V = R 2 ). Then any basis, β , for V will give a Jordan canonical basis for T (since [ T ] β = I is diagonal, and hence a Jordan matrix). (h) The dot diagrams of a linear operator on a finite-dimensional vector space are unique. True. This is the Corollary on page 500. Problem ( § 7.2 # 13) . Let T be a nilpotent operator on an n -dimensional vector space V , and suppose that p is the smallest integer for which T p = T 0 . Prove the following results. (a) N ( T i ) N ( T i +1 ) for every positive integer i . (b) There is a sequence of ordered bases β 1 , β 2 , . . . , β p such that β i is a basis for N ( T i ) and β i +1 contains β i for 1 i p - 1 . (c) Let β = β p be the ordered basis for N ( T p ) = V in (b). Then [ T ] β is an upper triangular matrix with each diagonal entry equal to zero. (d) The characteristic polynomial of T is ( - 1) n t n . Hence the characteristic polyno- mial of T splits, and 0 is the only eigenvalue of T . Proof. Part (a): This is just Exercise 7(a) from Section 7.1. Part (b): Given β i , a basis for N ( T i ), just extend it to a basis β i +1 for N ( T i +1 ) N ( T i ). (We can do this because any linearly independent set can be extended to a basis.) Part (c): The point is that T ( β i +1 ) span ( β i ) (since given x N ( T i +1 ) then T ( x ) N ( T i )). So, clearly, [ T ] β will be upper-triangular. The diagonal entries will be 0 because given v β i +1 - β i , then T ( v ) span ( β i ), and hence T ( v ) can be written in terms of basis elements strictly before v itself.
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