Let α 1 x x 2 be the standard basis then a t α 2 1

Info icon This preview shows pages 4–6. Sign up to view the full content.

View Full Document Right Arrow Icon
Let α = { 1 , x, x 2 } be the standard basis. Then A = [ T ] α = 2 - 1 0 0 2 - 2 0 0 2 . So clearly the characteristic polynomial is - ( t - 2) 3 . We also see that A - 2 I = 0 - 1 0 0 0 - 2 0 0 0 has rank 2. This means that E 2 has dimension 1. Therefore, there is only one cycle in the basis for K 2 = P 2 ( R ). Just as in example 3 (on pages 493-494), we just test each element of α until we find an end vector for this cycle. Again it is x 2 . Note ( T - 2 I )( x 2 ) = - 2 x and ( T - 2 I )( - 2 x ) = 2. Therefore β = { 2 , - 2 x, x 2 } is the basis, consisting of a cycle, which we want. The Jordan form looks like J = [ T ] β = 2 1 0 0 2 1 0 0 2 . Part (b): V is the real vector space of functions spanned by the set of real valued functions { 1 , t, t 2 , e 2 , te t } , and T is the linear operator on V defined by T ( f ) = f . One quickly notices that { 1 , t, t 2 } are all generalized eigenvectors of λ = 0. In fact, β 1 = { 2 , 2 t, t 2 } is a cycle, and a basis for K 0 . One also notices that β 2 = { e 2 , te t } are generalized eigenvectors of λ = 1, and since ( T - 1 I )( te t ) = e t , we see that β 2 is already a cycle. Thus β 1 and β 2 are the bases we wanted to find. Further, the Jordan canonical form looks like 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 1 . Part (c): T is the linear operator on M 2 × 2 ( R ) defined by T ( A ) = 1 1 0 1 · A for all A M 2 × 2 ( R ). Let α = 1 0 0 0 , 0 1 0 0 , 0 0 1 0 , 0 0 0 1 be the standard basis for V = M 2 × 2 ( R ). Then [ T ] α = 1 0 1 0 0 1 0 1 0 0 1 0 0 0 0 1 . If we look at this matrix, we notice that it is very close to a Jordan canonical form. In fact, if we change the order of the elements in our basis, and take β = 1 0 0 0 , 0 0 1 0 , 0 1 0 0 , 0 0 0 1 then we find [ T ] β = 1 1 0 0 0 1 0 0 0 0 1 1 0 0 0 1
Image of page 4

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
HOMEWORK 13 SOLUTIONS 5 is the Jordan canonical form for this transformation. Thus, β is a union of two cycles (of length 2 each) for the eigenvalue 1. Problem ( § 7.1 # 7) . Let U be a linear operator on a finite-dimensional vector space V . Prove the following results. (a) N ( U ) N ( U 2 ) . . . N ( U k ) N ( U k +1 ) . . . . (b) If rank ( U m ) = rank ( U m +1 ) for some positive integer m , then rank ( U m ) = rank ( U k ) for any positive integer k m . (c) If rank ( U m ) = rank ( U m +1 ) for some positive integer m , then N ( U m ) = N ( U k ) for any positive integer k m . (d) Let T be a linear operator on V , and let λ be an eigenvalue of T . Prove that if rank (( T - λI ) m ) = rank (( T - λI ) m +1 ) for some integer m , then K λ = N (( T - λI ) m ) . (e) Second Test for Diagonalizability. Let T be a linear operator on V whose char- acteristic polynomial splits, and let λ 1 , λ 2 , . . . , λ k be the distinct eigenvalues of T . Then T is diagonalizable if and only if rank ( T - λI ) = rank (( T - λI ) 2 ) for 1 i k . (f) Use (e) to obtain a simpler proof of Exercise 24 of Section 5.4: If T is a di- agonalizable linear operator on a finite-dimensional vector space V and W is a T -invariant subspace of V , then T W is diagonalizable. Proof. Part (a): Let x N ( U k ) for some k 1. This means U k ( x ) = 0. Then U k +1 ( x ) = U ( U k ( x )) = U (0) = 0. So x N ( U k +1 ). Since x is arbitrary, N ( U k ) N ( U k +1 ) for every k 1.
Image of page 5
Image of page 6
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern