# Linear Indepedence of Egienvectors and Jordon blocks

by crazygrey
Tags: blocks, egienvectors, indepedence, jordon, linear
 P: 7 Hi everyone, I had couple questions: 1) If I want to proof that egienvectors are linearly indpendent by induction, how do so? I do understand that I can start with a dimension of 1 and assume v1 to be a non zero vector so hence a linear indepedent, what do I do after that for other cases? 2) What the power of jordon block? for example (J)^k general case Thanks
 Sci Advisor HW Helper P: 9,396 1) Since eigenvectors have no reason to be linearly indepedendent then you cannot do this. Perhaps you should check the wording of the question? There are some extra hypotheses you've omitted. So what are they and how can you use them? 2) What have you done? Any examples? Have you tried computing some powers of jordan blocks? If not why not? If you just do it (try putting ones on the diagonal), you should be able to spot some patterns and then formulate an easy to prove conjecture.
 P: 7 1) Basically the idea is that having different eigenvlaues will results in independent eigenvectors. I just want to prove this thoerm. Let v=[v1 v2 v3.......vn] set of eigenvectors with distinct eigenvalues w1,w2,w3,......,wn. By induction: if n=1, v1 has to be linearly independent since it is a non zero vector. if n=2, a1v1+a2v2=0------1) Multiply 1) by A knowing that Av=wv we have a1w1v1+a2w2v2=0 ----2) Multiply 1) by w1 a1w1v1+a2w1v2=0 ----3) 2)-3) a2(w2-w1)v2=0 , thus a2=0 I want to prove that this holds for all eigenvectors, how do I do that? 2) Knowing that Jordan Block is a representation of a square matrix A with repeated eigenvalues. I understand the pattern of the jordan block where the repeated eigenvalues are allocated in the diagonal and 1's in the superdiagonal. I know if I square the jordan block, the 1s starts shifting location, but I'm unable to obtain a general forumla of (J)^k knowing that A=(q)^-1*J*q where q is the set of eigenvectors. I hope I clarified what I meant. Appreicate your help
 Sci Advisor HW Helper P: 9,396 Linear Indepedence of Egienvectors and Jordon blocks 1) You cannot prove it holds for all eigenvectors because it is trivially false. (if v is an eigenvector so is 2v). 2) Sorry, but this is quite an easy question once you guess the formula and I strongly urge you to try a couple of examples, like working out the first few powers of [1 1 0] [0 1 1] [0 0 1] because the numbers tou will see are very very well known and will show you what to do. You can even do that in your head. In fact doing [k 1 0] [0 k 1] [0 0 k] is probably just as easy and instructive. If you just do this you will learn a lot more than being told the answer.
 P: 7 1) I wanted to prove that all eigenvectors in the vector space are linearly independent, not a multiplication of it with a scalar. If I have a set of vectors [v1,v2,v3,--vn] 1,2,3,..,n are indices...this set of vectors belong to a vector space , and I want to prove that all of them are linearly independent... 2) Thanks I will work it through ...
Math
Emeritus