Powers of a Matrix and Eigenvalues proof

muzziMsyed21
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Homework Statement



Prove that if A is an nxn matrix with eigenvector v, then v is an eigenvector for Ak where kε(all positive integers)


Homework Equations



Av=λv

The Attempt at a Solution



Av=λv
A(Av)=A(λv)
Akv=λ(Av)

i know i may not be doing it right but this is what i can think of so far
 
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muzziMsyed21 said:
Av=λv
A(Av)=A(λv)

What if I rewrote that as
A(Av) = λ(Av)
 
If you want to show that v is an eigenvector of a power of A (given that v is an eigenvector of A itself), you need to know what happens when you multiply v by some power of A. Consider the second power to begin with - the pattern you see continues for higher powers. So,

if Av = λv, what do you get when you examine

A(Av) = A(λv)

* and simplify the left side?
* and simplify the right side (remember that A(λv) = λ(Av)) when you simplify the right?

If you correctly simplify these two pieces you should be able to show that v IS an eigenvector of A-squared, AND you will know the associated eigenvalue. The latter is important, because it will give you a major hint about the link between A^k, v, and the corresponding eigenvalue. Once you see that link, you will know how to answer your question.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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