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Let $A$ be and $n\times n$ matrix over $\mathbb{C}$. If $A$ is diagonalizable and $k$ is a positive integer, prove that $A^k$ is diagonalizable. Matrix $A$ $n\times n$ is diagonal, if $a_{ij}=0$, when $i \neq j$. $n\times n$ matrix $A$ is diagonalizable if $A$ is similar to some diagonal matrix. In other words, a matrix is diagonalizable if there is a diagonal matrix $D$ and an invertible matrix $P$ such that $P^{-1}AP$= $D$. How are the eigenvalues and eigenvectors of $A^k$ related to those of $A$?

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Have you attempted this? What have you tried so far? –  Robin Hoode Nov 1 '10 at 11:48
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Another question, phrased in the imperative mood, asking for some standard theory. :-( –  Robin Chapman Nov 1 '10 at 11:50
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Hint: $\left(\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\right)^{k}=\left(\mathbf{P}\mathbf{D}‌​\mathbf{P}^{-1}\right)\left(\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\right)\dots\left(‌​\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\right)\left(\mathbf{P}\mathbf{D}\mathbf{P}^{-‌​1}\right)$ and note that you have products of the form $\mathbf{P}^{-1}\mathbf{P}$. –  J. M. Nov 1 '10 at 12:06
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This user is asking the exact same questions at Yahoo Answers, getting complete answers there, and never replying. answers.yahoo.com/my/qa/… –  Samuel Nov 1 '10 at 13:21
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I really want a mod to step in and give this guy a cool-off period. –  J. M. Nov 1 '10 at 14:35
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2 Answers

I think robinhoode and Robin are right: is it so difficult to ask things nicely?

Anyway, here you have some hints:

  1. Let $v$ be an eigenvector of eigenvalue $\lambda$. That is, $Av = \lambda v$. Compute $A^k v$.
  2. Compute $(P^{-1}AP)^k$.
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Roig: Of course it is difficult; it requires doing something more than mindlessly copying the homework assignment onto the box and waiting for a solution... –  Arturo Magidin Nov 1 '10 at 14:18
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Here is a second approach. A matrix is diagonalizable if and only if it has a basis of eigenvectors.

Suppose $v$ is an eigenvector of a matrix $A$ with eigenvalue $\lambda$. Then $$A^nx = \lambda^n v.$$ Hence any eigenvector of $A$ is an eigenvector of $A^n$. Hence, any basis of eigenvalues of a matrix $A$ is a basis of eigenvalues of $A^n$ for any nonnegative integer n.

More generally if $f$ is a polynomial in the scalar field of the matrix, $f(A)$ is diagonalizable too. This is pretty easy to see.

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