prove that $ A^k$ is diagonalizable

Question

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 $A$?

Answer 1

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 $(P^{-1}AP)^k$.

Answer 2

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.