Powered matrices and its eigenvalues

Question

Let $A$ be a n by n matrix, with eigenvalues $\lambda_1$, ...$\lambda_n$. I want to prove that the eigenvalues of $A^k$ are eigenvalues of $A$, raised to power $k$.

I managed to prove that if $\lambda$ is an eigenvalue of $A$, then $\lambda^k$ is an eigenvalue of $A^k$.
However, we need to consider the algebraic multiplicities. For instance, if $A$ has eigenvalues $1, 1, 2$, then $A^2$ has eigenvalues $1$ and $4$, but we do not know the multiplicities to conclude whether they are $1, 1, 4$, or $1, 4, 4$. Any ideas on how to prove?

Edit: I did some research and apparently triangulation could help to prove.

Answer 1

Let's consider the case of a diagonalizable matrix.

Let $A$ a $n\times n$ matrix and $\lambda_1,\ldots,\lambda_n$ its eigenvalues, suppose $A$ is diagonalizable, then there exist $P$ an invertible matrix such that $$A=P^{-1}\Lambda P,$$ where $$\Lambda=\begin{pmatrix}\lambda_1&0&\cdots&0\\0&\lambda_2&\cdots&0\\\vdots&\vdots&\ddots&\vdots\\0&0&\cdots&\lambda_n\end{pmatrix}.$$ Thus $$A^k=(P^{-1}\Lambda P)^k=\underbrace{(P^{-1}\Lambda P)\cdot(P^{-1}\Lambda P)\cdots(P^{-1}\Lambda P)}_{\mbox{k times}}=P^{-1}\Lambda^kP.$$ Hence the eigenvalues of $A^k$ are $(\lambda_1)^k,(\lambda_2)^k,\ldots,(\lambda_n)^k$.

Thsi solves the problem to this special case.

Answer 2

First off, this is only true if you talk about complex eigenvalues, counted with their algebraic multiplicities. The matrix $A=(\begin{smallmatrix}0&-1\\1&0\end{smallmatrix})$ has no real eigenvalues, yet $A^2$ has a real eigenvalue $-1$ with multiplicity$~2$.

Assuming this is about complex square matrices, the correct formula follows from a base change:

For every square matrix $A$ over$~\Bbb C$, there exist an invertible complex matrix $P$ such that the matrix $T=P^{-1}AP$ (obtained after a base change given by the columns of $P$) is upper triangular.

Now since the characteristic polynomial is invariant under base change, you can compute the characteristic polynomials of $A$ and $A^k$ as those of $T=P^{-1}AP$ respectively $P^{-1}A^kP=T^k$. Since those matrices are triangular, it is clear that of $\chi_A=(X-\lambda_1)\ldots(X-\lambda_n)$ (where $\lambda_1,\ldots,\lambda_n$ are the diagonal entries of$~T$), then $\chi_{A^k}=(X-\lambda_1^k)\ldots(X-\lambda_n^k)$ (where $\lambda_1^k,\ldots,\lambda_n^k$ are the diagonal entries of$~T$).