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Suppose $A$ is a $n\times n$ matrix with real eigenvalues, and that the characteristic polynomial of $A$ is $(x-\lambda_1)^{m_1}(x-\lambda_2)^{m_2}$. I must prove that the characteristic polynomial of $A^k$ is $(x-\lambda_1^k)^{m_1}(x-\lambda_2^k)^{m_2}$.

I know that if $\lambda$ is an eigenvalue of $A$ then $\lambda^k$ is the eigenvalue of $A^k$. But here I must prove that $\lambda_1^k$ and $\lambda_2^k$ are the only eigenvalues of $A^k$ (and that the algebraic multiplicities are preserved). I think it can be proven using Jordan's theorem, but I don't know how.

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Besides the solution in the other post I can not help myself to give a quit different approach. The characteristic polynomial of $A$ is the same polynomial for any matrix of the form $P^{-1} A P$. Here is the key lemma:

Claim: there is a matrix $P$ such that $P^{-1} A P = T$ is upper triangular.

So the above claim allows us to reduce the problem to upper triangular matrices. But if $$T =\begin{bmatrix}u_{1,1} & u_{1,2} & u_{1,3} & \ldots & u_{1,n} \\ & u_{2,2} & u_{2,3} & \ldots & u_{2,n} \\ & & \ddots & \ddots & \vdots \\ & & & \ddots & u_{n-1,n}\\ 0 & & & & u_{n,n}\end{bmatrix}$$ then the characteristic polynomial of $T$ is $\Pi_{j=1}^n (u_{j,j} - x)$. Observe that also $T^k$ is upper triangular: $$ T^k =\begin{bmatrix}u_{1,1}^k & * & *& \ldots & * \\ & u_{2,2}^k & * & \ldots & * \\ & & \ddots & \ddots & \vdots \\ & & & \ddots & *\\ 0 & & & & u_{n,n}^k\end{bmatrix}$$ so the characteristic polynomial of $T^k$ is $\Pi_{j=1}^n (u_{j,j}^k - x)$.

The claim is not difficult to show. Just use the existence of a eigenvalue & eigenvector and proceed by induction by passing to the quotient space. If anybody want I can edit the answer and add a detailed proof of the claim.

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    $\begingroup$ Your claim is essentially the Schur decomposition. $\endgroup$ Commented Jun 21, 2015 at 15:50
  • $\begingroup$ Thank you very much. I was not aware that the claim is called Schur decomposition. $\endgroup$
    – Holonomia
    Commented Jun 21, 2015 at 16:02

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