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$A\in M_n(\mathbb C)$ invertible and non-diagonalizable matrix. I need to prove that $A^{2005}$ is not diagonalizable as well. I am asked as well if Is it true also for $A\in M_n(\mathbb R)$. (clearly a question from 2005).

This is what I did: If $A\in M_n(\mathbb C)$ is invertible so $0$ is not an eigenvalue, We can look on its Jordan form, Since we under $\mathbb C$, and it is nilpotent for sure since $0$ is not an eigenvalue, and it has at least one 1 in it's semi-diagonal. Let $P$ be the matrix with Jordan base, so $P^{-1}AP=J$ and $P^{-1}A^{2005}P$ but it leads me nowhere.

I tried to suppose that $A^{2005}$ is diagonalizable and than we have this $P^{-1}A^{2005}P=D$ When D is diagonal and we can take 2005th root out of each eigenvalue, but how can I show that this is what A after being diagonalizable suppose to look like, for as contradiction?

Thanks

3 Answers 3

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If $A^m$ is diagonalizable, then $A^m$ is cancelled by its minimal polynomial $P$, which has simple roots. Therefore $A$ is cancelled by $P(X^m)$ which has simple roots because $P(0)\neq 0$ ($A$ is invertible).

Indeed, if $P(X)=\prod (X-\lambda_i)$, then $P(X^m)=\prod (X^m-\lambda_i)$ whose roots are all the $m$-roots of $\lambda_i$ which differ one frome another (if $\mu_i$ is a $m$-root of $\lambda_i$, then $\mu_i\neq \mu_j$ else $\lambda_i=\mu_i^m=\mu_j^m=\lambda_j$).

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    @Jozef: Over $\mathbb{R}$ the result is false. Take the rotation by $90^{\circ}$, A=\left(\begin{array}{rr}0&1\\-1&0\end{array}\right). Then $A^2=-I$ is diagonalizable, but $A$ is not. Even with all eigenvalues positive for $A^m$ the result does not follow for $m\gt 2$, because the factors $x^m-\lambda$ no longer split, since the real numbers have at most two roots for $x^m-\alpha$. It's true if $A^2$ is diagonalizable$and$has all positive eigenvalues.2011-08-22
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As luck would have it, the implication: $A^n$ diagonalizable and invertible $\Rightarrow A$ diagonalizable, was discussed in XKCD forum recently. See my answer there as well as further dicussion in another thread.