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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?


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up vote 12 down vote accepted

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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I don't understand. A is cancelled by which minimal polynomial? of $A^m$? – Jozef Aug 22 '11 at 16:02
Sorry, I made a misprint. Now it's corrected. – Henri Aug 22 '11 at 16:04
@user14829: It means that if $\mu(x)$ is the minimal polynomial of $A^m$, then $\mu(A^m)$. Write out $\mu(x)=x^k + a_{k-1}x^{k-1}+\cdots+a_1x + a_0$. Plug in $A^m$. You get $A^{mk} + a_{k-1}A^{m(k-1)} + \cdots+a_1A^{m} + a_0I$. This is the same as you would get if you evaluate $p(x) = x^{mk} + a_{k-1}x^{m(k-1)} + \cdots + a_1x^m + a_0$ at $A$, so $p(A)=0$, so the minimal polynomial of $A$ must divide $p(x)$. – Arturo Magidin Aug 22 '11 at 16:06
@Arturo:You meant $\mu(A^m)=0$? – Jozef Aug 22 '11 at 16:10
@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. – Arturo Magidin Aug 22 '11 at 16:48

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.

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