$A\in M_n(\mathbb C)$ invertible and non-diagonalizable matrix. Prove $A^{2005}$ is not diagonalizable $A\in M_n(\mathbb C)$ 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
 A: 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.
A: 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 $\lambda_i=\mu_i^m=\mu_j^m=\lambda_j$).
A: This is just a rewriting of this other answer adding some more details in the derivation.
Suppose $A^m$ is diagonalisable. This is equivalent to the minimal polynomial of $A^m$ splitting into linear factors. Denoting with $\mu$ the minimal polynomial of $A^m$ and with $\lambda_i$ its eigenvalues (where $\lambda_i\neq\lambda_j$ for $i\neq j$), this means that the polynomial
$$\mu(x) = \prod_i (x - \lambda_i)\tag A$$
is the one with the smallest degree such that $\mu(A^m)=0$.
Define now the polynomial $p$ as the one such that $p(x)=\mu(x^m)$. Clearly, this satisfies $p(A)=0$. Using the expression (A), we can write $p$ as
$$p(x)=\prod_i (x^m - \lambda_i).\tag B$$
Now, if we are working in $\mathbb C$, then all polynomials of the form $x^m-c$ with $c\neq0$ have $m$ distinct roots and therefore split linearly as
$$x^m - c = \prod_j (x - c_j), \qquad c_j^m = c.$$
Note that if $A$ is non-invertible, this leaves open the possibility that $\lambda_i=0$ for some $i$, and then $p$ contains a factor of the form $x^m$, which would imply $A$ is not diagonalisable (it doesn't have to imply this, because the minimal polynomial of $A$ only has to divide $p$, and therefore could still not contain these terms).
If, on the other hand, $A$ is invertible, then we know that $\lambda_i\neq0$, therefore $p$ splits linearly, and therefore the minimal polynomial of $A$ does as well.
