Let $ A\in M_n(\mathbb{C}) $ be an invertible and non-diagonalizable matrix. Prove that for all $k\ge 1 \Rightarrow A^k$ is not diagonalizable. 
Let $ A\in M_n(\mathbb{C}) $ be an invertible and non-diagonalizable matrix. Prove that for all $k\ge 1 \Rightarrow A^k$ is not diagonalizable.

Hi all. Since $A$ is over $\mathbb{C}$ then $A$ must have a Jordan normal form which is not a diagonal matrix. Therefor at least one of its basic Jordan blocks is of the form $\begin{pmatrix}
    \lambda & 1 & 0 & \dots & 0 \\
    0 & \lambda & 1 & \ddots  & \vdots \\
    0 & 0 & \ddots & \ddots & 0 \\
0 & 0 & \ddots & \ddots & 1 \\
0 & 0 & \dots & 0& \lambda \\
    \end{pmatrix} = J_n(\lambda) $
($\lambda \neq 0$, for $A$ is invertible.)
Since $A$ is similar to some $J_A$ Jordan matrix then $P^{-1}AP=J_A \implies P^{-1}A^kP=(J_A)^k$ then $A^k$ is similar to $(J_A)^k$ which is a block-diagonal matrix with one of its blocks being $(J_n(\lambda))^k$.
I had already proven that
$$(J_n(\lambda))^k = \begin{pmatrix}
\lambda^k & \binom{k}{1}\lambda^{k-1} & \binom{k}{2}\lambda^{k-2} & \cdots & \cdots & \binom{k}{j-1}\lambda^{k-j+1} \\
 & \lambda^k & \binom{k}{1}\lambda^{k-1} & \cdots & \cdots & \binom{k}{j-2}\lambda^{k-j+2} \\
 &  & \ddots & \ddots & \vdots & \vdots\\
 &  & & \ddots & \ddots & \vdots\\
 &  & &  & \lambda^k & \binom{k}{1}\lambda^{k-1}\\
 &  &  &  &  & \lambda^k
\end{pmatrix}$$
and we notice that $$(J_n(\lambda))^k=f(J_n(\lambda))=\begin{pmatrix}f(\lambda)&f^\prime(\lambda)&\cdots&\frac{f^{(n-1)}(\lambda)}{(n-1)!}\\&f(\lambda)&\ddots&\vdots\\&&\ddots&f^\prime(\lambda)\\&&&f(\lambda)\end{pmatrix}$$ where $f$ is the polynomial $f(t)=t^k$
I wanted to prove this statement using this, with the differentiation operator $D$ (for example, I know that $f\in \ker(D-\lambda I) \iff D^k(e^{-\lambda t}f)=0 $, and I thought of using that- but I am not sure how).
Please notice:
I know how to prove this using the minimal polynomial, but I would like to prove this using the Jordan-normal form since this is the direction our prof' wants us to go. 
I would love to hear your thoughts, thank you :)
 A: We can assume that $A$ is in the Jordan form.
$A$ is not diagonalizable so $\exists\lambda \in \sigma(A)$ such that the  $A$ has a block $J_j(\lambda)$ with  $j \ge 2$, which is equivalent to $\dim\ker (A - \lambda I) < \dim\ker (A - \lambda I)^2$.
We have 
$$J_j(\lambda) = \begin{pmatrix}
    \lambda & 1 & 0 & \dots & 0 \\
    0 & \lambda & 1 & \ddots  & \vdots \\
    0 & 0 & \ddots & \ddots & 0 \\
0 & 0 & \ddots & \ddots & 1 \\
0 & 0 & \dots & 0& \lambda \\
    \end{pmatrix} \implies 
J_j(\lambda)^k = \begin{pmatrix}
    \lambda^k & k\lambda^{k-1} & * & \dots & * \\
    0 & \lambda^k &  k\lambda^{k-1} & \ddots  & \vdots \\
    0 & 0 & \ddots & \ddots & * \\
0 & 0 & \ddots & \ddots &  k\lambda^{k-1} \\
0 & 0 & \dots & 0& \lambda^k \\
    \end{pmatrix}$$
so $$(J_j(\lambda)^k - \lambda^kI)^2 = J_j(\lambda)^k = \begin{pmatrix}
    0 & 0 & k^2\lambda^{2k-2} & \dots & * \\
    0 & 0 &  0 & \ddots  & \vdots \\
    0 & 0 & \ddots & \ddots & k^2\lambda^{2k-2} \\
0 & 0 & \ddots & \ddots &  0 \\
0 & 0 & \dots & 0& 0 \\
    \end{pmatrix}$$
having nonzero terms $k^2\lambda^{2k-2}$ on the diagonal two places above the main diagonal. So $$\dim\ker (J_j(\lambda)^k - \lambda^k I)^2 = 2 > 1 =\dim\ker (J_j(\lambda)^k - \lambda^k I)$$
A block-diagonal matrix retains its block-diagonal structure when taking powers, so $\dim\ker (A^k - \mu^k I)$ is simply the sum of dimensions of kernels of its blocks (which are of the same sizes as those for $A$). Same holds for $\dim\ker (A^k - \mu^k I)^2$.
Hence $\dim\ker (A^k - \lambda^k I)^2 > \dim\ker (A^k - \lambda^k I)$ which implies that $A^k$ has a Jordan block of size $\ge 2$ associated to $\lambda^k$.
We conclude that $A^k$ is not diagonalizable.
