# 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 :)

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.

• That is a very nice proof! Thank you very much! – Noy Perel Jun 2 '18 at 18:20