Suppose $A\in \mathbb{R}^{n\times n}$ is a non-diagonalizable matrix and, if $A\in \mathbb{R}^{n\times n}$ is singular, zero eigenvalue is simple, is it always possible to find a diagonal matrix $D\in \mathbb{R}^{n\times n}$ with nonzero diagonal entries such that $DA$ is diagonalizable?
I thought the answer was yes, but I cannot prove it, nor find exact statements in the textbook that support my thought. If the answer is yes, please provide some resources to retrieve the anwser or some hints to prove it.
Thanks and regards!
The answer is negative. Take $A=\left(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right)$. Its only eigenvalue is $0$. If $D=\left(\begin{smallmatrix}a&0\\0&b\end{smallmatrix}\right)$, with $a,b\neq0$, then$$DA=\begin{pmatrix}0&a\\0&0\end{pmatrix},$$which is not diagonalizable.