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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!

Xianwei
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1 Answers1

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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.

José Carlos Santos
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  • Thanks! Here is a prerequisite, i.e., $A$ only has one zero eigenvalue, so the example is not the one I mean. BTW, I still require the diagonal entries of $D$ to be nonzero, which is missing in the original question. I have added it. – Xianwei Oct 15 '17 at 12:53
  • @Xianwei I don't understand. My matrix $A$ has one and only one eigenvalue, which is $0$. What's the problem? – José Carlos Santos Oct 15 '17 at 13:04
  • Thanks for your further your clarification. Quite sorry for my inexact statement. I mean 0 is a simple eigenvalue, that is, the algebraic multiplicity of the zero eigenvalue is 1. I have made the correction in the question. – Xianwei Oct 15 '17 at 13:21
  • If all eigenvalues have algebraic multiplicity of $1$ then the matrix is diagonlizable. [if and only if, actually] – Oria Gruber Oct 15 '17 at 13:25
  • @Oria Gruber Thanks! Possibly relating to this fact, my question can be rephrased as: Is it always possible to find a diagonal matrix $D$ with nonzero diagonal entries such that all the eigenvalues of $DA$ have algebraic multiplicity of 1? – Xianwei Oct 15 '17 at 13:53