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I'm interested to know if it's true because I saw that if $A^2$ is diagonalizable then $A$ is not necessary Diagonalizable.
I have a feeling it's true but I'm not sure how to prove it.
This is my proof but I'm not sure if it's good to assume that :
My try
By Diagonalizable we know that exist invertible matrix $P$ such that :
$P^{-1}AAP=diag(\lambda_1,...,\lambda_n)$
$AA=Pdiag(\lambda_1,...,\lambda_n)P^{-1}$
Set $D=Pdiag(\sqrt{\lambda_1},...,\sqrt{\lambda_n})P^{-1}$
So : $D^2=(Pdiag(\sqrt{\lambda_1},...,\sqrt{\lambda_n})P^{-1})(Pdiag(\sqrt{\lambda_1},...,\sqrt{\lambda_n})P^{-1})=Pdiag(\lambda_1,...,\lambda_n)P^{-1}=AA$
Here I wonder - $A=D=Pdiag(\lambda_1,...,\lambda_n)P^{-1} \rightarrow A$ is diagonalizable.

Roach87
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  • Have you learnt theorems concerning the minimal polynomial? – PseudoNeo Feb 18 '21 at 20:28
  • Yes, Jordan forms , Cayley-Hamilton , Nilpotent matrix... – Roach87 Feb 18 '21 at 20:29
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    Note that your candidate proof is not using the fact that $A$ is invertible. For example $\begin{pmatrix} 0 & 1 \ 0 & 0\end{pmatrix}$ squares to a diagonalizable matrix (the zero matrix) but is not itself diagonalizable. – leslie townes Feb 18 '21 at 20:33
  • @leslietownes Yes I know I didn't use the fact that A is invertible which makes me think I did a mistake somewhere – Roach87 Feb 18 '21 at 20:35

2 Answers2

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Here is a relatively quick proof. Note that a matrix is diagonalizable if and only if its minimal polynomial is a product of distinct linear factors. Because $A^2$ is diagonalizable, its minimal polynomial can be written as $$ p(x) = (x - \lambda_1) \cdots (x - \lambda_n) $$ where the eigenvalues $\lambda_i \in \Bbb C$ are distinct. Notably, each $\lambda_j$ is non-zero because $A$ is invertible. Thus, $A$ satisfies $q(x) = 0$, where $$ q(x) = p(x^2) = (x^2 - \lambda_1) \cdots (x^2 - \lambda_n) \\ = (x - \sqrt{\lambda_1})(x + \sqrt{\lambda_1}) \cdots (x - \sqrt{\lambda_n})(x + \sqrt{\lambda_n}). $$ The polynomial $q(x)$ has no repeating factors. The minimal polynomial of $A$ must divide $q(x)$, so the minimal polynomial of $A$ cannot have any repeating factors. Thus, $A$ must be diagonalizable.

Ben Grossmann
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  • Nice proof , also $\lambda \ne 0$ because $A$ is invertible. I though about it yet didn't know how to prove it. so by diagonalization the minimal polynomial has distinct eigenvalues with 1 multiplicity. since its the minimal polynomial of $A^2$ we can exchange $x$ for $A^2$ by Cayley-Hamilton $\rightarrow$ the minimal polynomial of $A$ also has distinct roots with 1 multiplicity $\rightarrow A$ is diagonalizable. Right? – Roach87 Feb 18 '21 at 20:41
  • @Roach87 Right, I've edited that in now – Ben Grossmann Feb 18 '21 at 20:43
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a simple approach is to argue by contradiction. Suppose $A$ is invertible and defective but $A^r$ is diagonalizable for some natural number $r$.

This means there is some non-zero eigenvalue $\lambda_1$ that has algebraic multiplicity $d$ strictly greater than geometric multiplicity. Via re-scaling $A$ we can assume WLOG that $\lambda_1=1$. (i.e. re-run the argument on $A':=\lambda_1^{-1}\cdot A$ if you prefer.)

via any triangularization (Jordan, Schur, or others) we have

$S^{-1} A S = \displaystyle \left[\begin{matrix}I_d + N & * & \\ \mathbf 0 & *\end{matrix}\right]$
$S^{-1} A^r S = \displaystyle \left[\begin{matrix}(I_d + N)^r & * & \\ \mathbf 0 & *\end{matrix}\right]=\displaystyle \left[\begin{matrix} I_d & * & \\ \mathbf 0 & *\end{matrix}\right]$
where $N\neq \mathbf 0$ is strictly upper triangular.

Thus $I_d = (I_d + N)^r = I_d + \binom{r}{1}N^1+ \binom{r}{2}N^2+....+ \binom{r}{r-1}N^{r-1}+ \binom{r}{r}N^{r}$
$\implies \mathbf 0 = \binom{r}{1}N^1+ \binom{r}{2}N^2+....+ \binom{r}{r}N^{r}$
but nilpotent $N\neq \mathbf 0$ and its powers are linearly dependent, a contradiction.

user8675309
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