Multiply by orthogonal matrices and obtain diagonalizability

Question

It is possible to show that if a $ n \times n $ real matrix $ A $ verifies that $ PA $ is diagonalizable for all real invertible matrix $ P $ , then $ A=0 $ , because all the matrices that are equivalent to $ A $ are diagonalizable. I tried it with keeping only orthogonal matrices in the initial hypothesis (instead of simply invertible), and using SVD, it still works : the result is the same. Now, I want to check if it is still true keeping only special orthogonal matrices. When $ n $ is odd, it works because $ \mathcal{O}(n) = \mathcal{SO}(n) \cup (-\mathcal{SO}(n)) $ , but I have no solution for $ n $ even.

Answer 1

Consider $n=2$; then $\mathcal{SO}(2)$ is the set of rotations and $\mathcal{O}(2)\setminus\mathcal{SO}(2)$ is the set of reflections. Choose $A$ to be any reflection (eg $\mathrm{diag}(1,-1)$). If $P$ is a rotation then $PA$ is a reflection, which is diagonalizable over $\mathbb R$. Thus the claim fails for $n=2$.

The claim does hold for $n>2$. Indeed write $A=UDV^{-1}$ where $U,V\in\mathcal{SO}(n)$ and $D$ is diagonal. For any $O\in\mathcal{SO}(n)$, we have $VOU^{-1}\in\mathcal{SO}(n)$, so $VOU^{-1}A$ is diagonalizable by assumption. Hence $$ V^{-1}(VOU^{-1}A)V=OD $$ is diagonalizable.

Write $D=\mathrm{diag}(\lambda_1,\ldots,\lambda_n)$. First suppose the rank of $D$ is at least $2$. wlog suppose $\lambda_1,\lambda_2\neq0$. Let $$ O'=\begin{bmatrix} 0&\mathrm{sgn}(\lambda_2)&0\\ -\mathrm{sgn}(\lambda_1)&0&0\\ 0&0&\mathrm{sgn}(\lambda_2)\mathrm{sgn}(\lambda_1) \end{bmatrix}\in\mathcal{SO}(3), $$ $$ O=\begin{bmatrix}O'&0\\0&I_{n-3}\end{bmatrix}\in\mathcal{SO}(n). $$ Then $$ OD=\begin{bmatrix}X&0\\0&*\end{bmatrix} $$ where $$ X=\begin{bmatrix}0&|\lambda_2|\\-|\lambda_1|&0\end{bmatrix}. $$ Now $X$ is not diagonalizable over $\mathbb R$, so neither is $OD$, a contradiction.

Next suppose $D$ has rank $1$. wlog suppose $\lambda_1\neq0$ and $\lambda_i=0$ for $i>1$. Let $$ O'=\begin{bmatrix}0&1\\-1&0\end{bmatrix}\in\mathcal{SO}(2), $$ $$ O=\begin{bmatrix}O'&0\\0&I_{n-2}\end{bmatrix}\in\mathcal{SO}(n). $$ Now $OD$ is nilpotent and nonzero, again contradicting $OD$ being diagonalizable. Hence $D=0$, so $A=0$ as required.