Let $A\in \mathbb R^{n\times n}$. Is it true that
$A$ similar to a symmetric matrix $\implies $ $A$ symmetric ?
Let $B$ symmetric s.t. $A=PBP^{-1}$. Then $$A^T=(P^{-1})^TB^T P^T=(P^{-1})^T BP^T.$$
For me there is no reason that $P$ is orthogonal, so I would say it's false a priori. But in the same time, this theorem should be true since operator is self adjoint $\iff$ it's diagonalizable. I also know that matrices in any basis of Self Adjoint operator are symmetric. But if A is similar to a symmetric matrix, then it's diagonalizable and thus self adjoint, and thus, it should be symmetric in any basis… this is wrong ? If yes, why ?
Best Answer
Take any $A$ which is diagonizable but not symmetric. So $A = TDT^{-1}$ for a diagonal matrix $D$ and invertible $T$. For any orthogonal matrix $O$ $A$ is similar to $ODO^\intercal$, which is symmetric, but $A$ is not.