Real eigenvalues of $B = P^T A P$ with orthogonal and tall $P$

Question

I have a real square non-symmetric matrix $A\in\mathbb{R}^{n\times n}$ which has real eigenvalues. Another matrix $B\in\mathbb{R}^{m\times m}$ is defined as $B = P^T A P$ where $P\in\mathbb{R}^{n\times m}$ is a tall and real orthogonal matrix, i.e. $P^TP=I$.

From what I test on a simple script, it seems that if the eigenvalues of $A$ are real, then the eigenvalues of $B$ are also real. Does this apply generally? If yes, is there any way to proof it?

Answer 1

This is not true. E.g. $A=\pmatrix{0&-1&1\\ 1&0&2\\ 1&1&1}$ has a real spectrum $\{-1,0,2\}$ but $$ B=\pmatrix{1&0&0\\ 0&1&0} \pmatrix{0&-1&1\\ 1&0&2\\ 1&1&1}\pmatrix{1&0\\ 0&1\\ 0&0} =\pmatrix{0&-1\\ 1&0} $$ has only non-real eigenvalues.