Prove that if symmetric $A$ and skew-symmetric $B$ are similar, then $A = B = O$

Question

Let $A$, $B$ and $Q$ be square matrices with real entries. Show that if $A$ is symmetric, $B$ is skew-symmetric, and $Q$ is invertible such that $Q^{-1} A Q=B$, then $A=B=0$.

Answer 1

From the condition $Q^{-1}AQ = B$ we have $A = QBQ^{-1}$ which can be viewed as normal Jordan form of matrix $A$, however the Jordan blocks are upper triangular matrices and the only possibility for them to be skew-symmetric is that matrix $A$ is diagonolizable and each Jordan block has size of $1\times 1$. Hence, $B$ is diagonal and from the skew-symmetry it follows that the diagonal elements are zeros as well. Hence, matrix $A$ is also zero-matrix.