Let the eigen values of $A$ be distinct, $A=P\Lambda P^{-1}$ and $A^{T}=Q\Lambda Q^{-1}$,show that $Q^{T}P=I$

Question

Let the eigen values of $A$ be distinct, $A=P\Lambda P^{-1}$ and $A^{T}=Q\Lambda Q^{-1}$,show that $PQ^{T}=I$ where $\Lambda$ is a diagonal matrix.

We know that $A$ and $A^T$ have the same eigen values, but how to show this?

Answer 1

This isn't true. Suppose it is true. Let $D\ne I$ be an arbitrary nonsingular diagonal matrix and $P_1=PD$. Then $A=P_1\Lambda P_1^{-1}$, but $Q^TP_1=Q^TPD=D\ne I$.

The conditions $A=P\Lambda P^{-1}$ and $A^T=Q\Lambda Q^{-1}$ together imply that $Q^TP\Lambda=\Lambda Q^TP$. Since the only matrices that commute with a diagonal matrix with distinct diagonal entries are diagonal matrices, all you can say is that $Q^TP$ is a diagonal matrix. You cannot infer that $Q^TP=I$ without imposing further conditions.