If $A$ is a real diagonalizable matrix and $B$ is real symmetric and positive definite, is the matrix $B^{-1}A$ real diagonalizable?

Question

This problem is from a generalized eigenvalue problem. I can only proof it when $A$ is symmetric:

Let $G$ be the square root of $B$ and $G = G^T$, then $B^{-1}A = G^{-1}(G^{-1}AG^{-T})G$ is similar to $G^{-1}AG^{-T}$, which is again symmetric and thus diagonalizable.

However, this method fails when $A$ is just real diagonalizable. Are there any tricks to modify the proof, or is there a counterexample?

Answer 1

The statement is not true. Take \begin{align*} A = \begin{pmatrix} 0 & 1 \\ 0 & -1/2 \end{pmatrix}, \quad B = \begin{pmatrix} 1 & -1/2 \\ -1/2 & 1 \end{pmatrix} > 0, \end{align*} then \begin{align*} B^{-1}A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \end{align*} is not diagonalizable. But $A$ is clearly diagonalizable because it has two distinct eigenvalues.

Answer 2

Hint: for a counterexample, start with a matrix $C$ that is not diagonalizable, and see if you can find a symmetric positive definite matrix $B$ so that $A = BC$ is diagonalizable. Almost any one will do.