Related to symmetric, diagonal and invertible matrices

Question

While solving a problem I came across a specific question:

Given $A,B$ as $2$ real, symmetric, matrices with $B$ positive definite, does there exist a matrix (invertible) $P$ such that both $P^TAP$ and $P^TBP$ are diagonal matrices?

Answer 1

This is in the first edition of Horn and Johnson, Matrix Analysis paperback 1990. There are two conditions that are needed to be sure this can be accomplished. Given symmetric $A,B$ symmetric, first we require $A$ invertible. Second, defining $C = A^{-1} B,$ we require that $C$ be diagonalizable with some $R$ invertible and $R^{-1} C R = \Lambda$ diagonal. It is allowed to have $\Lambda$ complex, this may happen as $C$ need not be symmetric.

Example

$$ A = \left( \begin{array}{rr} 165 & -117 \\ -117 & 83 \end{array} \right) $$ and $$ B = \left( \begin{array}{rr} 1047 & -747 \\ -747 & 533 \end{array} \right) $$

Next $$ C = A^{-1} B = \left( \begin{array}{rr} -83 & 60 \\ -126 & 91 \end{array} \right) $$ has eigenvalues $1,7.$ As these are distinct, we can diagonalize.

$$ R = \left( \begin{array}{rr} 5 & 2 \\ 7 & 3 \end{array} \right) $$ has determinant $1,$ and $$ \Lambda = R^{-1} C R = \left( \begin{array}{rr} 1 & 0 \\ 0 & 7 \end{array} \right). $$

The construction arranges that $$ R^T B R = R^T A R \Lambda, $$ which finishes the problem when $\Lambda$ has a diagonal elements distinct. Indeed, $$ R^T AR = \left( \begin{array}{rr} 2 & 0 \\ 0 & 3 \end{array} \right) $$ and $$ R^T BR = \left( \begin{array}{rr} 2 & 0 \\ 0 & 21 \end{array} \right) $$

An example where $C$ has a repeat eigenvalue, but can be diagonalized, is at

Congruence and diagonalizations