# Related to symmetric, diagonal and invertible matrices

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?

• See simultaneous diagonalization under en.wikipedia.org/wiki/Diagonalizable_matrix – user251257 Apr 17 '16 at 16:47
• @user251257 :Thanks.. – Qwerty Apr 17 '16 at 16:49
• @user251257 no, I'm afraid that selection discusses $P^{-1} A P.$ With quadratic forms, or symmetric matrices, the correct expression is in the question, $P^T A P.$ And, as is in Horn and Johnson, this can be done once one of the symmetric matrices is guaranteed positive definite. – Will Jagy Apr 17 '16 at 16:54
• see math.stackexchange.com/questions/1697846/… and the first edition of Horn and Johnson, table 4.5.15T on page 229, then detail for case II on pages 231-232. – Will Jagy Apr 17 '16 at 16:56
• Oh I oversaw that $P$ need not be orthogonal/unitary. Sorry – user251257 Apr 17 '16 at 17:00

## 1 Answer

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

• Beautiful reference and answer. – Qwerty Apr 17 '16 at 20:12