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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?

Qwerty
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  • See simultaneous diagonalization under https://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 http://math.stackexchange.com/questions/1697846/simultaneous-diagonalization-of-two-bilinear-forms/1697985#1697985 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
  • looked up, it is enough that one of them, $B,$ be nonsingular. This follows from strict definiteness, as stated in the question. – Will Jagy Apr 17 '16 at 17:00
  • @user251257 natural. Many of the people asking about this are not aware of various distinctions, and ask for orthogonal diagonalization, either for one matrix where the eigenvalues cannot be found explicitly, or in pairs where that is impossible. I think I will try to make a jpeg with just the table from H+J (I) and the short discussion of what to do for this case. – Will Jagy Apr 17 '16 at 17:03
  • @WillJagy : So can you give a specific answer to my problem? Or refer me some article? And if good enough, please consider giving an upvote. – Qwerty Apr 17 '16 at 17:54

1 Answers1

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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

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