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I have two symmetric matrices $A$ and $B$, which are both invertible. Their eigenvalues are obviously real, but not necessarily positive. I know that if one matrix were positive definite, we could use (Product of two symmetric matrices is similar to a symmetric matrix) to show that $AB$ is similar to a symmetric matrix (with real eigenvalues). In my cases, the eigenvalues of $AB$ are generally complex, but can one adjust the proof to show that $AB$ is always diagonalizable?

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2 Answers 2

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No. Here is a counterexample that works not only over $\mathbb R$ but also over any field: $$ \pmatrix{1&1\\ 0&1}=\pmatrix{1&1\\ 1&0}\pmatrix{0&1\\ 1&0}. $$ In fact, it is known that every square matrix in a field $F$ is the product of two symmetric matrices over $F$. See Olga Taussky, The Role of Symmetric Matrices in the Study of General Matrices, Linear Algebra and Its Applications, 5:147-154, 1972 and also Positive-Definite Matrices and Their Role in the Study of the Characteristic Roots of General Matrices, Advances in Mathematics, 2(2):175-186, 1968.

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  • $\begingroup$ I'm glad to see that the OP decided at last to mark your answer as the accepted one, since it is mucher better than mine. $\endgroup$ Mar 17, 2022 at 11:44
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In general, no. Take$$A=\begin{bmatrix}1 & 1 \\ 1 & \frac{15}{64}\end{bmatrix}\quad\text{and}\quad B=\begin{bmatrix}1&-4\\-4&0\end{bmatrix}.$$Then both $A$ and $B$ are symmetric and invertible. But$$AB=\begin{bmatrix}-3&-4\\\frac1{16}&-4\end{bmatrix},$$which is not diagonalizable: its only eigenvalue is $-\frac72$, and the corresponding eigenspace is $1$-dimensional.

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