# Is the product of two invertible symmetric matrices always diagonalizable?

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?

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.
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.