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?
Is the product of two invertible symmetric matrices always diagonalizable
diagonalizationlinear algebra
Best Answer
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.