Is the product of two symmetric positive definite matrices and some rotation matrix positive definite?

Question

I am aware that the product of two symmetric positive definite matrices, $\mathbf{A}$ and $\mathbf{B}$, may not be a symmetric positive matrix. However, if both $\mathbf{A}$ and $\mathbf{B}$ are 2x2, then my own derivations indicate that one can always find a 2x2 rotation matrix $\mathbf{R}(v)$ with rotation angle $v\in (-\pi/2,\pi/2)$, such that $\mathbf{ABR}(v)$ is a symmetric positive definite matrix. Is this correct? If so, I also wonder whether this result can be generalized to nxn matrices?

Sketch of proof: I prove this using brute force for the 2x2 case. Each symmetric positive definite matrix can be described by three parameters. I multiply the three matrices and note that the resulting matrix $\mathbf{ABR}(v)$ is symmetric for a specific rotation angle $v^{*}\in (-\pi/2,\pi/2)$. My derivations actually indicate that symmetry of $\mathbf{A}$ and $\mathbf{B}$ is sufficient to get the weaker result that $v^{*}\in [-\pi/2,\pi/2]$. In the next step, I prove that the resulting symmetric matrix $\mathbf{ABR}(v^{*})$ is positive definite if $\mathbf{A}$ and $\mathbf{B}$ are both positive definite.

Answer 1

Let $A$ be any $n\times n$ matrix. Then $A=U\Sigma V^{*},$ for some unitary matrices $U$ and $V,$ and a diagonal matrix $\Sigma$ such that $\Sigma_{i,i}\geq0$ for all $1\leq i\leq n.$ This means that $A(VU^{*})=U\Sigma U^{*}$ is positive semidefinite, and $VU^{*}$ is clearly unitary. In general, $VU^{*}$ will not be a rotation matrix, however. In particular, this clearly holds if $A$ is of the form $A'B$ for some positive definite matrices $A'$ and $B.$

In the case $n=2,$ unitary matrices are quite restricted, and are essentially all rotation matrices (possibly with some rows or columns multiplied by $e^{i\theta}$ for some values of $\theta$), which explains your calculations in that case.