Rotating a matrix to become symmetric

Question

Given an $n \times n$ matrix Q having eigenvalues in $(0,1],$ is it possible to find an $n \times n$ orthogonal matrix $U$ such that $$(QU)^T = QU$$ holds and the eigenvalues of $QU$ also fall in $(0,1]$? If so, is the orthogonal matrix unique?

Answer 1

What you found is the Polar Decomposition. Any square matrix $Q$ can be written as $Q=UM$, where $U$ is unitary (orthogonal if $Q$ is real) and $M$ is positive-definite and symmetric (actually, $M=(Q^*Q)^{1/2}$, i.e. $(Q^TQ)^{1/2}$ if $Q$ is real).

Answer 2

Apologies for not trying harder before asking the question, but I have a solution, at least to finding one such $U.$

Let $Q = S\Sigma T$ be the singular value decomposition of $Q$ and set $U = (ST)^T.$ Then

\begin{align*} QU &= S\Sigma T (ST)^T \\ &= S \Sigma T T^T S^T\\ &= S \Sigma S^T \\ \end{align*} which is symmetric.