Let $A$ be a real square and invertible matrix. I would like to find
$$ s(A) = \min_U \rho(U A), $$ Where $U$ is orthogoal, i.e. $U U^T = I$ and $\rho(A)$ is the spectral radius, i.e. the largest eigenvalue of $A$ in absolute values.
I am interested in a numerical solution to determine $s(A)$.
For an $n\times n$ matrix, the answer is $$|\det A|^{\frac1n}.$$ Explanation: on the one hand, $\rho(UA)\ge|\det (UA)|^{\frac1n}=|\det A|^{\frac1n}$. On the other hand, singular value decomposition gives $A=PDQ$ where $P,Q$ are orthogonal and $D>0$ is diagonal. Then $\rho(UA)=\rho(QUPD)$ and $$\min_U\rho(UA)=\min_V\rho(VD)$$ where $V$ runs over the orthogonal group. Take for $V$ the permutation matrix associated with the cycle $(12\ldots n)$. We have $$VD=\begin{pmatrix} 0 & \cdots & \cdots & 0 & s_1 \\ s_2 & \ddots & & & 0 \\ 0 & \ddots & \ddots & & \vdots \\ \vdots & \ddots & \ddots & \ddots & \vdots \\ 0 & \cdots & 0 & s_n & 0 \end{pmatrix},$$ whose spectrum is made of the $n$-roots of $s_1\cdots s_n=\det D=|\det A|$.
Remark that the answer is valid for an arbitrary matrix, singular or not. It is valid for complex matrices too, provided one takes the minimum over unitary matrices $U$.
