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 $s(A)$.



Minimize spectral radius with orthogonal matrix
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 $U$.


