# Are $\log(\sigma(A(z))$ subharmonic functions?

Let $$A$$ be a matrix-valued entire function. It is then well-known that $$\log \Vert A(z)\Vert$$ is subharmonic. In particular, the operator norm is just the largest singular value of $$A$$.

Is it therefore also true that for any singular value $$\sigma$$ of $$A$$, in a domain where they are simple, turn $$\log \sigma(A(z))$$ into a subharmonic function?

Let $$F$$ be an analytic function from an open subset $$\Omega$$ of the complex plane into the algebra of $$n\times n$$ matrices. Denoting by $$s_1,\dots,s_n$$ the decreasing sequence of singular values of a matrix, we prove that the functions $$\log s_1(F(\lambda))+\dots + \log s_k(F(\lambda))$$ and $$\log^+ s_1(F(\lambda)) + \dots+\log^+s_k(F(\lambda))$$ are subharmonic on $$\Omega$$ for $$1\leq k\leq n$$.
In the same Studia article, an example is given just before Theorem 1 which shows that the answer to your stated question is negative. The example is quite easy to describe: take the entire matrix-valued function $$F(z) = \begin{pmatrix} 1 & 1 \\ 0 & z \end{pmatrix}$$ Then the smallest singular value satisfies $$s_2(z)=\min(1,|z|)$$, which is not subharmonic. Consequently $$\log s_2(z)$$ also cannot be subharmonic.