Does a matrix of minimum norm in an affine subspace of $M_n(\mathbb R)$ have minimum spectral radius? Let $\mathcal U \in M_n(\mathbb R)$ be a subspace defined by declaring certain entries to be $0$. More precisely, let $\Lambda, \Theta \subset \{1, \dots, n\}$, $\mathcal U$ is defined as
\begin{align*}
\mathcal U = \{ A \in M_n(\mathbb R): a_{ij} = 0 \text{ for } (i, j) \in \Lambda \times \Theta \text{ when } i \neq j \text{ and } a_{ii} = 0 \text{ for } i = \{1, \dots, n\} \}.
\end{align*}
I hope this is clear. Essentially $\mathcal U$ is a subspace with certain zero patterns on its entries and in particular, the diagonal entries are $0$.
Now let $A \in \mathcal U$ and we consider the affine subspace $\mathcal S := A + \mathcal U^{\perp}$ where $\mathcal U^{\perp}$ would be the matrix with zero pattern opposite to $\mathcal U$. It is clear with respect to the inner product defined by $\langle M, N \rangle = \text{tr}(M^TN)$, $A$ is of minimum norm in $\mathcal S$.
My question: suppose $\rho(A) \ge a$ where $a$ is some scalar in $\mathbb R$ and $\rho$ denotes spectral radius, is it possible for any $B \in \mathcal S$, we have $\rho(B) \ge a$. In general, I know matrix norm is an upper bound of spectral radius and we should not expect such inequality. But I failed to construct a counterexample or prove it.
 A: If $A$ is nilpotent, $a$ has to be non-positive. Hence $\rho(B)$ is always $\ge a$ and the answer to your question is yes in this case.
If $\lambda=\arg\max_{|\lambda_i(A)|=\rho(A)}|\Re(\lambda_i(A))|$ and $-\lambda$ is not an eigenvalue of $A$, then provided that $a$ is sufficiently close to $\rho(A)$, there always exists some scalar matrix $tI\in U^\perp$ such that $\rho(A-tI)<a$ and hence the answer to your question is no in this case.
E.g. suppose $n=3,\ \Lambda=\{1\}$ and $\Theta=\{3\}$. Then $A\in\mathcal U$ if and only if $a_{13}=0$ and $A$ has a zero diagonal. Let
$$
A=\pmatrix{0&1&0\\ 2&0&2\\ 2&2&0}.
$$
Its spectrum is $\{1+\sqrt{3},\ -2,\ 1-\sqrt{3}\}$ and $\rho(A)=1+\sqrt{3}\approx2.732$. Let $a\in(2,\ \rho(A))$. Then for any $t\in(\rho(A)-a,\ \rho(A)-2)$, we have $tI\in\mathcal U^\perp$ but $\rho(A-tI)=\rho(A)-t<a$.
So, the only interesting case is when both $\lambda=\arg\max_{|\lambda_i(A)|=\rho(A)}|\Re(\lambda_i(A))|$ and $-\lambda$ are eigenvalues of $A$, for which I haven't any answer yet.
