Ytukyg

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.

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 $lambda=argmax_{|lambda_i(A)|=rho(A)}|Re(lambda_i(A))|$ and $-lambda$ is not an eigenvalue of $A$, then provided that $a$ is sufficiently close to $rho(A)$, there always exists some scalar matrix $tIin U^perp$ 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 $Ainmathcal U$ 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 $ain(2, rho(A))$. Then for any $tin(rho(A)-a, rho(A)-2)$, we have $tIinmathcal U^perp$ but $rho(A-tI)=rho(A)-t<a$.

So, the only interesting case is when both $lambda=argmax_{|lambda_i(A)|=rho(A)}|Re(lambda_i(A))|$ and $-lambda$ are eigenvalues of $A$, for which I haven't any answer yet.