# 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.

• This may not be essential, but just to be clear: does "$i\in\Lambda, j\in\Theta$" mean "$(i,j)\in\Lambda\times\Theta$" or "$(i,j)\in(\Lambda\times\{1,2,\ldots,n\})\cup(\{1,2,\ldots,n\}\times\Theta)$"? Commented Nov 21, 2018 at 10:13
• @user1551: Yes. I have a problem in formulating formally. Ideally, I want to guarantee the diagonal entries are $0$. In addition, the other entries can be zeros at will. In writing down the question, I was thinking that $\Lambda \cap \Theta$ might not be empty so I chose the way in the question. Thanks. Commented Nov 21, 2018 at 16:50

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 $$\rho(A-tI) and hence the answer to your question is no in this case.
E.g. suppose $$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.
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.