Quick question: matrix with norm equal to spectral radius

Question

For $A\in \mathcal{M}_n(\mathbb{C})$, define:

the spectral radius

$$ \rho(A)=\max\{|\lambda|:\lambda \mbox{ is an eigenvalue of } A\} $$

and the norm

$$ \|A\|=\max_{|x|=1}|A(x)| $$ where |.| is the Euclidean norm on $\mathbb{C}^n$.

Problem: Find all $A\in \mathcal{M}_n(\mathbb{C})$ such that $\rho(A)=\|A\|$.

Thank you very much!

Answer 1

A Note: Note that $||A||=\sigma_1$, the highest singular value of $A$. Thus, all hermitian matrices (and skew hermitian) readily satisfies your requirement. But I am not sure if other matrices can be added.

Answer 2

I don't think this gives the largest possible class of such matrices but it is still quite large:

Consider the Schur decomposition of $A$: $A=UTU^*$ where $U$ is unitary and $T$ is upper triangular with the eigenvalues of $A$ on the diagonal of $T$. Note that the eigenvalues of $A$ can be on the diagonal of $T$ in any order. If $T$ can be partitioned as $$ T=\begin{bmatrix}D & 0 \\ 0 & R\end{bmatrix}, $$ where $D$ is diagonal such that $\rho(D)=\|D\|\geq\|R\|$ and $R$ is upper triangular, then $\rho(A)=\|A\|$.

First, $\rho(A)=\max\{\rho(D),\rho(R)\}$, and since $\rho(R)\leq\|R\|$ (which holds for any matrix norm), we have $\rho(R)\leq\|R\|\leq\|D\|=\rho(D)$. Hence $\rho(A)=\rho(D)$. Second, the 2-norm is unitarily equivalent and hence $\|A\|=\|T\|=\max\{\|D\|,\|R\|\}=\|D\|$ since $\|R\|\leq\|D\|$. But $\|D\|=\rho(A)$ and hence $\|A\|=\rho(A)$.

I'm not sure this characterises all the matrices $A$ such that $\|A\|=\rho(A)$ but the class is still quite large since it contains all normal matrices (matrices with no block $R$ above) and yet something more which can be squeezed in the $R$-block, e.g., a block diagonal matrix with one block being normal and the other arbitrary with a small enough norm and eigenvalues.