Quick question: matrix with norm equal to spectral radius 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!
 A: 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.
A: This is a solved problem. A complex square matrix $A$ is called radial if its spectral radius is equal to its spectral norm $\|A\|_2=\max_{\|v\|_2=1}\|Av\|_2$. A complete characterisation of radial matrices was found in

M. Goldberg and G. Zwas (1974), On matrices having equal spectral radius and spectral norm, Linear Algebra and Its Application, 8: 427-434.

The main result in this paper is that $A$ is radial if and only if $A$ is unitarily similar to a matrix of the form
$$
\pmatrix{D&0\\ 0&L}
$$
for some diagonal matrix $D$ and some lower triangular matrix $L$ such that $\|L\|_2\le\rho(A)$.
A: The answers so far have been excellent. Perhaps it is interesting to add singular values to the discussion.
Every square matrix $A \in \mathbb C^{n \times n}$ can be written as $A = U \Sigma V$, where $U, V \in \mathbb C^{n \times n}$ are unitary, and where $\Sigma \in \mathbb C^{n \times n}$ is a diagonal matrix with real entries. Without loss of generality, we let those be $\sigma_n(A), \dots, \sigma_1(A)$ in descending order.
The spectral norm of $A$ is precisely the largest singular value of $A$. So $\| A \|_2 = \sigma_n(A)$. 
This means, of course, that $\rho(A) \leq \sigma_n(A)$.
This is an equality for an important example of matrices, namely the symmetric matrices. In that case, the singular value decomposition is just the spectral decomposition. 
The (non-zero) nilpotent matrices are perhaps the worst example: their spectral radius is zero, whereas their largest singular value can be very large.
A: 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. 
