# Quick question: matrix with norm equal to spectral radius

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

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

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!

This is actually 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 when $n\ge3$, $A$ 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 $\rho(A)^2I-L^\ast L$ is positive semidefinite.

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.

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 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.

• I think the answer should be all the diagonalizable matrices with eigenvalues having equal magnitudes. We can reduce to $n=2$ to prove. – lnth Jan 25 '14 at 8:28
• @lnth counterexample: $$A = \pmatrix{1 & 1\\0 & -1}$$ $A$ has eigenvalues $\pm 1$, but $\|A\| = \sqrt{\frac{3 + \sqrt{5}}{2}} \approx 1.6$ – Omnomnomnom Jan 25 '14 at 16:49

Let $A$ have the Schur decomposition $A = UTU^*$, where $T$ is and upper-triangular matrix whose diagonal entries (eigenvalues) are in order from greatest absolute value to least. We note that $$A^*A = (UTU^*)^*(UTU^*) = UT^*TU^*$$ It follows that $A$ will be a matrix with $\|A\|>\rho(A)$ (one of the matrices we don't want) iff there is a vector $y$ for which $$y^*(T^*T)y >\rho(A)^2 y^*y$$ I'm not yet sure where to proceed, but this may be a helpful first step.