If $A$ is simple, does $\|A\|_{\rm op}=\max_{\lambda\in\sigma(A)}|\lambda|$ imply that $A$ is normal?

Question

We know that, in general, the operator norm of a matrix is larger than its spectral radius: $\|A\|_{\rm op}\ge r(A)\equiv \max_{\lambda\in\sigma(A)}\lvert \lambda\rvert$, where $\sigma(A)$ denotes the spectrum of $A$.

If $A$ is normal, these quantities are clearly the same, as $\|A\|_{\rm op}$ equals the largest singular value of $A$, which equals the largest $|\lambda|$ for $\lambda$ an eigenvalue of $A$.

I can also find examples of $\|A\|_{\rm op}=r(A)$ for $A$ not normal, by building it as direct sum of a non-normal matrix and a matrix with a large eigenvalue. E.g. $$\begin{pmatrix}2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 1\end{pmatrix}.$$ This is not diagonalisable, but satisfies the equality.

Are there such examples with simple matrices? In other words, given a simple $A$ (i.e. a matrix composed of a single Jordan block), can we have $\|A\|_{\rm op}=r(A)$?

Answer 1

Yes, but only when $A$ is $1\times1$. Note that every complex square matrix is unitarily triangulable. Suppose $B=U^\ast AU$ is such a triangularisation. Since all eigenvalues of $B$ are the same, we have $\rho(B)=|b_{jj}|\le\|\mathbf b_j\|_2\le\|B\|_2=\rho(B)$ for every $j$, where $\mathbf b_j$ denotes the $j$-column of $B$. It follows that $|b_{jj}|=\|\mathbf b_j\|_2$ for every $j$, i.e. $B$ is a diagonal matrix. As $B$ is similar to $A$ and the Jordan form of $A$ has only one Jordan block, $B$ and $A$ must be $1\times1$.

More generally, $\|A\|=\rho(A)$ for some submultiplicative matrix norm $\|\cdot\|$ if and only if every eigenvalue of $A$ with maximum modulus is semi-simple. For a proof, please see here.

Answer 2

Yes, $A$ must be normal. Otherwise, given that $A$ consists of a single Jordan block, that block is necessarily nontrivial (i.e. it is of size at least 2) so there are two nonzero vectors $u$ and $v$ such that $Au=\lambda u+v$ and $Av=\lambda v$. Therefore, for every scalar $b$ one has that $$ \|A(u+bv)\|^2 = \|\lambda (u+bv) + v\|^2 = $$$$ = |\lambda |^2\|u+bv\|^2 + \|v\|^2 + 2 \Re\lambda \langle u+bv, v\rangle = $$$$ = |\lambda |^2\|u+bv\|^2 + \|v\|^2 + 2 \Re\lambda \langle u, v\rangle + 2 \Re \lambda b \langle v, v\rangle . $$

Assuming that $λ≠0$ we may choose $b$ such that $$ \|v\|^2 + 2 \Re\lambda \langle u, v\rangle + 2 \Re\lambda b\langle v, v\rangle >0, $$ so we see that $$ \|A(u+bv)\| > |\lambda |\|u+bv\|, $$ whence $\|A\|>|\lambda | = r(A)$, contradicting the hypothesis.

On the other hand, if $λ=0$ we also get a contradiction because $\|A\|>0 =r(A)$.