Let's denote

$$\Vert{A}\Vert := \max_{x\neq0}\frac{x^* Ax}{x^*x}$$

and let $\rho(A)$ denote the largest absolute value of the eigenvalues of matrix $A$. From basic linear algebra, one could characterize normal matrices as those unitarily diagonalizable ones, namely, for $A^*A=AA^*$, there exist $Q \in \mathsf{SU}(n)$ and $\Lambda \in \mathsf{diag}(n)$ such that

$$ A = Q\Lambda Q^*. $$

Therefore, the spectral norm is exactly the spectral radius, $\rho(A)=\Vert A\Vert$. On the other hand, when $A$ is not normal, even if it is diagonalizable, it is easy to construct some matrices with small spectral radius but large spectral norm, e.g., for


we clearly have $\rho(A)=2|\epsilon|\rightarrow 0$ but $\Vert A\Vert\geq\frac{1}{|\epsilon|}\rightarrow +\infty$ as $\epsilon\rightarrow 0$.

It seems natural that, if we quantify the obstruction from a not-normal matrix to normal, we might be able bound the gap between spectral radius/norm. So here is my question:

Suppose for $A\in\mathbb{C}^{n\times n}$ there is some $Q\in\mathsf{SU}(n)$ that $\Vert A-QA^*\Vert\leq\epsilon$ is small, could we have some upper bound on either the multiplicative gap $\Vert A\Vert/\rho(A)$ or additive gap $\Vert A\Vert-\rho(A)$ between its spectral norm and radius?

  • $\begingroup$ @RodrigodeAzevedo I agree with your edits $\endgroup$ Commented Dec 29, 2019 at 15:12
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    $\begingroup$ basically you want to understand the relation between the largest singular value and the largest eigenvalue. one direction is clear: the largest singular value is larger than the largest eigenvalue, it is not so clear what is konwn about the other direciton $\endgroup$ Commented Dec 30, 2019 at 21:35
  • $\begingroup$ @SandeepSilwal indeed that is obvious. the whole point here is about the converse (under some relaxed condition) $\endgroup$ Commented Dec 31, 2019 at 13:26


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