# "Almost Normal" Matrix and Gap between Spectral Radius/Norm

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

$$A=\pmatrix{\epsilon&0\\\frac{1}{\epsilon}&2\epsilon}$$

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?

• @RodrigodeAzevedo I agree with your edits Commented Dec 29, 2019 at 15:12
• 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 Commented Dec 30, 2019 at 21:35
• @SandeepSilwal indeed that is obvious. the whole point here is about the converse (under some relaxed condition) Commented Dec 31, 2019 at 13:26