# Add a multiple of $I$ to a matrix to minimize its operator norm

Given $A\in\mathbb{C}^{n\times n}$, what is $s_* = \arg\min \|A-sI\|$?

Here $\|A\|$ is the operator norm, $\|A\|=\rho(A^*A)^{1/2}$, and $I$ is the identity.

The corresponding problem for the Frobenius norm $\|A\|_F = \left(\sum A_{ij}^2\right)^{1/2}$ has a simple solution: reduce to the triangular case with a Schur factorization $A=QTQ^*$ (which does not alter $\|A\|_F=\|T\|_F$), then the problem becomes the one of minimizing $\sum_i |\lambda_i-s|^2$, with $\lambda_i=T_{ii}$ the eigenvalues of $A$, and then the solution is the arithmetic mean $s_{*,F}=\frac{1}{n}\sum \lambda_i$.

I do not see a simple solution for the case of the operator norm, though; I have tried applying the Schur factorization or the SVD, but the resulting problem is not immediate. Do you have any ideas? Or is this problem a known one?

• in the case of A normal, I guess one gets s = the center of the minimum disk containing spec(A), right? – Pietro Majer Feb 21 '16 at 9:31
• Conjugation by a unitary matrix doesn't affect the operator norm either, so you may assume $A$ is 70034-triangular. If $A$ is diagonal it's operator norm is the largest entry, so indeed in the normal case take $s$ to be the circumcentre of the spectrum. This may hold in general. – Lior Silberman Feb 21 '16 at 11:09
• @PietroMajer Yes, if $A$ is normal then that is the solution (via a diagonalization argument). This also shows that in general the Frobenius-norm minimizer is different from the operator-norm minimizer: just take a $3\times 3$ diagonal matrix whose diagonal entries form a triangle with centroid different from its circumcenter. – Federico Poloni Feb 21 '16 at 11:48