Let $A$ be a real square matrix of size $n \times n$. Is there an upper bound on the minimum spectral norm under diagonal similarity, i.e.,

$$ s(A) = \min_{D} \lVert D^{-1} A D\rVert_2, $$ where $D$ is a non-singular, diagonal real matrix. Also, is there are a relation between $s(A)$ and the spectral radius $\rho(A)$?

For the numerical radius $r(A) = \max_{\lVert x \rVert_2 = 1} \lVert \langle Ax, x\rangle \rVert$, it is

$$\rho(A) \leq r(A) \leq \lVert A\rVert_2 \leq 2 r(A).$$

I was hoping that it may be possible to minimize $r(D^{-1} A D)$.

  • $\begingroup$ Do you have any reason to expect a simpler expression for $s(A)$ than the definition? $\endgroup$ – Mark Meckes Jul 23 '18 at 13:56
  • $\begingroup$ @MarkMeckes There are few: 1) special case of a triangular matrix $T$, where $s(T) = \rho(T)$. 2) there should be at least bounds related to the numerical radius. $\endgroup$ – Sebastian Schlecht Jul 23 '18 at 13:59
  • $\begingroup$ 1) It's hard to get anywhere from the triangular case for non-unitarily invariant quantities. 2) Okay, bounds should be findable. Phrasing the question as "What is ...?" makes it sound like you want an exact expression. $\endgroup$ – Mark Meckes Jul 23 '18 at 14:02
  • $\begingroup$ @MarkMeckes Sorry, for the implication. I'm happy with a bound. $\endgroup$ – Sebastian Schlecht Jul 23 '18 at 14:04
  • $\begingroup$ Denis's answer gives you $\rho(A) \le s(A) \le \| A \|_2$. Is that good enough for your purposes? $\endgroup$ – Mark Meckes Jul 23 '18 at 14:08

You cannot get an upper bound in general in terms of the spectral radius $\rho(A)$. Counterexample: if $$ A = \begin{bmatrix} x & 1 \\ -x^2 & -x \end{bmatrix} $$ then $\rho(A) = 0$ and $s(A) = 2|x|$. (This $A$ is essentially the most general $2\times 2$ matrix whose eigenvalues are both $0$.)

  • $\begingroup$ That is very insightful! Ok, so the upper bound is futile. Any good idea on how to compute $s(A)$ numerically? But, this is then a new question, I guess. $\endgroup$ – Sebastian Schlecht Jul 23 '18 at 15:23
  • $\begingroup$ @SebastianSchlecht: That would be both a new question, and one that is outside my expertise. $\endgroup$ – Mark Meckes Jul 23 '18 at 16:47

The norm $A\mapsto s(A)$ has the flaw of not being unitarily invariant. However, if you think that every $P\in{\bf GL}_n({\mathbb R})$ can be factorized out $P=UDV$ where $U$ and $V$ are orthogonal and $D$ is diagonal, you obtain $$\inf\{s(U^TAU)\,;\,U\in {\bf O}_n\}=\inf\{\|P^{-1}AP\|_2\,;\,P\in{\bf GL}_n({\mathbb R})\}=\rho(A).$$ The second inequality is a rather well-known fact. It can be used in the proof of Householder Theorem. See my book on matrices (Springer-Verlag GTM216).

  • $\begingroup$ Thank you very much for this. As discussed with Mark Meckes above, I need rather an upper bound for $s(A)$. I have adjusted the question accordingly. $\endgroup$ – Sebastian Schlecht Jul 23 '18 at 14:26

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