I'm looking for a criterion on symmetric, positive definite matrices $A$ which ensures the constant $M$ in the lower bound of the norm equivalence $M\|A\|_\infty \le \|A\|_2 \le \|A\|_\infty$ does not depend on the dimension of the matrix.

EDIT: I was thinking that it sufficient to constrain the eigenvalues of $A$, but so far I can't make this work.

So far, I have only a quite unsatisfying attempt based on matrix sensitivity: supposing $A = Id + \Delta$, the matrix $\Delta$ should have $\|\Delta\|_2 \le \frac M{\sqrt n}$. One can generalize a bit by taking any diagonal (positive definite) matrix (EDIT: with bounded eigenvalues) in the place of the identity, but I'd be interested in a more general criterion.

UPDATE: OK, something simple I missed: if an eigenvector of $A$ is $e = (1,...,1)$, then $$ \|A\|_\infty = \max_{\|v\|_\infty =1}\|Av\|_\infty = \|Ae\|_\infty = \lambda_{e}(A) \le \lambda_{max}(A) = \|A\|_2. $$ However, I need more generality than this.

UPDATE 2: Just an idea, not sure that it helps: what one needs is a constant $M$ independent of $n$ such that $$ \|A\|_\infty = \max_{i} \sum_{j=1}^n |a_{ij}| \le \frac {\|A\|_2}{M} $$ for all matrices in the class.

  • $\begingroup$ From the first update, I think the right direction is to say that the vector $e$ is sufficiently close to some eigenvalue of $M$, and that the conditioning number of $M$ should be reasonably to one. $\endgroup$
    – Plamen
    Apr 3, 2015 at 15:42
  • $\begingroup$ Do you have a reason (e.g., some special assumptions) why to think that the lower bound could be independent of $n$? This is certainly not true, e.g., for $A=\mathrm{tridiag}(-1,2,-1)$. It can be true, e.g., if $A$ is strictly diagonally dominant provided that the "strictness" of the dominance is $n$-independent. $\endgroup$ Apr 3, 2015 at 20:05
  • $\begingroup$ Actually, it's the special assumption I'm looking for :-) I'm thinking about the asymptotic $n\to\infty$. I think strict diagonality is a subclass of update 2: If the entries of each row are elements of an absolutely converging series, and every series was bounded by $M$, then it works. $\endgroup$
    – Plamen
    Apr 3, 2015 at 21:05
  • $\begingroup$ This problem seem somewhat related to the Grothendieck constant. $\endgroup$
    – Surb
    Nov 27, 2019 at 11:32


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