MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Consider two square matrices $A, B \in \mathbb{R}^{n \times n}$ and let $\| \cdot\|_1$ and $\|\cdot\|$ be, respectively, the trace norm (the sum of singular values) and the usual operator norm (the maximum of singular values).

Is there a known bound for the following quantity? $$ \sup\{\alpha > 0: \; \alpha \, \text{tr}(A^TB) \le \| A+B\|_1, \, \forall B \; \text{s.t.} \;\|B\| \le 1\} $$

share|cite|improve this question
Cross-posted on MSE. – 1015 Jul 7 '13 at 20:29
Maybe I'm missing something. If $A = 0$, then $\alpha$ becomes unbounded....? – Suvrit Jul 9 '13 at 17:42
@suvrit: the MSE version is already edited. It is annoying with all these cross-posts... – András Bátkai Jul 9 '13 at 18:57
@András: ah, ok. You are right, too much cross-posting! – Suvrit Jul 9 '13 at 22:08
up vote 4 down vote accepted

Since I cant comment, I will leave this thought here. Since $||\cdot||_1$ and $||\cdot||$ (as you defined them) are dual norms, it must be that tr$((A+B)^TX)\leq||A+B||_1$ for any $X$ such that $||X||\leq 1$. Therefore, tr$(A^TB)\leq ||A+B||_1 - ||B||^2<||A+B||_1$ (since tr$(B^TB)\geq||B||^2$).

(edit: fixed typo and replaced $||B||$ with $||B||^2$ in the final step)

share|cite|improve this answer
It is rather $\|B\|^2=\|B^TB\|=\rho(B^TB)\leq \mbox{tr}(B^TB)$. And since $\|B\|\leq 1$... But you still have $\|A+B\|_1-\mbox{tr}(B^TB)\leq \|A+B\|_1$, though. – 1015 Jul 7 '13 at 20:46
@julien. You're right. I fixed the typo. Thanks! – Skoro Jul 7 '13 at 21:03
@Skoro, thanks. Your argument shows that the supremum is at least one. But can it be bigger? For example, are there matrices $A\neq 0$ for which that supremum is at least, say 3? I am interested in bounds relating that quantity to the spectrum of $A$. – passerby51 Jul 8 '13 at 18:53
Interesting. It looks like one can get an upper bound on $\alpha$ by substituting $B = A/||A||$. This results in the following bound: $\alpha \leq \frac{||A||_1(||A||+1)}{tr(A^TA)} = \frac{(\sum \sigma_i)(\max \sigma_i+1)}{\sum \sigma_i^2}$. maybe this could be controlled using the condition number of $A$. – Skoro Jul 8 '13 at 21:07

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.