Let $A \in {M_n}$ and $\left\| {\left| . \right|} \right\|$ be a matrix norm on ${M_n}$.Why does ${\left\| {\left| A \right|} \right\|_2} \le \left\| {\left| A \right|} \right\|_1^{\frac{1}{2}}\left\| {\left| A \right|} \right\|_\infty ^{\frac{1}{2}}$?


marked as duplicate by Algebraic Pavel, TravisJ, user147263, Daniel Robert-Nicoud, Jonas Meyer Jun 1 '15 at 23:59

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  • $\begingroup$ See also "Hölder inequality". $\endgroup$ – mvw Jun 1 '15 at 15:46
  • $\begingroup$ Dear @Kavir I see that, although you have already asked 5 question in this site and received answers in most of them, you have not mark a best answer in any of them. You can do it so by clicking on the checkmark next to the answer that you think is the one that helped you the most. Please read here for more detail. $\endgroup$ – Leo Sera Jul 2 '15 at 21:31

Let $r(A)=max\{|\lambda|:\lambda\ is\ eigenvalue\ of\ A\}$. Then $||A||_2=\sqrt{r(A^TA)}$

Now, for any matrix norm $||\cdot||$ we have $r(A)\leq||A||$.

Thus $||A||_2=\sqrt{r(A^TA)}\leq\sqrt{||A^TA||_\infty}\leq\sqrt{||A^T||_\infty||A||_\infty}=\sqrt{||A||_1||A||_\infty}$

  • 1
    $\begingroup$ You can bound $r(A^TA)$ directly by $\|A^TA\|_\infty$. For any matrix norm $\|\cdot\|$, which is consistent with some vector norm, $r(X)\leq\|X\|$. $\endgroup$ – Algebraic Pavel Jun 1 '15 at 16:51
  • $\begingroup$ Yes, thank you! $\endgroup$ – Maximilian M. Jun 1 '15 at 22:23

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