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

  • $\begingroup$ See also "Hölder inequality". $\endgroup$ – mvw Jun 1 '15 at 15:46
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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 $||A||_2=\sqrt{r(A^TA)}\leq\sqrt{||A^TA||_\infty}\leq\sqrt{||A^T||_\infty||A||_\infty}=\sqrt{||A||_1||A||_\infty}$

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    $\begingroup$ 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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