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I am trying to show that

$$\frac{1}{\sqrt{n}}\|A\|_2\leq \|A\|_1\leq \sqrt{n}\|A\|_2.$$ I start with the definition $$ \|A\|_2=\max_{v\neq 0}\frac{\|Av\|_2}{\|v\|_2} .$$ I know that $\forall v\in\mathbb{R}^n$, $\|v\|_2\leq\|v\|_1$. But I do not know how to relate this property to the question at hand. For instance, is $\|Av\|_2\leq\|Av\|_1$?

M B
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  • See also https://math.stackexchange.com/questions/901233/proving-result-on-matrix-norms – Arnaud D. Mar 09 '20 at 07:40
  • @ArnaudD. I am still a bit confused. I don't know how $$\max_{v\neq 0}\frac{|Av|2}{|v|_2}\leq \max{v\neq 0}\frac{\sqrt{n}|Av|_1}{|v|_1}\implies|A|_2\leq\sqrt{n}|A|_1,$$ which establishes the first part of the inquality $\frac{1}{\sqrt{n}}|A|_2\leq|A|_1$. – M B Mar 09 '20 at 07:54

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