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Let $A$ be a square Matrix and $||\cdot ||_p$ the induced Matrix norm for $1 \leq p \leq \infty$. Is it true that $$||A||_p\leq \max(||A||_1,||A||_{\infty})?$$ For $p=2$ the answer is yes because $||A||_2^2\leq ||A||_1||A||_{\infty}$.

The motivation is that I have a family of matrices where I can bound the infinity and 1-Norm but I dont know how to bound the $p$-Norm.

Thx for any help in advance

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2 Answers 2

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Yes, by the Riesz-Thorin theorem.

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For the Shatten $p$-norm (i.e. $\|A\|_p=tr_n(|A|^p)^{1/p}$) it is true. This follows from the positivity of the trace: $tr_n(|A|^p)^{1/p}=tr_n(\sqrt{A^*A}^{p})^{1/p}=tr_n((A^*A)^{\frac{p-1}{2}}(A^*A)^{\frac{1}{2}})^{1/p}\leq (\|A\|_{\infty}^{p-1} tr_n(|A|))^{1/p}\leq{\max(\|A\|_1,\|A\|_{\infty})}$

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  • $\begingroup$ It is not clear to me what the OP means by $\|\cdot\|_p$. $\endgroup$ Jun 25, 2013 at 19:03
  • $\begingroup$ "Induced matrix norm" normally means the $\ell_p^n \to \ell_p^n$ operator norm. $\endgroup$ Jun 25, 2013 at 19:34
  • $\begingroup$ yes I meant the operator norm $\endgroup$
    – user35593
    Jun 25, 2013 at 19:53

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