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The $p$-norm of a matrix $A$ for $p \geq 1$ is defined as $$\|A\|_p = \max_{ x \in \mathbb{R}^n, \|x\|_p=1} \|Ax\|_p.$$ My question: does this equal $$ \max_{ x \in \mathbb{C}^n, \|x\|_p=1} \|Ax\|_p$$ The only difference is the replacement of $\mathbb R^n$ by $\mathbb C^n$. The answer is certainly yes for $p=1,2,\infty$ and the Wikipedia article on matrix norms appears to imply the answer is yes for any $p$.

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Note that this is not true for the norms $\|A\|_{p,q}$. For example, if $A = \pmatrix{1 & -1\cr 1 & 1\cr}$, $$\max_{x \in {\mathbb R}^2: \|x\|_\infty = 1} \|A x\|_1 = 2$$ but $x = \pmatrix{1\cr i\cr}$ has $\|x\|_\infty = 1$ and $\|Ax\|_1 = 2 \sqrt{2}$.

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A really nice example. –  robinson Feb 1 '12 at 7:04
    
+1: Nice. ${}{}{}{}$ –  copper.hat Apr 22 '13 at 15:21
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If the entries of $A$ are not all real, the answer may be no. For example, consider the $2 \times 1$ matrix $A = (1 \ i)$ with $p=2$. $$\max_{x \in {\mathbb C}^2, \|x\|_2 = 1} \|Ax\|_2 = \sqrt{2}$$ but $$\max_{x \in {\mathbb R}^2, \|x\|_2 = 1} \|Ax\|_2 = 1$$

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Re "If the entries of A are all real, then the answer is yes." - Is this easy to prove? Can you add a proof to the post? –  Srivatsan Feb 1 '12 at 2:08
    
Actually I removed that claim, because I don't see a way to prove it. –  Robert Israel Feb 1 '12 at 2:17
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Have you considered looking at the wikipedia article on Matrix norm? I guess your answer is well written there. In any case, if not, then yes, it is true regardless of whether $x$ is over $\mathbb{R}^n$ or $\mathbb{C}^n$.

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The wikipedia article does seem to imply the two are equal, but is there an argument there to that effect? If so, I must be overlooking it. –  robinson Jan 28 '12 at 3:03
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