Suppose $A_{n \times n}$ is a matrix and $A' = (|A_{ij}|)$ is its entry wise absolute form, can be give an upper bound and lower bound of the L_p norm $\|A\|_p$ using the L_p norm of the absolute matrix $\|A'\|_p$.
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1$\begingroup$ When you say the $L_p$ norm, do you mean as an operator from $L_p$ to itself? $\endgroup$– Yemon ChoiCommented Oct 22, 2012 at 6:20
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$\begingroup$ I do not understand the question, either. If for $L_p$-norm the norm $$ \| A\|_p := \sum_{i,j=1}^n |A_ij|^p $$ is meant, then it is clear that the two norms coincide by definition. $\endgroup$– Delio MugnoloCommented Oct 22, 2012 at 6:40
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$\begingroup$ Assuming that the question refers to the operator norm of $A$ and $A'$ as operators on the $N$ dimensional space $\ell^p _ N$, the bound $\|A\|_p\le\|A'\|_p$ is immediate by the monotonicity property of the $L_p$ norm (meaning that for vectors $x$ and $x'$ one has $\|x\|_p\le\|x'\|_p$ whenever $x\le x'$ componentwise). The difficult part of the quesion should be the other bound, $\|A'\|_p\le C\|A\|_p$, for some proper constant $C >0$. $\endgroup$– Pietro MajerCommented Oct 22, 2012 at 7:19
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3 Answers
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Okay, so it's established that $\| A \|_p$ means the induced norm. A few basic facts: $$ \| A \|_1 = \max_{j} \sum_{i} |a_{ij}| \le n^{1-1/p} \| A \|_p, $$ $$ \| A \|_\infty = \max_{i} \sum_{j} |a_{ij}| \le n^{1/p} \| A \|_p, $$ $$ \| A \|_p \le \|A\|_1^{1/p} \| A \|_\infty^{1-1/p}. $$ The first two lines are elementary (the inequalities following from standard comparisons of $\ell_p$ norms for vectors), and the third is a finite-dimensional version of the Riesz–Thorin theorem. Putting these together, $$ \|A'\|_p \le \| A' \|_1^{1/p} \| A' \|_\infty^{1-1/p} = \| A \|_1^{1/p} \| A \|_\infty^{1-1/p} \le n^{\frac{2}{p}(1- \frac{1}{p})} \|A\|_p. $$
When $p = 2$ and $A$ is a Hadamard matrix this is sharp, and of course it's sharp for $p=1$ or $p = \infty$. I'd guess it's sharp always but I haven't thought about it.
As noted by Pietro, $\| A \|_p \le \| A' \|_p$ always.
answered Nov 6, 2012 at 16:10
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The matrix L_p norm means $\max\limits_{\|x\|_p = 1} \|Ax\|_p $, here $x$ is an $n \times 1$ vector, and the $Ax$ is an $n \times 1$ vector too. so when $p=2$, it is the well known operator norm.
Thanks. I try to answer the question in a particular sense. It can be seen that even for the operator norm, that is $p=2$, when n is really large, a entry wise random $+1,-1$ Bernoulli matrix have the largest singular value similar to $\sqrt{n}$, but the absolute matrix have the largest singular value $n$, so asymptotically, when $n$ is really large, we cannot have a constant $C>0$ such that $C \cdot \|A \|_2 \geq \|A' \|_2$.
As @Pietro Majer has said, I think it is easy to get $\|A \|_p \leq \|A' \|_p$. but the other way around for a fixed n is still hard to me.
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$\begingroup$ Here matrix norm is the induced norm, not treated simply as a long vector. $\endgroup$– TerryCommented Oct 22, 2012 at 14:51
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$\begingroup$ You should be able to edit your original question so you can add the extra clarification there $\endgroup$ Commented Oct 23, 2012 at 0:23
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$\begingroup$ Thanks. I try to answer the question in a particular sense. It can be seen that even for the operator norm, that is $p=2$, when $n$ is really big, a entry wise random +1,-1 Bernoulli matrix have the largest singular value similar to $\sqrt{n}$, but the absolute matrix have the largest singular value $n$, so asymptotically, when n is really large, we cannot have a constant $C>0$ such that $C \times \|A\|_2 \geq \|A'\|_2$ As @Pietro Majer has said, I think it is easy to get $\|A\|_p \leq \|A'\|_p$ but the other way around for a fixed n is still hard to me. $\endgroup$– terryCommented Oct 23, 2012 at 1:29
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There is quite a bit of information on this in Chapter 5 of Horn and Johnson "Matrix Analysis" (Cambridge University Press 1985). Perhaps it is even an exercise there. :-)
answered Oct 22, 2012 at 9:04