Welcome to MathOverflow

Question

1

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 $\|A'\|_p$.

flag	asked Oct 22 at 4:12 terry 11●1

1

When you say the $L_p$ norm, do you mean as an operator from $L_p$ to itself? – Yemon Choi Oct 22 at 6:20

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. – Delio Mugnolo Oct 22 at 6:40

1

On a finite-dimensional spaces all norms are equivalent. – Jochen Wengenroth Oct 22 at 6:41

1

Jochen, yours is not really an answer. Surely all norms are equivalent, but it can still be very good to know how they relate to each other. Besides, terry's question was not about comparing two norms of the same objects, but the same norm applied to two slightly different objects. – Delio Mugnolo Oct 22 at 6:56

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 $C >0$. – Pietro Majer Oct 22 at 7:19

show 1 more comment

Mark Meckes · Answer 1 · 2012-11-06 16:10:46Z UTC

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.

Dima Pasechnik · Answer 2 · 2012-10-22 09:04:49Z UTC

0

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 at 9:04

Dima Pasechnik
3,080●3●14

however, the OP question is still unclear (in particular, what "L_p norm of A" means there) – Pietro Majer Oct 22 at 10:23

Terry · Answer 3 · 2012-10-23 01:38:10Z UTC

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.

Comparison of the L_p norm of a matrix and its entry-wise absolute value

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

3 Answers

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.