Comparison of the L_p norm of a matrix and its entry-wise absolute value - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T16:20:28Z http://mathoverflow.net/feeds/question/110299 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/110299/comparison-of-the-l-p-norm-of-a-matrix-and-its-entry-wise-absolute-value Comparison of the L_p norm of a matrix and its entry-wise absolute value terry 2012-10-22T04:12:28Z 2012-11-06T16:10:46Z <p>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 $\|A'\|_p$.</p> http://mathoverflow.net/questions/110299/comparison-of-the-l-p-norm-of-a-matrix-and-its-entry-wise-absolute-value/110310#110310 Answer by Dima Pasechnik for Comparison of the L_p norm of a matrix and its entry-wise absolute value Dima Pasechnik 2012-10-22T09:04:49Z 2012-10-22T09:04:49Z <p>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. :-)</p> http://mathoverflow.net/questions/110299/comparison-of-the-l-p-norm-of-a-matrix-and-its-entry-wise-absolute-value/110334#110334 Answer by Terry for Comparison of the L_p norm of a matrix and its entry-wise absolute value Terry 2012-10-22T14:43:38Z 2012-10-23T01:38:10Z <p>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. </p> <p>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 $C \cdot \|A \|_2 \geq \|A' \|_2$. </p> <p>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.</p> http://mathoverflow.net/questions/110299/comparison-of-the-l-p-norm-of-a-matrix-and-its-entry-wise-absolute-value/111658#111658 Answer by Mark Meckes for Comparison of the L_p norm of a matrix and its entry-wise absolute value Mark Meckes 2012-11-06T16:10:46Z 2012-11-06T16:10:46Z <p>Okay, so it's established that <code>$\| A \|_p$</code> means the induced norm. A few basic facts: <code>$$\| A \|_1 = \max_{j} \sum_{i} |a_{ij}| \le n^{1-1/p} \| A \|_p,$$</code> <code>$$\| A \|_\infty = \max_{i} \sum_{j} |a_{ij}| \le n^{1/p} \| A \|_p,$$</code> <code>$$\| A \|_p \le \|A\|_1^{1/p} \| A \|_\infty^{1-1/p}.$$</code> 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&ndash;Thorin theorem. Putting these together, <code>$$\|A'\|_p \le \| A' \|_1^{1/p} \| A' \|_\infty^{1-1/p} = \| A \|_1^{1/p} \| A \|_\infty^{1-1/p} \le n^{\frac{2}{p}(1- \frac{1}{p})} \|A\|_p.$$</code></p> <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.</p> <p>As noted by Pietro, $\| A \|_p \le \| A' \|_p$ always.</p>