# Bounding the 1-norm of a matrix

Let $M$ be a real $n\times n$ matrix and let $\|M\|_1 = \sum_{ij} |M_{ij}|$ ("the entry-wise 1-norm").

My question is: How well can we bound $\|M\|_1$ in terms of $|\det M|$ (from below)?

Here is an obvious bound: Let $m = \operatorname{max}_{ij} |M_{ij}|$.

Then it is obvious that $$|\det M| \leq \sum_{\sigma \in S_n} m^n = n! m^n \quad or \quad \|M\|_1 \geq m \geq (|\det M|/n!)^{\frac1n}.$$ Can this bound be improved?

(In fact, the above result is strong enough for my purposes, but this question came out of curiosity. I have played around for a while but I couldn't improve the bound, nor achieve it.)

-
Watch it ; you've forgotten to sum over all entries of your matrix when computing your lower bound ; you actually get $m \ge (|\det M|/n!)^{1/n}$, so that $\| M \| _1 \ge n^2 m \ge n^2 (| \det M | / n! )^{1/n}$. – Patrick Da Silva May 23 '11 at 20:47
@Patrick Da Silva: You think so? $m$ is the maximum, not the minimum of the entries, in fact $n^2 m$ should be larger than $\| M\|_1$. – Myself May 23 '11 at 20:52
Got tired here... haha good. – Patrick Da Silva May 23 '11 at 21:08

If we denote by $v_1,\ldots,v_n$ the column of $M$, we have $|\det M|\leq \prod_{j=1}^n\lVert v_j\rVert_2$. But $\lVert v_j\rVert_2^2 =\left(\sum_{k=1}^n|m_{kj}|\right)^2-2\sum_{j<k}|m_{jk}|\leq \left(\sum_{k=1}^n|m_{kj}|\right)^2$ hence $\lVert v_j\rVert_2\leq \sum_{k=1}^n|m_{jk}|\leq \lVert M\rVert_1$. We finally get that $|\det M|^{\frac 1n}\leq \lVert M\rVert_1$ which is a better bound.
That is a better bound! Is the first inequality well-known or obvious or something? The rest I agree with. (By the way, note that $\|v_j \| \leq \|M\|_1$ is obvious, in fact $\|M\|_1 = \sum_j \|v_j\|$.) – Myself May 23 '11 at 20:59
I should have remembered that, thanks. Apparently $\| v_i\| = \| v_i\|_2$ here so my last comment should be read as $\|M\|_1 = \sum_j \|v_j\|_1 \geq \|v_j\|_2$. – Myself May 23 '11 at 21:21
$|\det M| \le \prod_{j=1}^n \|v_j\|_2$ because $|\det M|$ is the $n$-dimensional volume of a parallelepiped whose edges correspond to the vectors $v_j$. Now by the AM-GM inequality we get $|\det M|^{1/n} \le \frac{1}{n} \sum_j \|v_j\|_2$. Finally, $\|v_j\|_2 \le \|v_j\|_1$, so the result is $\|M\|_1 \ge n |\det M\|^{1/n}$. And this is best possible, as seen by taking $M=I$. – Robert Israel May 23 '11 at 23:42