ohgodamnitw0

2022-09-09

Let A be a $n×n$ square matrix, and
$‖A{‖}_{1}:=\underset{j=1}{\overset{n}{max}}\sum _{i=1}^{n}|{A}_{i,j}|$
and
$‖A{‖}_{max}:=\underset{i,j}{max}|{A}_{i,j}|$
1)Does $‖A{‖}_{1}\le 1$ imply that $‖A{‖}_{max}\le 1$?
2)Is there any relation between $‖A{‖}_{1}$ and $‖A{‖}_{max}$?

Derick Ortiz

Since
$|{A}_{i,j}|\le \underset{1\le i\le n}{max}|{A}_{i,j}|,$
for each $1\le j\le n$, then
$\sum _{i=1}^{n}|{A}_{i,j}|\le \sum _{i=1}^{n}\underset{1\le i\le n}{max}|{A}_{i,j}|=n\underset{1\le i\le n}{max}|{A}_{i,j}|$
Therefore
$\underset{1\le j\le n}{max}\sum _{i=1}^{n}|{A}_{i,j}|\le n\underset{1\le i,j\le n}{max}|{A}_{i,j}|.$
That is
$‖A{‖}_{1}\le n‖A{‖}_{\mathrm{\infty }}.\phantom{\rule{2em}{0ex}}\left(1\right)$
The estimate (1) is sharp. Meaning the factor n cannot be replaced by a smaller positive constant. To see this consider the matrix A with ${A}_{i,j}=1$ for which (1) is actually an equality.
Now, does $‖A{‖}_{1}$ control $‖A{‖}_{\mathrm{\infty }}$?
We have that
$\underset{1\le i\le n}{max}|{A}_{i,j}|\le \sum _{i=1}^{n}|{A}_{i,j}|⟹\underset{1\le j\le n}{max}\underset{1\le i\le n}{max}|{A}_{i,j}|\le \underset{1\le j\le n}{max}\sum _{i=1}^{n}|{A}_{i,j}|$
So, it is true that
$‖A{‖}_{\mathrm{\infty }}\le ‖A{‖}_{1}\phantom{\rule{2em}{0ex}}\left(2\right)$
Again the constant one in (2) is sharp. When $A={I}_{n}$, the unit matrix, we have an equality in (2).

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