Does $1$ norm exist for non-square matrices? By $1$ norm I mean
$d (x,y)=\sum_{i=1}^{n} |x^i-y^i|, x=(x_1,\dots, x_n), y=(y_1,\dots, y_n)$
Suppose $A$ is $m\times n, (m\ne n)$ matrix what can we say about $\|A\|_1$? Also, can we say $\|A\|_1=\|A^T\|_1= \|A^TA\|_1=\|AA^T\|_1$? and $\|AB\|_1\le \|A\|_1\|B\|_1$ Thanks for helping.
There is no problem to define an "entrywise" matrix norm defined as follows: $$ \|A\|_1=\|vec(A)\|_1=\sum_{i,j}|A_{i,j}| $$ see wikipedia, Matrix norms. It is a norm and the induced distance is (if $A$, $B$ have same dimensions): $$ d(A,B)=\sum_{i,j}|A_{i,j}-B_{i,j}| $$ This norm is a sub-multiplicative norm (see here): $$ \|AB\|_1\leq\|A\|_1\|B\|_1 $$ but, attention, in general: $$ \|A^tA\|_1\neq\|AA^t\|_1 $$
Another example of "entrywise" matrix norm often encountered in practice is the Frobenius norm. This norm is defined as follows:
$$ \|A\|_F = \sqrt{\text{tr}(A^tA)}=\sqrt{\sum_{i,j}|A_{i,j}|^2} $$ Frobenius norm also fulfills the sub-multiplicative property (Cauchy-Schwarz in action, see here): $$ \|AB\|_F\leq \|A\|_F\|B\|_F $$
Compared to the previous case $\|.\|_1$, like we have $\text{tr}(A^tA)=\text{tr}(AA^t)$, Frobenius norm also fulfills the property:
$$ \|AA^t\|_F=\|A^tA\|_F $$
To be complete one must also say a word about Matrix norms induced by vector norms. One can define: $$ \|A\|_1=\sup_{x\neq 0}\frac{\|Ax\|_1}{\|x\|_1} $$ (attention despite identical the notation, this norm is different from the previously defined $\|vec(.)\|_1 $, here we have $\|A\|_1=\max_j \sum_i|A_{i,j}|$)
An immediate generalization, valid for any $1\leq p \leq \infty$, is: $$ \|A\|_p=\sup_{x\neq 0}\frac{\|Ax\|_p}{\|x\|_p} $$
Such norms are called Matrix norms induced by vector norms, they automatically fulfill the sub-multiplicative property: $$ \|AB\|_p \leq \|A\|_p \|B\|_p $$
Attention: however in general, $$ \|AA^t\|_p \neq \|A^tA\|_p $$
Also note this important fact, as we are in finite dimension all these previously defined matrix norms are equivalent in the sense that for any two matrix norm $\|.\|_\alpha$ and $\|.\|_\beta$ it exists $r$ and $s$ such that: $$ \forall A,\ r\|A\|_\alpha\leq \|A\|_\beta \leq s\|A\|_\alpha $$ in peculiar if a sequence $n\rightarrow (A)_n$ is convergent for a given matrix norm it is also convergent for all the other norms.
