I consider this matrix in $\mathbb{M}^{m\times n}$ $$A=a\otimes b,$$ with $b\in \mathbb{R}^m$ and $a\in\mathbb{R}^n$.
I would like to prove that $|A|_{nm}=|a|_{n} |b|_{m}$, where $|.|_{p}$ is the module on $\mathbb{R}^p$.
Is there a simple way to prove this?
Should I use some other matrix norms to prove that? Thanks for the help!
As a comment, the notation $|x|_p$ is really awckward. This is usually denoted $|x|_2$, irrespective of the size of the matrix. As this is standard, I'll use this notation.
Assuming that by $|x|_2$ you mean $(\sum_{k,j}|x_{kj}|^2)^{1/2}$, this follows from the equalities $$ |x|_2^2=\text{Tr}\,(x^Tx),\ \ \ \ \text{ and } \ \ \ \ \text{Tr}\,(a\otimes b)=\text{Tr}(a)\,\text{Tr}(b). $$ Indeed, $$ |A|_{2}^2=\text{Tr}((a\otimes b)^T(a\otimes b))=\text{Tr}(a^Ta\otimes b^Tb)=\text{Tr}(a^Ta)\,\text{Tr}(b^Tb)=|a|_2^2\,|b|_2^2 $$