Show $||A||_2=(\sum\limits_{i,j=1...n} a^2_{ij})^{(1/2)}$ defines a Matrix Norm

Question

Show $$\def\norm#1{\left\lVert{#1}\right\rVert_2{}}\norm A ={\left(\sum_{i,j=1\ldots n} a^2_{ij}\right)}^{1/2}$$ defines a Matrix Norm for $A\in\mathbb R^{n\times n}$

to show: $\norm{ AB}\le \norm A \norm B$

If i use Cauchy-Schwarz

$\sum a^2_{ij}\sum b^2_{ij}\ge(\sum a_{ij}b_{ij})^2$

but it is not always the case that;

$(\sum a_{ij}b_{ij})^2\ge\sum (a_{ij}b_{ij})^2 $

Answer 1

Note that if $C=AB$ then $c_{ij} = \sum_t a_{it} b_{tj}$. Hence

$$\|AB\|_2^2 = \sum_{ij} \Bigl(\sum_t a_{it} b_{tj} \Bigr)^2 {\color{Red}\leq} \sum_{ij} \sum_t a_{it}^2 \sum_t b_{tj}^2 = \sum_{it} a_{it}^2 \sum_{jt} b_{tj}^2 = \|A\|_2^2\|B\|_2^2, $$

where the red inequality follows from the Schwarz inequality.