Prove that $ AA^T=0\implies A = 0$

Let

$$A=(a_{ij})\implies A^t=(a_{ji})\implies AA^t=(b_{ij})=\left(\sum_{k=1}^ma_{ik}a_{jk}\right)$$

so that

$$0=\sum_{i=1}^nb_{ii}=\sum_{i=1}^n\sum_{k=1}^na_{ik}a_{ik}$$

Complete the proof now.

$||A||=\sqrt{\sum_{i,j=1}^n |a_{ij}|^2}$

i.e. $||A||=\sqrt {trace(A^t A)}$ .Now by the given condition $b_{11}+b_{22}+...+b_{nn}=0$

i.e. trace of $A^tA=0$

i.e. $||A||=0$ if and only if $A=0$

$\forall X \in \mathbb{R}^n$, $X^tAA^tX=0$, i.e. $\left<A^tX, A^tX\right>=0$, hence $A^tX=0, \forall X \in \mathbb {R}^n$, so $A^t=0$, thus $A=0$.