Prove that $ AA^T=0\implies A = 0$ Let $A$ be an $n \times n$ matrix with real entries, where $n\geq2$.
  Let $AA^T = [b_{ij}] $, where $A^T $ is the transpose of $A$. If
  $b_{11} + b_{22 }+\cdots+ b_{nn} = 0$, show that $A = 0$.


From what I've gleaned so far, $AA^T$ is a symmetric matrix, and the diagonals are zero. I can't figure out how to solve this question. Is there some property that exists that I'm missing for handling this question?
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.
I thought about this proof, but I think it is not totally correct:

Supose $A≠0$.

Then, for any $X$, $A+X≠X$.

Multiplying by $A^T$ at the right, we get:

$(A+X)A^T≠XA^T$

$AA^T+XA^T≠XA^T$

$XA^T≠XA^T$

Arriving in a contradiction, we get $A=0$.
I realise it may not be totally correct to multiply by $A^T$ without making some proper assumptions, but I know I can pick any $X$. Is there a way to formalise this proof?