If $AA^T$ is the zero matrix, then $A$ is the zero matrix 
Let $A$ be a $4 \times 4$ matrix. Show that if $A^TA$ or $AA^T$ is the zero matrix, then $A$ is the zero matrix.

I feel very close to solving the problem so far. I have said that
$$[0]_{ij}=\sum_{k=1}^4 [A]_{ik}[A^T]_{kj} =\sum_{k=1}^4 [A]_{ik}[A]_{jk} \qquad \text{for all }i,j \in \{1,2,3,4\} $$
and proven for the case if $i=j$. However, I cannot seem to find a good way to prove it to be true if $i\neq j$.
 A: Here's a more matrix algebraic, less coordinate dependent way to do it:  for any matrix $M$, we have $\langle M^Tx, y \rangle = \langle x, My \rangle$ for the standard inner product.  Then if $A^TA = 0$, $\langle x, A^TAx \rangle =0$, so $\langle Ax, Ax\rangle =0$, whence $Ax = 0$ for all $x$, whence $A = 0$.  Same idea works if $AA^T = 0$.  Cheers.
A: An idea: if we put $\;A=(a_{ij})_{1\le i,j\le n}\;$ , then $\,A^t=(b_{ij})\;$ , with $\,b_{ij}=a_{ji}\,$ , so by definition:
$$AA^t=\left(\sum_{k=1}^n a_{ik}b_{kj}\right)=\left(\sum_{k=1}^n a_{ik}a_{jk}\right)$$
If you now look at the main diagonal's general entry of the above, you get
$$\sum_{k=1}^n a_{ik}a_{ik}=\sum_{k=1}^n a_{ik}^2$$
So if $\,AA^t=0\;$ then the above diagonal's entries are zero, but a sum of squared real numbers is zero iff each number is zero, so...
The same result is true with complex matrices if instead we require $\,AA^*=0\;,\;\;A^*:=\overline{A^t}\;$
A: Look at matrix $A$ as column vectors
$$A = [c_1, c_2, \cdots, c_n]$$
Then the $i$-th diagonal element of $A^T A$ is $c_i^T c_i = 0$, this indicates $c_i = 0, i = 1, 2, \cdots, n$. Thus $A = 0$.
