If $AA^T$ is the zero matrix, then $A$ is the zero matrix

Solution 1:

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 $\,AA^*=0\;,\;\;A^*:=\overline{A^t}\;$

Solution 2:

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.