Proof for why a matrix multiplied by its transpose is positive semidefinite

The top answer to this question says

Moreover if $A$ is regular, then $AA^T$ is also positive definite, since $$x^TAA^Tx=(A^Tx)^T(A^Tx)> 0$$

Suppose $A$ is not regular. It holds that $$x^TAA^Tx=(A^Tx)^T(A^Tx)= \|A^Tx\|^2_2 \ge 0$$ Therefore $AA^T$ is positive semidefinite. Is this argument enough, or am I missing something?

Solution 1:

I don't see anything wrong with your proof. And the result is true even for complex matrices, where you'll consider the hermitian conjugate, instead of the transposed. This is the basis for the Polar Decomposition of complex matrices.

The part where you consider the non regular case, you could have been more clear anda say that, either x belongs to Ker(A), and then it will give zero. Or it has a component in the Im(A) and therefore it must be positive, since the internal product on a vector space is positive definite.