# 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?

• Yes, that's enough. Oct 3, 2015 at 23:35
• Two comments: 1) Usually, the definition of a positive semidefinite matrix includes the requirement that $A$ is symmetric (or hermitian for complex matrices). You did not check that. 2) Your argument shows that $A^T A$ is positive semidefinite. It does not show that $A^T A$ is not positive definite. Oct 4, 2015 at 10:07
• What is does it mean that "A is regular" in this context?
– Itay
Sep 17, 2016 at 9:32
• It means the same as invertible. So if $A$ is not invertible, then there are $x$ other than $0$ for which $Ax=0$ and thus strict inequality doesn't hold. On the other hand, if $A$ is invertible (thus regular), then $Ax=0$ only holds for $x=0$ and thus strict inequality (definiteness) holds for all $x \ne 0$. I think that more generally in this case regular means that the columns of $A$ are independent. So $A$ doesn't have to be square. Feb 5, 2017 at 20:50