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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?

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Lucas Alanis
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    Yes, that's enough. – Casteels Oct 03 '15 at 23:35
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    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. – PhoemueX Oct 04 '15 at 10:07
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    What is does it mean that "A is regular" in this context? – Itay Sep 17 '16 at 09:32
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    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. – Davor Josipovic Feb 05 '17 at 20:50
  • I think this piece of answer should be added to the top answer of the linked question! – Rémy Hosseinkhan Boucher Jan 07 '22 at 15:44
  • Instead of a matrix, can we conclude it for vectors? I mean if $x \in R^d$, then is it true that $x x^T$ is a PSD matrix? – Amin Oct 26 '22 at 20:28

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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.

Carl Edward Sagan
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