Is $AA^T$ positive semidefinite?

matrices

Solution 1:

Let $y=A^Tx$.
$x^T AA^Tx=y^Ty=\sum_{k=1}^N y_k^2\geqslant 0$

So your answer is yes.

Solution 2:

$AA^{T}$ is positively semidefinite $\Leftrightarrow$ it is obviously true that $A^{T}A$ is positively semidefinite. We'll prove the right.

It is true that $A^{T}A$ is symmetric. Let $x$ be a non-zero column vector. Then we have: $x^{T}A^{T}Ax = (Ax)^{T}(Ax)$.

Notice that $Ax$ is also a non-zero column vector so $(Ax)^{T}(Ax)$ is the square of the inner product of $Ax$

Then: $(Ax)^{T}(Ax) = ||Ax||^{2} \geq 0$. Done ! Sorry for my English

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