Singularity of a matrix times its transpose

Question

If $A$ is a $k \times n$ matrix of rank $k$, where A can be broken down into $UDV^T$ as per singular value decomposition, why is $AA^T$ non-singular?

Answer 1

$D$ is a $k \times n$ diagonal matrix where each row is non-zero. Its $(i,i)$-th entry, $d_i$ is positive.

$DD^T$ is a $k \times k$ matrix where each row is non-zero, its $(i,i)$-th entry, $d_i^2$ is positive. and hence it is nonsingular.

$$AA^T=UDV^TVD^TU^T=UDD^TU^T$$

Since $U$ is nonsingular, $DD^T$ and $U^T$ are nonsingular, $AA^T$ is nonsingular.

Answer 2

Since you are using transpose ($T$) rather than Hermitian conjugate, we're working over the reals.

Since $A$ has full row rank, it is surjective.

$\text{Ker}(A^T) = (\text{Ran}(A))^\perp = \{0\}$.

For any nonzero vector $x \in \mathbb R^k$, $x^T A A^T x = \|A^T x\|^2 > 0$, so $A A^T x \ne 0$. Thus $\text{ker}(A A^T) = \{0\}$, so $A A^T$ is nonsingular.