Proof Help : $A \in \mathbb{R}^{n \times n}, \sigma > 0$ if and only if the next matrix is singular.

Question

I've been stuck with the next proof. I've tried to approach the first implication using that if $\sigma$ is singular value of A, then is solution of $ P(\lambda) = \det(A^t A - \lambda I)$, but i don't know how to continue.

Problem: $A \in \mathbb{R}^{n \times n}, \sigma > 0$ singular value of A if and only if $\Big(\begin{matrix} A && -\sigma I \\ -\sigma I && A^t \end{matrix} \Big)$ is singular (non invertible).

Answer 1

Suppose $\sigma>0$ satisfies $\det (A^TA -\sigma^2I) = 0$ then there is some $v\neq 0$ such that $A^Tv = \sigma^2 v$. Let $u ={1 \over \sigma} Av$, then $Av-\sigma u = 0$ and $\sigma v - A^T u=0$ hence the above matrix is singular.

Conversely, if $Av = \sigma u$ and $A^T u = \sigma v$ wth $(u,v) \neq 0$ then $(A^TA -\sigma^TI) v = 0$ and so $\det (A^TA -\sigma^2I) = 0$.