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

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).

• How do you define a singular value? – copper.hat May 31 '20 at 4:13
• My bad for not specify, I'm talking about Singular Value Decomposition, this is better explained here link – Lisandro Di Meo May 31 '20 at 4:47
• I understand, but there are various ways of defining the singular values, one way is en.wikipedia.org/wiki/…, I was asking how you define a singular value for the purposes of this exercise. – copper.hat May 31 '20 at 4:53
• Tell me if it's wrong, but we (my class) define singular value $\sigma_i = \sqrt{\lambda_i}$, the i-th eigenvalue of $A^t A$. Sorry for the confusion. Let me know if it's not possible to continue the excersie with that definition, and would be great if you could tell me which definition I should take to try to solve the problem. Thanks! – Lisandro Di Meo May 31 '20 at 5:35

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 $$\det (A^TA -\sigma^2I) = 0$$.