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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).

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  • $\begingroup$ How do you define a singular value? $\endgroup$ – copper.hat May 31 at 4:13
  • $\begingroup$ My bad for not specify, I'm talking about Singular Value Decomposition, this is better explained here link $\endgroup$ – Lisandro Di Meo May 31 at 4:47
  • $\begingroup$ 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. $\endgroup$ – copper.hat May 31 at 4:53
  • $\begingroup$ 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! $\endgroup$ – Lisandro Di Meo May 31 at 5:35
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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 $\det (A^TA -\sigma^2I) = 0$.

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