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).

  • $\begingroup$ How do you define a singular value? $\endgroup$ – copper.hat May 31 '20 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 '20 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 '20 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 '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 $\det (A^TA -\sigma^2I) = 0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.