1
$\begingroup$

the singular value decomposition of an $m\times n$ random Gaussian matrix ${\displaystyle \mathbf {M} }$ is a factorization of the form ${\displaystyle \mathbf {U\Sigma V^\ast} }$, ${\displaystyle \mathbf {\Sigma } }$ is an ${\displaystyle m\times n}$ rectangular diagonal matrix with non-negative ordred real numbers on the diagonal, my question is:

What is the distribution of the singular values of $\Sigma $?

can I say that the singular value corresponds to the absolute value of Gaussian variable?

$\endgroup$

1 Answer 1

3
$\begingroup$

With fixed ratio $\lambda=m/n$ the Marchenko-Pastur distribution gives the asymptotic distribution of singular values for a Gaussian rectangular matrix.

The extreme singular values limiting distribution in particular is given by the Tracy-Widom distribution.

Also, the singular values do not have limiting distribution corresponding to the absolute value of a Gaussian distribution.

$\endgroup$
2
  • $\begingroup$ and what can I say if $m$ is fixed and $n$ is arbitrarily large? $\endgroup$
    – Student88
    Commented Nov 14, 2022 at 8:13
  • $\begingroup$ That's not quite true. The Marchenko Pastur distribution is the limiting distribution for the singular values squared. $\endgroup$ Commented Apr 6, 2023 at 2:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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