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?


1 Answer 1


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.

  • $\begingroup$ and what can I say if $m$ is fixed and $n$ is arbitrarily large? $\endgroup$
    – Student88
    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. $\endgroup$ Apr 6, 2023 at 2:37

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