I am interested in the following problem: Let $P$ be a $n\times n$ complex finite matrix such as $PP^\dagger =W$. Given $W$, what can I say about the spectrum of $P$?

This matrix "square-root" has of course no unique solution, for if $P$ is a solution, $PU$ is also a solution if $U$ is unitary. If we consider the space $M_n(\mathbb{C})/U(n)$ we can seek an unique answer $P_* $, but I could not determine the shape of the space spanned by the eigenvalues of the equivalence class of $P_*$, apart for $n=2$ and some pretty horrible non-linear equations involving too many variables. Spectral theory is not my strong point, so I was wondering if anyone knew an answer to this: can we say what happens to the eigenvalues of $P$ if we multiply it by $U$ unitary?

  • $\begingroup$ If $P$ is hermitian, then the spectral radius does not increase (and usually decreases) when you multiply on the left by a unitary transformation. More than that, I cannot say :( $\endgroup$ – Igor Rivin Oct 22 '12 at 13:50

By specifying $PP^{\dagger}=W$ you are prescribing the singular values $s_n$ ($n=1,2,\ldots N$) of the $N\times N$ matrix $P$. These are just the positive square roots of the eigenvalues of the Hermitian, nonnegative matrix $W$. So your question can be rephrased as, what is the relation between the eigenvalues $\lambda_n$ and the singular values $s_n$ of the matrix $P$. This is a classic problem studied by Horn (1954), who showed that the only relationships one can state in full generality are those obtained by Weyl (1949):

$\prod_{n=1}^{k} |\lambda_n| \leq \prod_{n=1}^{k} s_n,$ for $k\lt N$, and $\prod_{n=1}^{N} |\lambda_n| = \prod_{n=1}^{N} s_n,$

for the ordering $|\lambda_1|\gt|\lambda_2\gt\cdots\gt|\lambda_N|$ and $s_1\gt s_2\gt\cdots\gt s_N$. (The equality for $k=N$ follows trivially by equating the determinant of $PP^\dagger$ with the determinant of $W$.)

More can be said for random matrices. As shown by Guionnet, Krishnapur, and Zeitouni (arXiv:0909.2214) in a probabilistic sense for “typical matrices”, the singular values almost determine the eigenvalues. In particular, the "single ring theorem" relates the eigenvalue density to the density of singular values.

This presentation by Mark Rudelson gives an introduction.


A. Horn, On the eigenvalues of a matrix with prescribed singular values, Proc. Amer. Math. Soc. 5, 4–7, (1954).

H. Weyl, Inequalities between the two kinds of eigenvalues of a linear transformation, Proc. Nat. Acad. Sci. USA 35, 408–411, (1949).

  • 1
    $\begingroup$ This is exactly what I was looking for, thank you! My problem arises from Random Matrix Theory, so, tip of the hat for you, fellow physicist. $\endgroup$ – Ricardo Marino Oct 23 '12 at 9:51

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