MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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 :( – Igor Rivin Oct 22 '12 at 13:50
up vote 7 down vote accepted

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).

share|cite|improve this answer
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. – Ricardo Marino Oct 23 '12 at 9:51

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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