Measure of Subspace of Matrices with repeated Singular Values - MathOverflow

Measure of Subspace of Matrices with repeated Singular Values

2011-09-23T23:45:14Z

Hi All,

Let us consider a P x Q real matrix (P >= Q). It can be thought of as an element of $\mathbb{R}^{PQ}$. We are considering Lebesgue measure over that space. My question is whether the subspace of matrices with repeated singular values are of measure 0 or not.

Any suggestions would be welcome.

Thanks Ashin

Answer by Chris Godsil for Measure of Subspace of Matrices with repeated Singular Values

2011-09-24T00:24:57Z

The eigenvalues of $$ \widehat M = \begin{pmatrix}0&M\\ M^T&0\end{pmatrix} $$ are the squares of the singular values of $M$. A symmetric $n\times n$ matrix $N$ has a repeated eigenvalue if and only if the rank of $$ N\otimes I - I\otimes N $$ is less than $n^2-n$. So the set of matrices $\widehat M$ with a repeated eigenvalue is a proper subvariety of the set of matrices $\widehat M$, hence this set will have measure zero.

Answer by Ashin for Measure of Subspace of Matrices with repeated Singular Values

2011-09-24T03:33:41Z

Hi Chris,

Thanks a lot for your quick response. But I'm not that familiar with the results that you are using to arrive at the conclusion. Would you kindly be a bit more elaborate. Or maybe you can guide me to the resources where I could find the results that you are using here.

Thanks a lot for your help. Ashin

Answer by Denis Serre for Measure of Subspace of Matrices with repeated Singular Values

2011-09-24T07:59:25Z

The set of matrices with a repeated eigenvalue is defined by an algebraic equation ${\rm disc}(M)=0$. This is the discriminant in the eigenvalues $$\prod_{i\lt j}(\lambda_j-\lambda_i)^2,$$ which is a polynomial in the entries of $M$. Because this polynomial is non-trivial, your set is a non-trivial algebraic variety. In particular, it has zero measure and is closed.