1
$\begingroup$

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

Improve this question
$\endgroup$

2 Answers 2

Reset to default
3
$\begingroup$

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.

Improve this answer
$\endgroup$
3
  • $\begingroup$ To expand as asked below. The diagonal blocks of $\hat M^2$ are $MM^T$ and $M^TM$. This gives the relation between the singular values of $M$ and the eigenvalues of $\widehat{M}$. The eigenvalues of $N\otimes I-I\otimes N$ are of the form $\lambda-\mu$, where $\lambda$ and $\mu$ run over the eigenvalues of $N$ (because its eigenvectors have the form $x\otimes y$ for eigenvectors $x$ and $y$). My third step is the same as Denis's, which is a consequence of the implicit function theorem. $\endgroup$
    – Chris Godsil
    Commented Sep 24, 2011 at 14:22
  • $\begingroup$ But this proves that in the set of matrices $\hat{M}$ which is in $\mathbb{R}^{(P+Q)^2}$ equipped with the Lebesgue measure in that space. But how does that imply the result in $\mathbb{R}^{PQ}$ equipped with the corresponding Lebesgue measure. Best Ashin $\endgroup$
    – Ashin
    Commented Sep 26, 2011 at 3:17
  • $\begingroup$ The point is that the matrices with repeated singular eigenvalues form a subvariety of the variety of matrices, and therefore they form a set of measure zero. $\endgroup$
    – Chris Godsil
    Commented Sep 26, 2011 at 13:09
2
$\begingroup$

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.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Hi Denis, Thanks for your reply, this is really helpful. Just to clarify, are you using the result that any non-trivial algebraic variety has zero measure. Could you please suggest me some text or online link where I could read more about that result. Thanks Ashin $\endgroup$
    – Ashin
    Commented Sep 24, 2011 at 13:59

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .