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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

edited Sep 23 2011 at 23:55

asked Sep 23 2011 at 23:45

Ashin
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Chris Godsil · Accepted Answer · 2011-09-24 00:24:57Z UTC

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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.

answered Sep 24 2011 at 0:24

Chris Godsil
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	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. – Chris Godsil Sep 24 2011 at 14:22
	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 – Ashin Sep 26 2011 at 3:17
	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. – Chris Godsil Sep 26 2011 at 13:09

Theo Buehler · Answer 2 · 2011-09-24 11:57:07Z UTC

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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.

edited Sep 24 2011 at 11:57

Theo Buehler

answered Sep 24 2011 at 7:59

Denis Serre
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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 – Ashin Sep 24 2011 at 13:59

Ashin · Answer 3 · 2011-09-24 03:33:41Z UTC

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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

answered Sep 24 2011 at 3:33

Ashin
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If you do not have sufficient reputation points to leave comments, please add them as edits to your original question, and not as separate "answers" – Yemon Choi Sep 24 2011 at 4:09

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You should also consider registering an account so that the system will always be able to recognize you as the same person when you post. – jc Sep 24 2011 at 18:26

Measure of Subspace of Matrices with repeated Singular Values

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