3
$\begingroup$

Let $V$ be the space of $4$ by $4$ Hermitian matrices, that is a vector space of dimension $16$ over $\mathbb{R}$. Is the uniform measure of $$ \left\{ W\in Gr\left(5,V\right):W \text{contains no nonzero Hermitian matrix with at least 2 eigenvalues that are 0}\right\} $$ equal to $0$?

Does any of the experts know or have the answer to this question?

Improve this question
$\endgroup$
10
  • $\begingroup$ BTW, the definition of uniform measure on the Grassmannian can be found here, en.wikipedia.org/wiki/Grassmannian#Associated_measure. $\endgroup$
    – Ayna
    Commented Nov 22, 2013 at 18:48
  • $\begingroup$ @CarloBeenakker: This question actually has not been answered. Let me know if you know or have the answer to this question. Thanks! $\endgroup$
    – Ayna
    Commented Nov 22, 2013 at 19:19
  • $\begingroup$ @Ricardo: sorry, my bad for misreading it. it should indeed be $W \in$ .. because the subset in question is already enclosed in braces! but it is bad wording...because it says $W$ has no hermitian matrix...so $W$ is a set.... $\endgroup$
    – Suvrit
    Commented Nov 22, 2013 at 22:18
  • 1
    $\begingroup$ @Ayna, there is no problem. I apologize for misinterpreting your question. When you say that a matrix has two eigenvalues equal to zero, do you then mean that zero is an eigenvalue with algebraic multiplicity at least two? If so, then note that the zero matrix is in any linear subspace $W$ of $V$. So you need to correct for that in your question, as I did in its previous version. $\endgroup$
    – Ricardo Andrade
    Commented Nov 23, 2013 at 17:32
  • 3
    $\begingroup$ I'll just point out that the 'measure theoretic' aspects of this question are irrelevant. If the set is nonempty, it has positive uniform measure because it is open, i.e., if $W\in\mathrm{Gr}(5,V)$ is a subspace with the desired property then all sufficiently near subspaces $W'$ will have this property as well. Thus, the real question is whether the set in question is nonempty. $\endgroup$
    – Robert Bryant
    Commented Nov 23, 2013 at 20:45

1 Answer 1

Reset to default
4
$\begingroup$

It appears that the answer is 'no', based on the answer to the OP's previous question, A question on eigenvalues. In the answers to that question, it is pointed out that there is a 1965 paper by Adams, Lax, and Phillips that implies that there exists a 5-dimensional subspace $W_0\subset V$ such that its nonzero elements are, in fact, nonsingular (which is considerably stronger than what the OP required).

If you let $\hat C\subset V$ denote the $12$-dimensional cone of Hermitian matrices with at least two zero eigenvalues, then the projectivization of $\hat C$ is a closed algebraic subvariety $C\subset\mathbb{P}(V)\simeq\mathbb{RP}^{15}$ of dimension $11$. By construction $\mathbb{P}(W_0)\subset \mathbb{P}(V)$ does not meet $C$. The set of subspaces $W\in\mathrm{Gr}(5,V)$ such that $\mathbb{P}(W)\cap C=\emptyset$ is therefore nonempty and it is clearly open in $\mathrm{Gr}(5,V)$ (because $C$ is closed). Therefore, in particular, it has nonzero uniform measure.

NB: The OP asked what I meant by 'near' in my comment above. If one fixes an inner product $q$ on $V$, say the obvious $\mathrm{U}(4)$ invariant one (but any positive definite inner product will do), then there is induced on each $\mathrm{Gr}(k,V)$ a natural metric, unique up to scale, that is invariant under the the orthogonal group of $q$. By 'near', I meant 'close in such a metric'.

Improve this answer
$\endgroup$
7
  • $\begingroup$ Thank you for your answer, but why is $\hat{C}$ 11-dimensional. Thanks! $\endgroup$
    – Ayna
    Commented Nov 27, 2013 at 2:08
  • $\begingroup$ @Ayna: Sorry. You are right; I miscounted, it's actually $12$-dimensional. The dimension doesn't affect the answer in any way, though. It's $12$-dimensional because a generic element of it is determined by picking complex lines in $\mathbb{C}^4$ that are Hermitian orthogonal (the first eigenspace is specified by $6$ parameters and the second is specified by $4$, once the first has been specified) and then choosing $2$ (real) eigenvalues for these two eigenspaces, which gives you a total of $12$ parameters. $\endgroup$
    – Robert Bryant
    Commented Nov 28, 2013 at 2:53
  • $\begingroup$ Thanks for the explanation, but can you get the dimension of $\hat{C}$ by looking at the eigenvalues which are real numbers and the dimension of the unitary group $U(4)$? Thanks a lot! $\endgroup$
    – Ayna
    Commented Dec 5, 2013 at 17:03
  • $\begingroup$ @Anya: Well, what you really need is not just the dimension of $\mathrm{U}(4)$, which is $16$, but also the dimension of its subgroup $G$ that preserves a pair of orthogonal complex lines $L_i\subset\mathbb{C}^4$, which is $6$. That means that the space of pairs of orthogonal complex lines is the homogeneous space $\mathrm{U}(4)/G$, a space of dimension $16-6 = 10$. Then, once the lines are specified, you fix the (real) eigenvalues, which gives you $2$ more parameters, making a total of $10 + 2 = 12$. $\endgroup$
    – Robert Bryant
    Commented Dec 5, 2013 at 18:22
  • $\begingroup$ Thanks for the further explanation. But my question asks whether $$ \mathcal{X}:=\left\{ W\in Gr\left(5,V\right):W\text{contains a nonzero Hermitian matrix with at least 2 zero-eigenvalues}\right\} $$ is dense in $Gr\left(5,V\right)$. Since $\hat{C}$ is the set of all Hermitian matrices with at least 2 zero-eigenvalues, $\mathcal{X}\supset\left\{ span\left(W_{1},\cdots,W_{5}\right):W_{1},\cdots,W_{5}\in\hat{C}\right\} $ and $\mathcal{X}$ is different from your $C$ that is $\mathbb{P}\left(\hat{C}\right)$. So do you know if $\mathcal{X}$ is dense in $Gr\left(5,V\right)$? Thanks a lot! $\endgroup$
    – Ayna
    Commented Dec 6, 2013 at 19:17

You must log in to answer this question.

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