1
$\begingroup$

For the vector space $M_{n,n}(\mathbb{C})$ of $n\times n$ matrices we know that the subset

$$M_{2r}:= \{A\in M_{n,n}(\mathbb{C}) \mid \mbox{rank} (A) = 2r \}$$

is a manifold of dimension $2n(2r)-(2r)^2$ (see this question).

If I now take a look at Hermitian $n \times n$ matrices of rank $2r$ with exactly $r$ positive and $r$ negative eigenvalues, how can I determine the dimension of this manifold?

Improve this question
$\endgroup$
5
  • 2
    $\begingroup$ Why should the dimension be any different? This partitions the manifold of dimension $2n(2r) - (2r)^2$ into a finite number (15) of submanifolds. $\endgroup$
    – Peter Shor
    Commented Jul 3, 2018 at 19:08
  • $\begingroup$ I thougth that the restriction of $r$ postivie and $r$ negative eigenvalues reduces the dimension. I don't got your point in total. Could you explain it with a few more words? $\endgroup$
    – Alpha001
    Commented Jul 3, 2018 at 20:48
  • $\begingroup$ Given a Hermitian matrix with $r$ positive and $r$ negative eigenvalues, any sufficiently close Hermitian matrix of rank $2r$ will also have $r$ positive and $r$ negative eigenvalues. On the other hand, the restriction to Hermitian matrices does change the dimension. $\endgroup$
    – Robert Israel
    Commented Jul 3, 2018 at 22:31
  • 1
    $\begingroup$ @RobertIsrael: Oops. I completely missed the Hermitian requirement. $\endgroup$
    – Peter Shor
    Commented Jul 3, 2018 at 22:35
  • $\begingroup$ @RobertIsrael Is it clear how it would change the dimension? $\endgroup$
    – Alpha001
    Commented Jul 3, 2018 at 23:17

1 Answer 1

Reset to default
5
$\begingroup$

A Hermitian matrix of rank $r$ can be represented uniquely as $U D U^\ast,$ where $U$ is an $n\times r$ matrix with orthogonal rows of unit length, and $D$ is an $r\times r$ matrix (this is the singular value decomposition, if you want to think of it that way). The (complex) dimension of the space of $U$ is $n-1 + n-2 + \dotsc+ n-r$ the (real) dimension of the space of $D$s is $r$, for a total of $$(2n -r -1)r + r = 2n r - r^2$$ real dimensions, near as I can tell.

Improve this answer
$\endgroup$
4
  • $\begingroup$ This is a nice simple approach, but it's not exactly what the OP is asking. Also, I'm not sure I agree with the dimension count of the $U$'s, for example the unit vectors (or $n\times 1$ matrices $U$ as above) have real dimension $2n-1$, not $2n-2$. $\endgroup$
    – Christian Remling
    Commented Jul 4, 2018 at 1:22
  • $\begingroup$ But this is has the same dimension like in the case of $M_{2r}$? Could this be since we did not assume that the matricies in $M_{2r}$ are hermitian. $\endgroup$
    – Alpha001
    Commented Jul 4, 2018 at 10:21
  • $\begingroup$ @IgorRivin : Look at the example $n=3$ and $r=2$ then your formula results in a dimension of $2nr-r^2 = 8$ but the space of symmetric matricies is only of dimension $\frac{n(n+1)}{2}=6$? Since the manifold is subset of the space of symmetric/hermitian matricies something went wrong? $\endgroup$
    – Alpha001
    Commented Jul 5, 2018 at 15:03
  • $\begingroup$ @Alpha001: Hermitian matrices are not always symmetric matrices. The dimensionality of the space of symmetric matrices is 6. For Hermitian matrices, it's 9. $\endgroup$
    – Peter Shor
    Commented Jul 8, 2018 at 0:53

You must log in to answer this question.

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