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Let $H$ be a linear subspace of the space of Hermitian $n\times n$ matrices. Is there a good characterization of those $H$ such that every $A\in H$ has at least $k$ positive and $k$ negative eigenvalues?

For $k=1$ a nice characterization is the following: there is a positive definite matrix $B$ orthogonal to $H$ (w.r.t. the scalar product $(A,B)=\mathrm{tr}(AB)$), or equivalently there exists a basis of $\mathbb C^n$ such that all matrices in $H$ have zero trace.

Even for $k=2$ I was not able to find any good characterization.

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Nitpick: You need to restrict attention to the nonzero matrices in H. And a trivial observation: Since H is closed under negation you only need a condition that ensures at least k positive eigenvalues. – Harald Hanche-Olsen Nov 9 '09 at 13:21
Thanks for the remark about the zero matrix. I preferred to formulate the question with k positive and k negative because it seems to me more intuitive (although equivalent). – user175348 Nov 9 '09 at 14:10

In an 2n-dimensional space the space of block matrices of the form

0 A*

A 0

have n positive and n negative eigenvalues.

They are plus or minus the singular values of A. (Meaning eigenvalues of |A|=(A*A)^(1/2)).

(This fact is in Bhatia's matrix analysis book.)

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Only if A is nondegenerate. Indeed in this case the proof is immediate: matrices of this form are nondegenerate, and the associated hermitian form admits an isotropic subspace of dim n. – user175348 Nov 9 '09 at 16:41
Since H is a subspace, it is implied by the question that the zero matrix satisfies the desired property. – Jon Nov 10 '09 at 16:24

There might be something in this paper:


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