Suppose we have a finite square n x n matrix of complex numbers H that is Hermitian and skew-symmetric:
$H^\dagger = H$ and $H^T = -H$.
(T denotes transpose, $\dagger$ denote conjugate transpose. I know that these conditions mean that H is a purely imaginary skew-symmetric matrix.)
It is a textbook result that these two conditions ensure that it's eigenvalues are real, its rank is even and its eigenvalues appear in positive and negative pairs.
If the normalised, orthonormal eigenvectors associated to non-zero eigenvalues are denoted by $u_i$ and $v_i$ and the positive eigenvalues are denoted $\lambda_i$ then we have:
$H u_i = \lambda_i u_i$ and $H v_i = -\lambda_i v_i$
where i=1,2,…,s where 2s is the rank of H. The eigenvectors can be chosen to be complex conjugates of each other: $u_i = v_i^*$.
In terms of these eigenvectors and eigenvalues we can write the eigen-decomposition of H as:
$H = \sum_{i=1}^s \lambda_i u_i u_i^\dagger – \lambda_i v_i v_i^\dagger$
We can now define a new matrix Q as just the "positive eigenvalue part" of H:
$Q := \sum_{i=1}^s \lambda_i u_i u_i^\dagger$
This Q is Hermitian, positive semi-definite and satisfies $H = Q – Q^T = Q – Q^*$. Apparently this Q is also the "closest Hermitian positive semi-definite matrix" to H, as measured in the Frobenius norm (and possibly other norms too).
This all goes through smoothly for finite n x n matrices H.
My question is, if H is now a countably infinite dimensional matrix $H_{xy}$ for x,y=1,2,…,$\infty$ which satisfies the same conditions as before:
$H_{xy} = H_{yx}^*$ and $H_{xy} = -H_{yx}$
can we do the same construction and obtain the matrix Q?
It seems like the rigorous way to do this requires treating the infinite matrix as an operator on the $\ell^2$ Hilbert space of infinite square-summable sequences. For general H it seems like this will be an unbounded operator that is probably not defined on the whole Hilbert space (due to issues of convergence when doing the infinite matrix multiplication).
In a best-case scenario we'd like H to define a self-adjoint operator on $\ell^2$. We could then (presumably) apply the spectral theorem and sum the positive eigenvalue part to get a Q operator/infinite-matrix.
What conditions do we have to impose on the infinite H matrix entries to ensure that the Q matrix exists? or to ensure that H defines a self-adjoint operator?
Can we sidestep the use of the eigenvectors and eigenvalues when defining Q and instead seek Q as the "closest positive semi-definite matrix" to H? Does this even make sense in the infinite matrix case?
Can we calculate Q from H if H satisfies the right conditions?
Any thoughts greatly appreciated.
Best Answer
Not an answer really, but a collection of several comments.
The "skew-symmetric" condition is not really natural for an operator on a complex Hilbert space, since it isn't preserved by unitary transformations.
Do you have a reference for the statement that Q is the closest Hermitian positive semidefinite matrix to H in Frobenius norm? Does this rely in an essential way on H being skew-symmetric?
The "positive part" construction would apply to any (possibly unbounded) self-adjoint operator, using an appropriate version of the spectral theorem; self-adjoint operators can be "diagonalized" in a certain general sense. (One version says that, up to a unitary transformation, your Hilbert space is $L^2(X,\mu)$ for some measure space $(X,\mu)$, and your operator is multiplication by some real-valued function $h$ on $X$. So the positive part corresponds to multiplication by the positive part of $h$.)
I am not aware of a condition on the entries of an infinite matrix that's equivalent to the corresponding operator being self-adjoint (though I'd be interested to know if there is). Self-adjointness is a fairly delicate property, in general; it requires the domain of the operator to be neither too large nor too small.
You could certainly seek the nearest positive semidefinite Hermitian operator to a given one, with respect to some norm. However you are restricting yourself to those operators for which that norm is finite. The Hilbert-Schmidt norm might be natural as it generalizes the Frobenius norm; the operator norm is another choice. The positive semidefinite Hermitian operators are closed under both norms, so looking for a "nearest" one makes sense. Also, Hilbert-Schmidt operators, being compact, are diagonalizable in the more usual sense (there is an orthonormal basis of eigenvectors).