# Infinite hermitian matrix

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.

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1. The "skew-symmetric" condition is not really natural for an operator on a complex Hilbert space, since it isn't preserved by unitary transformations.

2. 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?

3. 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 $h$.)

4. 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.

5. 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).

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The reference for Q being the closest Hermitian positive semidefinite matrix to H is Theorem 9, p324 in "The electrical engineering handbook" By Richard C. Dorf available <a href="books.google.co.uk/…;. I'm not an electrical engineer, I just came across this reference on google. The skew-symmetric property is not needed explicitly, only the fact that H is Hermitian (the "negative part" is then the closest Hermitian negative semi-definite matrix to H). –  StevenJ Jun 4 '10 at 9:27

The answer is "yes," provided you can make a good self-adjoint operator out of $H_{x,y}$. The skew-symmetric structure you mention is actually quite natural in some contexts. In quantum mechanics complex conjugation usually represents the time reversal symmetry. The structure here involves three things: a Hilbert space, a conjugation $J$ on the Hilbert space, and a symmetric operator $H$. Your Hilbert space is $\ell^2$ and the conjugation $J$ is $J\psi = \overline{\psi}$ where $\overline{\psi}$ is complex conjugation. (Note that $J$ is a real linear operator, but is complex "skew" linear that is $J\alpha \psi = \overline{\alpha} J\psi$ for a scalar $\alpha$.)

To begin suppose that the symmetric operator $H$ defines a bounded operator. That is suppose $$H \psi(x) = \sum_{y} H_{x,y}\phi(y),$$ makes sense for all $\psi \in \ell^2$ and there is a $C <\infty$ such that $\|H\psi\|_2 \le C \|\psi\|_2$. Your assumption of skew-symmetry for $H$ shows that $$JHJ = -H.$$ It follows that $J p(H) J= \overline{p}(-H)$ for any polynomial $p$, where $\overline{p}$ is the polynomial with conjugate coefficients. (To see this it helps to note that $J^2=1$.) It follows then that $$Jf(H)J=\overline{f}(-H)$$ for any continuous function $f$, with $f(H)$ defined by the functional calculus. Now we can define $Q$, namely $$Q = g(H)$$ where $g(x)= x$ for $x >0$ and $g(x)=0$ for $x\le 0$. Note that $Q$ is positive, $Q$ and $JQJ$ have orthogonal ranges, and $$H= Q - JQJ.$$

However, $H$ as defined above may or may not make sense on the whole Hilbert space. As Jiri Lebl pointed out we must at least assume that $$\sum_{x} |H_{x,y}|^2 <\infty$$ for each $y$. Then we may certainly define an operator on the space $C_f$ of sequences with finite support. Let us call this operator $H_f$ to remind ourselves that is defined on $C_f$. To apply the functional calculus to get $Q$ we need a self-adjoint extension $H$.

The condition for a self adjoint extension to exist is $$\dim \ker (H_f^\dagger +i)=\dim \ker (H_f^\dagger -i)$$ where $H_f^\dagger$ is the adjoint: $$(H_f^\dagger\phi, \psi)= (\phi, H_f\psi)$$ on the domain $\mathcal D^\dagger$ of sequences $\phi$ such that $|(\phi,H_f\psi)|\le C \| \phi \|$. If this condition holds you now have to pick a self-adjoint extension $H$, and pick it carefully so as that $J H J =-H$, and then proceed as in the bounded case. (The extension is unique only if both dimensions are $0$.) I am not aware of a general condition in terms of $H_{x,y}$ for the existence of a self-adjoint extension, but there is a nice review article of Bary Simon from the late 90's in which, among other things, he analyzes under which conditions tri-diagonal matrices ("Jacobi matrices") have self-adjoint extensions and when they are unique.

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See Akhizer and Glazman's Theory of Linear Operators in Hilbert Space (can be had cheaply as a dover book). See in particular Section 47 on unbounded operators (and 26 for the everywhere defined compact case). I am not touching the skew symmetry thing, I'll just answer about the self-adjointness and spectral theorem application in $\ell^2$.

So I'm assuming that the matrix $H$ is conjugate symmetric (Hermitian).

1) If the entries in the matrix are square summable (sufficient but not necessary) then $H$ defines an everywhere-defined compact self-adjoint operator on $\ell^2$. You can apply the spectral theorem and everything is all wonderful. Then you simply need to worry about that that skew symmetry thing. See page 53 for a necessary and sufficient condition on the matrix to define a bounded operator.

2) (Theorem 4 on page 102) If the columns of the matrix are square summable, then $H$ does define a closed self-adjoint operator. However: the matrix does not transform natural under unitary operators. That is, if you take a unitary matrix and "do" the product formally, the new matrix may not define an operator and even if it does define an operator it may turn out to be a different operator then when you compose the operators. So any sort of formal manipulations on this matrix are just that. They do not necessarily correspond to the same operations in $\ell^2$. Therefore it is unlikely that you can apply Hilbert space theory results to the matrices.

3) If the columns of the matrix are not square summable, there is no hope of using Hilbert spaces at all.

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Theorem 4 on page 102 ensures that the infinite matrix defines a closed symmetric operator. There is no reason for the operator to be self-adjoint without extra assumptions. –  Samuel Monnier Dec 16 '13 at 14:19