Positive definite kernel

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In operator theory, a positive definite kernel is a generalization of a positive matrix.

1 Definition
2 Characterization
- 2.1 Motivation
- 2.2 Kolmogorov decomposition
3 First consequences
4 References

[edit] Definition

Let

$\{ H_n \}_{n \in {\mathbb Z}}$

be a sequence of (complex) Hilbert spaces and

$\mathcal{L}(H_i, H_j)$

be the bounded operators from H_i to H_j.

A map A on Z × Z where A(i, j) lies in

$\mathcal{L}(H_i, H_j)$

is called a positive definite kernel if for all m > 0, the following positivity condition holds:

$\sum_{-m \leq i,j \leq m} \langle A(i,j) h_j, h_i \rangle \geq 0.$

[edit] Characterization

[edit] Motivation

Consider a positive matrix A ∈ C^{n × n}, whose entries are complex numbers. Every such matrix A has a "square root factorization" in the following sense:

A = B*B where B: Cⁿ → H_A for some (finite dimensional) Hilbert space H_A.

Furthermore, if C and G is another pair, C a operator and G a Hilbert space, for which the above is true, then there exists an unitary operator U: G → H_A such that B = UC.

The can be shown readily as follows. The matrix A induces a denegerate inner product <·, ·>_A given by <x, y>_A = <x, Ay>. Taking the quotient with respect to the degenerate subspace gives a Hilbert space H_A, a typical element of which is an equivalence class we denote by [x].

Now let B: Cⁿ → H_A be the natural projection map, Bx = [x]. One can calculate directly that

<x, B*By> = <Bx, By>_A = <[x], [y]>_A = <x, Ay>.

So B*B = A. If C and G is another such pair, it is clear that the operator U: G → H_A that takes [x]_G in G to [x] in H_A has the properties claimed above.

If {e_i} is a given orthonormal basis of Cⁿ, then {B_i = Be_i} are the column vectors of B. The expression A = B*B can be rewritten as A_{i, j} = B_i*B_j. By construction, H_A is the linear span of {B_i}.

[edit] Kolmogorov decomposition

This preceding discussion shows that every positive matrix A with complex entries can expressed as a Gramian matrix. A similar description can be obtained for general positive definite kernels, with an analogous argument. This is called the Kolmogorov decomposition:

Let A be a positive definite kernel. Then there exists a Hilbert space H_A and a map B defined on Z where B(n) lies in

$\mathcal{L}(H_n, H_A) \quad \mbox{such that} \quad A(i,j) = B^*(i)B(j) \quad \mbox{and} \quad H_A = \bigvee_{n \in {\mathbb Z}} B(n) H_n \;.$

The condition that H_A = ∨B(n)H_n is referred to as the minimality condition. Similar to the scalar case, this requirement implies unitary freedom in the decomposition:

If there is a Hilbert space G and a map C on Z that gives a Kolmogorov decomposition of A, then there is an unitary operator

$U: G \rightarrow H_A \quad \mbox{such that} \quad UC(n) = B(n) \quad \mbox{for all} \quad n \in {\mathbb Z}.$

[edit] First consequences

Positive definite kernels provide a framework that encompasses some basic Hilbert space constructions.

[edit] Reproducing kernel Hilbert space

Further information: Reproducing kernel Hilbert space

The definition and characterization of positive kernels extend verbatim to the case where the integers Z is replaced by an arbitrary set X. One can then give a fairly general procedure for constructing Hilbert spaces that is itself of some interest.

Consider the set F₀(X) of complex-valued functions f: X → C with finite support. With the natural operations, F₀(X) is called the free vector space generated by X. Let δ_x be the element in F₀(X) defined by δ_x(y) = δ_xy. The set {δ_x}_{x ∈ X} is a vector space basis of F₀(X).

Suppose now K: X × X → C is a positive definite kernel, then the Kolmogorov decomposition of K gives a Hilbert space

$({\mathcal H}, \langle \;, \; \rangle)$

where F₀(X) is "dense" (after possibly taking quotients of the degenerate subspace). Also, <[δ_x], [δ_y]> = K(x,y), which is a special case of the square root factorization claim above. This Hilbert space is called the reproducing kernel Hilbert space with kernel K on the set X.

Notice that in this context, we have (from the definition above)

$\{ H_n \}_{n \in {\mathbb Z}}$

being replaced by

$\{ {\mathbb C} \}_{x \in X}.$

Thus the Kolmogorov decomposition, which is unique up to isomorphism, starts with F₀(X).

One can readily show that every Hilbert space is isomorphic to a reproducing kernel Hilbert space on a set whose cardinality is the Hilbert space dimension of H. Let {e_x}_{x ∈ X} be an orthonormal basis of H The then kernel K defined by K(x, y) = <e_x, e_y> = δ_xy reproduces a Hilbert spaceH' . The bijection taking e_x to δ_x extends to an unitary operator from H to H' .

[edit] Direct sum and tensor product

This section is a stub. You can help by expanding it.

Let H(K, X) denote the Hilbert space corresponding to a positive kernel K on X × X. The structure of H(K, X) is encoded in K. One can thus describe, for example, the direct sum and the tensor product of two Hilbert spaces via their kernels.