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A finite positive semidefinite kernel Gram matrix has a Euclidean feature factorization

Statement

Let $k$ be a real positive semidefinite kernel on a set $X$ , let $n \geq 1$ , and let $x_{1}, \dots, x_{n} \in X$ . There are vectors $z_{1}, \dots, z_{n} \in R^{n}$ such that $k (x_{i}, x_{j}) = z_{i}^{T} z_{j}$ for every $1 \leq i, j \leq n$ . The vectors may be zero and need not be linearly independent.

Statement references

Sheldon Axler, Linear Algebra Done Right, 4th ed., §7B, Theorem 7.29

Hastie, Tibshirani, and Friedman, The Elements of Statistical Learning, §12.3.1

Facts & Assumptions

Given: A real positive semidefinite kernel $k$ , an integer $n \geq 1$ , and points $x_{1}, \dots, x_{n} \in X$ .

[A1]

The Gram matrix $K = (k (x_{i}, x_{j}))_{i, j = 1}^{n}$ is real symmetric and satisfies $c^{T} K c \geq 0$ for every $c \in R^{n}$ (Positive semidefinite kernels and reproducing kernel Hilbert spaces).

[F1]

Every self-adjoint endomorphism of a finite-dimensional real inner-product space has an orthonormal eigenbasis (Real spectral theorem: a self-adjoint endomorphism of a finite-dimensional real inner product space has an orthonormal eigenbasis).

Proof

1.1givenA1F1construct

Regard $K$ as an endomorphism of $R^{n}$ with its usual dot product. By [A1], $K$ is symmetric, hence self-adjoint. By [F1], there are an orthonormal eigenbasis $q_{1}, \dots, q_{n}$ and real eigenvalues $λ_{1}, \dots, λ_{n}$ with $K q_{ℓ} = λ_{ℓ} q_{ℓ}$ .

2.1step 1.1A1

Since $∥ q_{ℓ} ∥_{2} = 1$ , [A1] gives $λ_{ℓ} = q_{ℓ}^{T} K q_{ℓ} \geq 0$ for each $ℓ$ . Thus each $\sqrt{λ_{ℓ}}$ is a real number, including when $λ_{ℓ} = 0$ .

3.1step 1.1step 2.1constructalgebra∎

For $1 \leq i \leq n$ , define $z_{i} = (\sqrt{λ_{1}} (q_{1})_{i}, \dots, \sqrt{λ_{n}} (q_{n})_{i}) \in R^{n}$ . The spectral decomposition $K = \sum_{ℓ = 1}^{n} λ_{ℓ} q_{ℓ} q_{ℓ}^{T}$ then gives $z_{i}^{T} z_{j} = \sum_{ℓ = 1}^{n} λ_{ℓ} (q_{ℓ})_{i} (q_{ℓ})_{j} = K_{i j} = k (x_{i}, x_{j})$ for all $i, j$ .

Depends on

Used by

Soft-margin SVM duality and KKT conditions Theorem

Dependency tree · two levels

9 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Sheldon Axler, Linear Algebra Done Right, 4th ed., §7B, Real Spectral Theorem (standard reference, not scraped)
Hastie, Tibshirani, and Friedman, The Elements of Statistical Learning, 2nd ed., corrected 12th printing (standard reference, not scraped)