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Nyström is a square-root orthogonal-projection compression

Statement

Let $F \in {R, C}$ , $n \geq 1$ , $s \geq 0$ , $A \in F^{n \times n}$ be Hermitian positive semidefinite, $S \in F^{n \times s}$ , and $B = A^{1 / 2} S$ . If $P_{Ran (B)}$ denotes the orthogonal projection onto the range of $B$ , then

${\hat{A}}_{n y s} = A^{1 / 2} P_{Ran (B)} A^{1 / 2} = A^{1 / 2} B (B^{*} B)^{†} B^{*} A^{1 / 2} = A^{1 / 2} B B^{†} A^{1 / 2} .$

Facts & Assumptions

Given: A finite Hermitian positive-semidefinite matrix $A$ , a sketch $S$ , and $B = A^{1 / 2} S$ .

[F1]

The positive-semidefinite Nyström approximation: ${\hat{A}}_{n y s} = A S (S^{*} A S)^{†} S^{*} A$ .

[F2]

The Moore-Penrose products are orthogonal projectors: for any finite $B$ , $B B^{†}$ and $B^{†} B$ are orthogonal projections onto $Ran (B)$ and $Ran (B^{†})$ .

[F3]

Every finite real or complex matrix has a unique Moore--Penrose pseudoinverse: every finite matrix has a unique Moore-Penrose inverse.

[F4]

A non-negative operator has a unique non-negative square root: $A$ has a unique Hermitian positive-semidefinite square root $A^{1 / 2}$ satisfying $(A^{1 / 2})^{2} = A$ .

[F5]

The Moore--Penrose pseudoinverse $A^{+}$ as the solution of the four Penrose equations: a Moore-Penrose inverse is specified by the four Penrose equations.

Proof

1.1F1F4algebra

Set $D = B^{*} B = S^{*} A S$ . From $(A^{1 / 2})^{2} = A$ and $A^{1 / 2} = (A^{1 / 2})^{*}$ , [F1] rewrites as ${\hat{A}}_{n y s} = A^{1 / 2} B D^{†} B^{*} A^{1 / 2}$ .

1.2F3F5algebra

Since $D = D^{*}$ , taking adjoints of the four defining equations [F5] shows $(D^{†})^{*}$ also satisfies them for $D$ . Uniqueness [F3] gives $(D^{†})^{*} = D^{†}$ . Thus $P = B D^{†} B^{*}$ is self-adjoint, and $P^{2} = B D^{†} (B^{*} B) D^{†} B^{*} = B D^{†} D D^{†} B^{*} = P$ by the Moore-Penrose equation $D^{†} D D^{†} = D^{†}$ . Hence $P$ is an orthogonal projection.

1.3F2F3F5algebra

Put $Q = D^{†} D$ . By [F2], $Q$ is an orthogonal projection. The equations [F5] $D D^{†} D = D$ and $D^{†} D D^{†} = D^{†}$ imply $D Q = D$ ; therefore $Q x = 0$ iff $D x = 0$ . Also $x^{*} D x = ∥ B x ∥_{2}^{2}$ , so $\ker D = \ker B$ . Since $Q$ is an orthogonal projection, $Ran (I - Q) = \ker Q = \ker B$ , and consequently $B (I - Q) = 0$ , or $B Q = B$ .

2.1F2step 1.1step 1.3algebra∎

Step 1.3 gives $P B = B D^{†} D = B Q = B$ . Since $P$ has range contained in $Ran (B)$ and fixes every vector in $Ran (B)$ , its range is exactly $Ran (B)$ . Thus $P = P_{Ran (B)}$ . The projector $B B^{†}$ has the same range by [F2]. For each vector, orthogonal projection decomposes it uniquely into a component in $Ran (B)$ and one in its orthogonal complement, so these two projectors coincide. Substituting this identity into step 1.1 proves all displayed formulas.

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Sources

Per-Gunnar Martinsson and Joel A. Tropp, Randomized Numerical Linear Algebra: Foundations & Algorithms, §14.2 (standard reference, not scraped)
Alex Gittens and Michael W. Mahoney, Revisiting the Nyström Method, Lemma 1 and §4.1 (standard reference, not scraped)