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Skeleton CUR is exact for a full-rank cross

Statement

Let $F \in {R, C}$ , let $m, n \geq 1$ , and let $A \in F^{m \times n}$ . Choose an integer $0 \leq k \leq \min (m, n)$ and ordered lists $I$ and $J$ of $k$ distinct valid row and column indices, respectively. Let $C = A (:, J)$ , $R = A (I, :)$ , and $W = A (I, J) \in F^{k \times k}$ as in Skeleton and projected least-squares CUR cores. If $rank (A) = rank (W) = k$ , then the skeleton CUR reconstruction satisfies $C W^{†} R = A$ . For $k = 0$ , use the empty-matrix convention: a rank-zero $A$ is zero and the empty product reconstructs the zero $m \times n$ matrix.

Facts & Assumptions

Given: A finite matrix $A$ , ordered row and column selections of common size $k$ , and $rank (A) = rank (W) = k$ .

[F1]

Skeleton and projected least-squares CUR cores: $C = A (:, J)$ , $R = A (I, :)$ , $W = A (I, J)$ , and the skeleton reconstruction is $C W^{†} R$ .

[F2]

The Moore--Penrose pseudoinverse $A^{+}$ as the solution of the four Penrose equations: a Moore-Penrose inverse satisfies $W W^{†} W = W$ , $W^{†} W W^{†} = W^{†}$ , and both square products are self-adjoint.

[F3]

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

[F4]

Rank and nullity of a linear map with finite-dimensional domain: the rank of a finite-dimensional linear map is the dimension of its image.

[F5]

Rank-nullity: $\dim_{F} V = nullity T + rank T$ : for a linear map with finite-dimensional domain, dimension of the domain is nullity plus rank.

Proof

1.1givenF4algebra

If $k = 0$ , the hypothesis $rank (A) = 0$ and [F4] imply $Ran (A) = {0}$ , so $A = 0$ . The stated empty product is also zero. Assume henceforth $k > 0$ .

1.2givenF4F5

The square matrix $W$ has rank $k$ in a $k$ -dimensional domain and codomain. By [F4] and [F5] its kernel has dimension zero and its image has dimension $k$ , so it is both injective and surjective and has an inverse $W^{- 1}$ .

1.3F2F3algebra

The matrix $W^{- 1}$ satisfies all four equations in [F2]: $W W^{- 1} W = W$ , $W^{- 1} W W^{- 1} = W^{- 1}$ , and $W W^{- 1} = W^{- 1} W = I_{k}$ are self-adjoint. By uniqueness [F3], $W^{†} = W^{- 1}$ .

2.1givenF4F5step 1.2algebra

If $C x = 0$ , restricting to rows $I$ gives $W x = 0$ , and invertibility of $W$ from step 1.2 gives $x = 0$ . Hence $C$ is injective and has rank $k$ by [F4] and [F5]. Since the columns of $C$ are columns of $A$ , $Ran (C) \subseteq Ran (A)$ ; both spaces have dimension $k$ , so they are equal: if the inclusion were proper, a basis of $Ran (C)$ together with a vector in $Ran (A) ∖ Ran (C)$ would give $k + 1$ linearly independent vectors in a $k$ -dimensional space.

3.1F1step 1.3step 2.1constructalgebra∎

Equality of these column spaces supplies a matrix $X$ with $A = C X$ . Restricting to rows $I$ yields $R = W X$ , so $X = W^{- 1} R$ . Therefore $A = C W^{- 1} R = C W^{†} R$ by step 1.3 and [F1].

Depends on

Used by

An exact rank-two CUR reconstruction from a nonsingular cross Example

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Sources

Per-Gunnar Martinsson and Joel A. Tropp, Randomized Numerical Linear Algebra: Foundations & Algorithms, §13.1 (standard reference, not scraped)
Nathan Halko, Per-Gunnar Martinsson, and Joel A. Tropp, Finding Structure with Randomness, §3.2.3 (standard reference, not scraped)