如何用CUR分解替代SVD分解？特性对比与降维方法咨询

Great question—CUR and SVD are both foundational for low-rank approximation, but their differences in interpretability and implementation make them suited for different scenarios. Let’s tackle your three questions one by one:

1. How to Replace SVD with CUR Decomposition #

SVD gives you an optimal (but abstract) low-rank approximation of your matrix A = UΣV^T, while CUR provides a data-driven, interpretable approximation A ≈ CUR, where:

• C is a subset of k columns from A
• R is a subset of k rows from A
• U is a small k×k matrix computed as U = (C^T C)^{-1} C^T A R (R R^T)^{-1}

To use CUR instead of SVD, follow this workflow:

• Step 1: Select k columns for C and k rows for R. The standard method uses statistical leverage scores (a measure of how "important" a row/column is to the matrix's structure) to pick these subsets—this ensures the approximation is as accurate as possible.
• Step 2: Compute the intermediate matrix U using the formula above.
• Step 3: Use CUR for your task instead of UΣV^T. For example:
  • For matrix reconstruction: Use CUR to approximate A directly.
  • For dimensionality reduction: Use the C/R matrices (with their pseudo-inverses) to project data (we’ll cover this in question 3).

The key reason to choose CUR over SVD is interpretability: since C and R are actual rows/columns from your original data, you can trace back the reduced dimensions to real features/samples, which is impossible with SVD’s abstract U/V bases.

2. Do C/R Have the Same Properties as SVD’s U/V? #

Short answer: No—they’re designed for different goals, so their properties differ significantly:

• Interpretability:
  • C = actual columns from A, R = actual rows from A → you can directly map each column/row in C/R to a real feature/sample in your dataset.
  • U/V = orthogonal bases derived from linear combinations of A’s rows/columns → these vectors have no direct correspondence to your original data, making them black boxes for interpretation.
• Orthogonality:
  • SVD’s U and V are strictly orthogonal matrices (U^T U = I, V^T V = I), which simplifies many mathematical operations.
  • C and R are almost never orthogonal (unless you explicitly select orthogonal subsets, which is rare in practice).
• Approximation Optimality:
  • SVD provides the optimal low-rank approximation of A (minimizes the Frobenius norm of the error ||A - UΣV^T||).
  • CUR is a near-optimal approximation—its error is bounded by a small multiple of the SVD error, but it trades off a tiny bit of accuracy for interpretability.
• Sparsity:
  • If your original matrix A is sparse, C and R will also be sparse, which makes CUR much more memory-efficient for large datasets.
  • U and V from SVD are almost always dense, even if A is sparse.

3. Which Matrix to Use for Reducing Dimensions from n to k? #

This depends on whether you’re reducing the dimensionality of samples (rows) or features (columns):

• Reducing sample (row) dimensionality (n → k):
  For a row vector a (1×n) representing a sample, project it to k dimensions using:

  a_projected = a * C * (C^T C)^{-1}
  

  Here, C acts as an interpretable basis for the column space of A, and (C^T C)^{-1} adjusts for the non-orthogonality of C. This replaces the SVD step of multiplying by U_k (the first k columns of U).

• Reducing feature (column) dimensionality (n → k):
  For a column vector x (n×1) representing a feature, project it to k dimensions using:

  x_projected = (R R^T)^{-1} R^T * x
  

  This uses R as a basis for the row space of A, replacing the SVD step of multiplying by V_k^T (the first k rows of V^T).

Alternatively, if you want a symmetric approach for the full matrix, you can approximate the reduced-rank version of A as C * U * R, where each row of R gives a k-dimensional representation of the original rows, and each column of C gives a k-dimensional representation of the original columns.

内容的提问来源于stack exchange，提问作者Prathamesh Raut