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The rank of a linear map is the number of its nonzero singular values
Statement
Let be a linear map between finite-dimensional real or complex inner product spaces. Then equals the number of positive singular values of .
Facts & Assumptions
Given: A linear map between finite-dimensional real or complex inner product spaces, and an SVD with .
Every linear map admits a singular value decomposition (Every linear map between finite-dimensional real or complex inner product spaces admits a singular value decomposition).
Rank-nullity holds for linear maps with finite-dimensional domain (Rank-nullity: ).
Proof
By [L1], one has exactly when for , so and therefore .
By [L2], . Combining this with step 1.1 gives , the number of positive singular values.
Depends on
Used by
- If A has full column rank, then A^+=(A^*A)⁻¹A^* Proposition
- In the spectral norm, numerical rank counts the singular values above the declared threshold Proposition
- AA^+ and A^+A are the orthogonal projections onto imA and imA^* Theorem
- Exact rank is locally constant at full-rank matrices and discontinuous at nonmaximal-rank matrices Theorem
- For full-column-rank A, the normal equations square the spectral condition number Theorem
- Spectral and Frobenius norms are unitarily invariant, are given by singular values, and satisfy the sharp rank comparison Theorem
- The Moore--Penrose pseudoinverse is continuous on each fixed-rank stratum and is not continuous across rank loss Theorem
- The reciprocal spectral condition number is the relative distance to singularity Theorem
Dependency tree · two levels
11 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, fourth edition (standard reference, not scraped)