proof of existence and uniqueness of singular value decomposition

Proof of existence and uniqueness of SVDFernando Sanz Gamiz


To prove existence of the SVD, we isolate the direction of the largest action of Am×n, and then proceed by inductionMathworldPlanetmath on the dimensionMathworldPlanetmathPlanetmathPlanetmath of A. We will denote hermitian conjugation by T. Norms for vectors in n will be the usual euclideanPlanetmathPlanetmath 2-norm =2 and for matrix the induced by norm of vectors.

Let σ1=A. By a compactness argument, there must be vectors v1n,u1*m with v1=1,u1*=σ1 and u1*=Av1. Normalize u1* by setting u1=u1*/u1* and consider any extensionsPlanetmathPlanetmath of v1 to an orthonormal basis {vi} of n and of u1 to an orthonormal basis {uj} of m; let U1 and V1 denote the unitary matricesMathworldPlanetmath with columns {vi} and {uj} respectively. Then we have


where 0 is a column vectorMathworldPlanetmath of dimension m-1, wT is a row vector of dimension n-1, and B is a matrix of dimension (m-1)×(n-1). Now,


so that S(σ12+w2)1/2. But U1 and V1 are unitary matrix, hence S=σ1; it therefore implies w=0.

If n=1 or m=1 we are done. Otherwise the submatrixMathworldPlanetmath B describes the action of A on the subspaceMathworldPlanetmathPlanetmath orthogonalMathworldPlanetmathPlanetmath to v1. By the induction hypothesis B has an SVD B=U2Σ2V2T. Now it is easily verified that


is an SVD of A. completing the proof of existence.

For the uniqueness let A=UΣVT a SVD for A and let ei denote the i-th, i=1min(m,n) vector of the canonical base of n. As U and V are unitary, Aei=σi2, so each σi is uniquely determined.

Title proof of existence and uniqueness of singular value decomposition
Canonical name ProofOfExistenceAndUniquenessOfSingularValueDecomposition
Date of creation 2013-03-22 17:07:46
Last modified on 2013-03-22 17:07:46
Owner fernsanz (8869)
Last modified by fernsanz (8869)
Numerical id 12
Author fernsanz (8869)
Entry type Proof
Classification msc 65-00
Classification msc 15-00