Home Health

# Singular Value Decomposition

A singular value decomposition (SVD) is powerful for analyzing matrices and for analyzing linear mappings. Applying the SVD transforms an M x N real matrix A into the following form:

where U is a M x N matrix of which column vectors are unit vectors and are orthogonal to each other and V is an N x N orthonormal matrix such that

Here, ui (i = 1,2, ...,M) is a unit M-vector and ujuj = 0 if i ф j, and vi (i =

1,2 , , M) is a unit N-vector and v j Vj = 0 if i ф j. It should be noted that

{vi|i = 1,2 , ..., N} is an orthonormal basis of the N-dimensional space. E in (2.52) is an N x N diagonal matrix,

where the scalars, oi (i = 1,2 , ..., N), are called singular values. In the following equations, it is assumed that the singular values are in decreasing order: ст1 > o2 > •••> on .

Let y denote an M-vector generated from an N-vector by a linear mapping such that

where A is a M x N matrix. Applying the SVD to A, the range of the linear mapping and of its zero-space can be derived, as will be described below. Substituting A = UE VT produces the following equation:

The last two factors, VTx, are transforming the coordinates of x using the orthonormal basis {vi|i = 1,2 , ..., N}, as described in the Sect.2.2.2.1. Let the new coordinates be denoted by a N-vector, z = [zi, z2, ..., zN ]T = VTx. Now, the linear mapping shown in (2.56) is represented as

Substituting E = diag(o1,o2, ..., on) and U = [u1|u2|... |uN], (2.57) can be rewritten as follows:

Let a set of indexes that indicates the singular values of zeros be denoted by Z = { j|j e {1,2 ,..., N}, 0}, and let a set of indexes for nonzero singular values be denoted by N = {jj e {1,2 , ..., N}, oj- ф 0}, where Z U N = {1,2 , ..., N} and Z П N = ф. Then, the M-vectors, uj, that correspond to the nonzero singular values, j e N, determine the domain of the linear map shown in (2.58): for any input vector, x, the mapped vector,y, that is in the subspace spanned by {uj|j e N}. The N-vectors, vk, that correspond to the zero-singular values, k e Z, determine the zero-space of A: the mapped vector, y, is always zero if the input vector, x, is in a subspace that is spanned by the N-vectors that correspond to the nonzero singular values, that is, x e span(VZ) where VZ = {vl e Z}. This is because, in (2.58), a; = 0 for i e Z and z; = vfx = 0 for i e N.

 Related topics