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In linear algebra, the singular value decomposition (SVD) is a factorization of a real or complex matrix into a rotation, followed by a rescaling followed by another rotation. It generalizes the eigendecomposition of a square normal matrix with an orthonormal eigenbasis to any m × n {\displaystyle m\times n} matrix.
Let A be a square n × n matrix with n linearly independent eigenvectors q i (where i = 1, ..., n).Then A can be factored as = where Q is the square n × n matrix whose i th column is the eigenvector q i of A, and Λ is the diagonal matrix whose diagonal elements are the corresponding eigenvalues, Λ ii = λ i.
The singular values are non-negative real numbers, usually listed in decreasing order (σ 1 (T), σ 2 (T), …). The largest singular value σ 1 (T) is equal to the operator norm of T (see Min-max theorem). Visualization of a singular value decomposition (SVD) of a 2-dimensional, real shearing matrix M.
Unit-Scale-Invariant Singular-Value Decomposition: =, where S is a unique nonnegative diagonal matrix of scale-invariant singular values, U and V are unitary matrices, is the conjugate transpose of V, and positive diagonal matrices D and E.
In machine learning, kernel functions are often represented as Gram matrices. [2] (Also see kernel PCA) Since the Gram matrix over the reals is a symmetric matrix, it is diagonalizable and its eigenvalues are non-negative. The diagonalization of the Gram matrix is the singular value decomposition.
The singular value decomposition of a matrix is = where U and V are unitary, and is diagonal.The diagonal entries of are called the singular values of A.Because singular values are the square roots of the eigenvalues of , there is a tight connection between the singular value decomposition and eigenvalue decompositions.
Truncated singular value decomposition (SVD) in numerical linear algebra can also use the Rayleigh–Ritz method to find approximations to left and right singular vectors of the matrix of size in given subspaces by turning the singular value problem into an eigenvalue problem.
The matrices R 1, ..., R k give conjugate pairs of eigenvalues lying on the unit circle in the complex plane; so this decomposition confirms that all eigenvalues have absolute value 1. If n is odd, there is at least one real eigenvalue, +1 or −1; for a 3 × 3 rotation, the eigenvector associated with +1 is the rotation axis.