The Singular Value Decomposition 2.3 2002/07/16 2002/07/23 00:00:00.003 GMT-5 Steven Cox cox@rice.edu Liqun Wang liqun@rice.edu Steven Cox cox@rice.edu singular value decomposition This module introduces the singular value decomposition. The singular value decomposition is another name for the spectral representation of a rectangular matrix. Of course if A is m-by-n and m n then it does not make sense to speak of the eigenvalues of A. We may, however, rely on the previous section to give us relevant spectral representations of the two symmetric matrices A A A A That these two matrices together may indeed tell us 'everything' about A can be gleaned from 𝒩 A A 𝒩 A 𝒩 A A 𝒩 A ℛ A A ℛ A ℛ A A ℛ A You have proven the first of these in a previous exercise. The proof of the second is identical. The row and column space results follow from the first two via orthogonality. On the spectral side, we shall now see that the eigenvalues of A A and A A are nonnegative and that their nonzero eigenvalues coincide. Let us first confirm this on the adjacency matrix associated with the unstable swing (see figure in another module) A 0 1 0 0 -1 0 1 0 0 0 0 1 The respective products are A A 1 0 0 0 2 0 0 0 1 A A 1 0 -1 0 0 1 0 0 -1 0 1 0 0 0 0 1 Analysis of the first is particularly simple. Its null space is clearly just the zero vector while λ1 2 and λ2 1 are its eigenvalues. Their geometric multiplicities are n1 1 and n2 2 . In A A we recognize the S matrix from the exercise in another moduleand recall that its eigenvalues are λ1 2 , λ2 1 , and λ3 0 with multiplicities n1 1 , n2 2 , and n3 1 . Hence, at least for this A, the eigenvalues of A A and A A are nonnegative and their nonzero eigenvalues coincide. In addition, the geometric multiplicities of the nonzero eigenvalues sum to 3, the rank of A. The eigenvalues of A A and A A are nonnegative. Their nonzero eigenvalues, including geometric multiplicities, coincide. The geometric multiplicities of the nonzero eigenvalues sum to the rank of A. If A A x λ x then x A A x λ x x , i.e., A x 2 λ x 2 and so λ ≥ 0. A similar argument works for A A . Now suppose that λj 0 and that x j , k k = 1 nj constitutes an orthogonal basis for the eigenspace ℛ Pj . Starting from A A x j , k λj x j , k we find, on multiplying through (from the left) by A that A A A x j , k λj A x j , k i.e., λj is an eigenvalue of A A with eigenvector A x j , k , so long as A x j , k 0 . It follows from the first paragraph of this proof that A x j , k λj , which, by hypothesis, is nonzero. Hence, 1 k nj y j , k A x j , k λj is a collection of unit eigenvectors of A A associated with λj. Let us now show that these vectors are orthonormal for fixed j. y j , i T y j , k 1 λj x j , i T A A x j , k x j , i T x j , k 0 We have now demonstrated that if λj 0 is an eigenvalue of A A of geometric multiplicity nj. Reversing the argument, i.e., generating eigenvectors of A A from those of A A we find that the geometric multiplicities must indeed coincide. Regarding the rank statement, we discern from that if λj 0 then x j , k ℛ A A . The union of these vectors indeed constitutes a basis for ℛ A A , for anything orthogonal to each of these x j , k necessarily lies in the eigenspace corresponding to a zero eigenvalue, i.e., in 𝒩 A A . As ℛ A A ℛ A it follows that dimℛ A A r and hence the nj, for λj 0 , sum to r. Let us now gather together some of the separate pieces of the proof. For starters, we order the eigenvalues of A A from high to low, λ1 λ2 … λh and write A A X Λn X where Xj x j , 1 … x j , n j X X1 … Xh and Λn is the n-by-n diagonal matrix with λ1 in the first n1 slots, λ2 in the next n2 slots, etc. Similarly A A Y Λm Y where Yj y j , 1 … y j , n j Y Y1 … Yh and Λm is the m-by-m diagonal matrix with λ1 in the first n1 slots, λ2 in the next n2 slots, etc. The y j , k were defined in under the assumption that λj 0 . If λj 0 let Yj denote an orthonormal basis for 𝒩 A A . Finally, call σj λj and let Σ denote the m-by-n matrix diagonal matrix with σ1 in the first n1 slots, σ2 in the next n2 slots, etc. Notice that Σ Σ Λn Σ Σ Λm Now recognize that may be written A x j , k σj y j , k and that this is simply the column by column rendition of A X Y Σ As X X I we may multiply through (from the right) by X and arrive at the singular value decomposition of A, A Y Σ X Let us confirm this on the A matrix in . We have Λ4 2 0 0 0 0 1 0 0 0 0 1 0 0 0 0 X 1 2 -1 0 0 1 0 2 0 0 1 0 0 1 0 0 2 0 Λ3 2 0 0 0 1 0 0 0 1 Y 0 1 0 1 0 0 0 0 1 Hence, Σ 2 0 0 0 0 1 0 0 0 0 1 0 and so A Y Σ X says that A should coincide with 0 1 0 1 0 0 0 0 1 2 0 0 0 0 1 0 0 0 0 1 0 -1 2 0 1 2 0 0 1 0 0 0 0 0 1 1 2 0 1 2 0 This indeed agrees with A. It also agrees (up to sign changes on the columns of X) with what one receives upon typing [Y, SIG, X] = scd(A) in Matlab. You now ask what we get for our troubles. I express the first dividend as a proposition that looks to me like a quantitative version of the fundamental theorem of linear algebra. If Y Σ X is the singular value decomposition of A then The rank of A, call it r, is the number of nonzero elements in Σ. The first r columns of X constitute an orthonormal basis for ℛ A . The n r last columns of X constitute an orthonormal basis for 𝒩 A . The first r columns of Y constitute an orthonormal basis for ℛ A . The m r last columns of Y constitute an orthonormal basis for 𝒩 A . Let us now 'solve' A x b with the help of the pseudo-inverse of A. You know the 'right' thing to do, namely reciprocate all of the nonzero singular values. Because m is not necessarily n we must also be careful with dimensions. To be precise, let Σ+ denote the n-by-m matrix whose first n1 diagonal elements are 1 σ1 , whose next n2 diagonal elements are 1 σ2 and so on. In the case that σh 0 , set the final nh diagonal elements of Σ+ to zero. Now, one defines the pseudo-inverse of A to be A+ X Σ+ Y In the case of that A is that appearing in we find Σ+ 2 0 0 0 1 0 0 0 1 0 0 0 and so A+ -1 2 0 1 2 0 0 1 0 0 0 0 0 1 1 2 0 1 2 0 1 2 0 0 0 1 0 0 0 1 0 0 0 0 1 0 1 0 0 0 0 1 therefore, A+ 0 -1 2 0 1 0 0 0 1 2 0 0 0 1 in agreement with what appears from pinv(A). Let us now investigate the sense in which A+ is the inverse of A. Suppose that b m and that we wish to solve A x b . We suspect that A+ b should be a good candidate. Observe by that A A A+ b X Λn X X Σ+ Y b because X X I A A A+ b X Λn Σ+ Y b by and A A A+ b X Σ Σ Σ+ Y b because Σ Σ Σ+ Σ A A A+ b X Σ Y b by A A A+ b A b that is A+ b satisfies the least-squares problem A A x A b .