The discussion about polar decomposition reminds me of an interesting interpretation of the SVD that isn't often mentioned: Let P, Q be two K-dimensional planes in R^L. To describe their relative position in R^L, it suffices to find the smallest angle theta_1 between any vector v in P and any vector w in Q; then to iteratively do the same for <v,w>^perp. (I.e., in the space <v,w>^perp we find the least angle between the subspaces (P \int <v,w>^perp) and (Q \int <v,w>^perp). This gives a sequence of angles theta_1,...,theta_K. (If P and Q intersect nontrivially, some of these will be 0.) Associated to these angles there are also orthonormal bases for P and Q consisting of the vectors that were found to be nearest. Now consider any orthonormal bases V = (v_1,...,v_K) of P and W = (w_1,...,w_K) of Q. Form their dot-product matrix: M = V^t W whose (i,j)th entry is v_i*w_j. Then by twiddling these bases by orthogonal matrices R, S, to get new bases VR and WS, it can be arranged that (VR)^t (WS) = D where D is diagonal with elements d_j = cos(theta_j) for all j = 1,...,K. Hence R^t V^t W S = D, so V^t W = R D W^t, i.e., M = R D W^t. The singular values -- the (necessarily nonnegative) cosines of the theta_j's that are the elements of D -- satisfy: M^t M = W D R^t R D W^t = W D^2 W^t and so the singular values are the square roots of the eigenvalues of the (symmetric positive semi-definite matrix) M^t M. --------------------------------------------------------- Thus the singular values are a complete set of invariants for the relative position of the K-planes P and Q in R^L. I.e., given any rotation O of R^L, then starting with the subspaces OP and OQ in the above will result in the same D, and conversely. --------------------------------------------------------- --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele