Hello, I have a matrix M of dimensions m * n, and I have the decomposition into singular values M = U * S * V, I want to know if it's possible to recover the matrix M without having U and V, but having only the S ? Thank you so much!
Unfortunately No.
The S matrix provides only the set of Singular values, but no information on Column space and Row space of M, which is essential to construct U and V matrices and hence Reconstruct M.
Debopam Ghosh please tell me, we must have all 3 informations at the same time and it is impossible to reconstruct with S or V or U only, even with 2 like U and V ?
In the Singular value decomposition of M= U*S*V', each of the three component matrices U, S and V are providing separate pieces of information which together is used to generate M.
A Real Matrix M represent a correlation beetween two bases, one in its Column space and the other in its Row space, this is represented by the Full Rank factorization of M :
M(m by n)= P(m by r)*Q(r by n), here P is the Basis matrix in Column space of M, Q' is the Basis matrix in Row space of M, correlated to P, here r is the rank of M.
By multiplying with an arbitrary Invertible matrix X(r by r) and invX(r by r) as P*X*invX*Q = P1*Q1 = M, we see that the bases themselves are not unique but the correlation is preserved and so is M, under such Basis transformations, here inv X stands for inverse of X.
However, if we take Y(r by r) as a different invertible matrix, Y not inverse of X, then P*X*Y*Q = N( m by n), here M and N are two different matrices having all four Fundamental subspaces ( Column space, Left Nullspace, Row space, Nullspace ) identical.
In general for any real M, we have M(m by n)= Q(m by r)*C(r by r)*R( r by n), where Q and R' are Orthonormal bases of Column space and Row space respectively of M, interrelated by the Invertible, non diagonal matrix C, it is a special case of the above full rank factorization, this special case becomes the SVD decomposition when we construct U and V in terms of eigenvectors of MM' and M'M respectively, which are real, symmetric and positive semidefinite/positive definite, and hence are Diagonalizable in Orthonormal eigenbases, since Column space of M = Column space of MM' and Rowspace of M = Column space of M'M, these orthonormal eigenbases can be used as Q and R matrices appropriately, with C matrix now becoming diagonal, the set of singular values of M.
Starting with S, unless you provide correct U and V, you will end up with a different matrix M1 having same set of singular values but different set of four Fundamental subspaces, likewise if you choose correct U and V but wrong S, you will get a matrix M1 with Identical set of the four Fundamental subspaces as that of M, but different set of singular values. Unless there is degeneracy in singular values, and that they are to be conventionally arranged in non increasing sequence from top to bottom in S, there is no flexibility in choosing the U and V matrices arbitrarily and still generating correct M.
Debopam Ghosh thank you very much for your very clear explanation, greetings.
You can have S by computing the
square rooto of the eigenvalues of M.M*.
Yes, Indeed, you can apply Gram-Schmidt Orthogonalisation Process to do so via
https://nptel.ac.in/content/storage2/courses/122104018/node49.html
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