Hello everybody!
Could somebody give me some advise how to solve the next problem?
(A + c*B)-1 , where A and B are symmetric positive semi definite matrices, c - is some constant value.
I need to rewrite this expression in such a way, that c will be factored out the brackets of inversion. Probably to do it linear with respect to constant c.
Thanks in advance!
I don't think you can have what you are looking for. (I of course hope that I did not misunderstand the question.) Even if A and B are ordinary numbers, not matrices, the expression 1/(A + cB) is not linear in c. You can of course solve for c (if (A + cB)−1 = X, cI = (X−1 − A)B−1 where I is the identity matrix) but I don't think this is what you seek and you probably knew it anyway.
Thanks for the answer!
I was also thinking in that way but I thought that probably there is some analog of Sherman-Morrison-Woodbury formula, but that can linearise this expression with respect to c (except Taylor expansion).
Do you probably know, is it possible to represent the matrix with the rank = r as the sum of r matrices with the rank=1 ?
Thanks!
How can you make a non-linear expression as linear unless you take some approximation ?
Andrii: I don't think that an analog of Sherman-Morrison applies here. Generally, it would mean that if you know the inverse A−1 of some matrix, you could "cheaply" compute the inverse of any other matrix C by taking c = 1, B = C − A. The only reason why I am hesitant is that you specified the matrices as symmetric and positive semi-definite, but I don't see how that would change things. In fact, perhaps I am mistaken but I think the above argument would still apply, namely that if A is symmetric positive semi-definite with its inverse known, the inverse of any other symmetric positive semi-definite C would be cheaply computable.
Answering your second question (almost forgot!) I believe the answer is yes, and an explicit form of that representation could be obtained just by playing the row operations that are used to reduce the matrix to row echelon form in reverse on selected rows.
Andrii: Sorry if this is already obvious to you but, formally, you can write (A+cB)-1 = A-1 - cA-1BA-1 + c2A-1(BA-1)2 + ... , so that to leading order the dependence on c is contained in the first two terms. This linearisation is a poor approximation unless (roughly speaking) cB is "small" compared with A, either in norm or some other measure eg weak topology.
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