Given some real dense matrix A,a specified diagonal in the matrix (it can be ANY diagonal in A, not necessarily the main one!), and a scalar constant c, is it possible to produce an expression that will give matrix B, which is all the same matrix as A, but with the elements on the specified diagonal multiplied by c? The expression should use only basic linear algebra operations (like multiplications by some special matrices that depend on c).
An example:
A=[1, 2, 3, 4
5, 6, 7, 8]
(bold numbers are the specified diagonal)
B=f(A,c)=[1, 2, 3*c, 4
5, 6, 7, 8*c]
P.S. What I actually need is a method to multiply each diagonal in A by some constant (i.e. if A is of size n*m then we have vector c of length (n+m-1)). But I think that if there exists a method to multiply one diagonal, then it will be pretty easy to generalize it for all diagonals.
It is possible, although the answer may not be one that you like. I'll show you the 2x2 version for the main diagonal.
Let L be the 2x2 matrix with rows 1,0 and 0,0, and R be the 2x2 matrix with rows 0,0 and 0,1.
Then for a 2x2 matrix M, to multiply the main diagonal by c, compute
c M - (c-1) ( L M R + R M L).
Essentially you are subtracting off from individual positions: L M R is the top right element of M, all others zeroed out. R M L is the bottom left element.
Neil, I like the idea to "isolate" and then subtract the off-diagonal elements, but I don't see how this approach can be generalized even for the main diagonal of a 3*3 matrix. Could you clarify it?
To obtain the entry in position i,j, multiply on the left by the matrix which has a 1 in the i,i position, and on the right by the matrix which has a 1 in the j,j position. These matrices are zero everywhere else. Thus for mxn matrices, we need m matrices of size mxm for the left, and n matrices of size nxn for the right. Does that help?
Dear Konstantin,
Answer: No it is not possible to multiply a diagonal in a matrix by a scalar using basic matrix operations.
Please see the attachment.
Regards,
Wiwat Wanicharpichat
Elementary matrices generate the group SL(n). Hence you can only produce diagonals in SL(n) as a product of elementary matrices nothing more.
Nail, that helps a lot, thanks. Now I do understand the idea and do understand that in order to "process" a diagonal of length n, I would need n items in the expression. And that does not look good for the analytical solution that I planned. It seems like I will have to stick to a numerical solution.
Wiwat, it is clear that it is impossible to obtain a diagonal of a matrix by one multiplication, but, as Nail has shown, it is possible via a series of multiplications and summation.
Anupam, what is "group SL(n)"? A simple Lie group?
Wiwat, I meant matrix multiplication, summation and other simple operations that are "reversible".
In fact, the problem that I want to solve is the following: given is some dense data matrix D of size n*m (disease incidence by year and age), I need to calculate its best approximation of the following form:
D=f_diag(a*(b'),c),
where a is a column vector of length n ("year coefficients"), b is column vector of length m ("default age-distribution"), and f_diag(*,c) is the above-mentioned diagonal-multiplying function with coefficients c (length (n+m-1)) that represents "cohort effect".
If this approximation model had a "compact enough" matrix form, I would try to find an analytical solution to the problem. But if there is no good expression for it, I can solve the problem numerically.
Konstantin, Wiwat, this is the reason that I said that "it is possible, although you may not like the answer". The method I outlined is not one that is amenable to reversing.
In the case of n*n matrices the elementary matrices of kind Id+e(i,j), i not equal to j and where e(i,j) is matrix with 1 at (ij)th place, generate all matrices of determinant 1. Thus one can produce diag(a, a^{-1}, 1, 1,....) matrices with elementary matrices. On main diagonal that is best you could hope. However I see that your matrix need not be square. In that case my answer does not apply.
Pasquale, summation is OK, but iterating is not. As it is explained in my previous post in this thread, I need a "reversible" expression.
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