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Asked: 2026-05-11T18:40:01+00:00

I have the matrix system: A x B = C A is a by

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I have the matrix system:

A x B = C

A is a by n and B is n by b. Both A and B are unknown but I have partial information about C (I have some values in it but not all) and n is picked to be small enough that the system is expected to be over constrained. It is not required that all rows in A or columns in B are over constrained.

I’m looking for something like least squares linear regression to find a best fit for this system (Note: I known there will not be a single unique solution but all I want is one of the best solutions)

To make a concrete example; all the a’s and b’s are unknown, all the c’s are known, and the ?’s are ignored. I want to find a least squares solution only taking into account the know c’s.

[ a11, a12 ]                                     [ c11, c12, c13, c14, ?   ]
[ a21, a22 ]   [ b11, b12, b13, b14, b15]        [ c21, c22, c23, c24, c25 ]
[ a31, a32 ] x [ b21, b22, b23, b24, b25] = C ~= [ c31, c32, c33, ?,   c35 ]
[ a41, a42 ]                                     [ ?,   ?,   c43, c44, c45 ]
[ a51, a52 ]                                     [ c51, c52, c53, c54, c55 ]

Note that if B is trimmed to b11 and b21 only and the unknown row 4 chomped out, then this is almost a standard least squares linear regression problem.

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1 Answer

  1. Editorial Team

    I have no idea on how to deal with your missing values, so I’m going to ignore that problem.

    There are no unique solutions. To find a best solution you need some sort of a metric to judge them by. I’m going to suppose you want to use a least squares metric, i.e. the best guess values of A and B are those that minimize sum of the numbers [C_ij-(A B)_ij]^2.

    One thing you didn’t mention is how to determine the value you are going to use for n. In short, we can come up with ‘good’ solutions if 1 <= n <= b. This is because 1 <= rank(span(C)) <= b. Where rank(span(C)) = the dimension of the column space of C. Note that this is assuming a >= b. To be more correct we would write 1 <= rank(span(C)) <= min(a,b).

    Now, supposing that you have chosen n such that 1 <= n <= b. You are going to minimize the residual sum of squares if you chose the columns of A such that span(A) = span(First n eigen vectors of C). If you don’t have any other good reasons, just choose the columns of A to be to first n eigen vectors of C. Once you have chosen A, you can get the values of B in the usual linear regression way. I.e. B = (A’A)^(-1)A’ C

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