Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have a spectroscopy problem that boils down to a matrix equation where X*A=C. I take N observations each consisting of 3 detector readings and my detectors suffer from some amount of cross-talk (some percentage of signal from detector 1 spills over into detector 2, etc.) In my specific case, C is an Nx3 matrix of detector readings, X is an Nx2 matrix of my unknown true signals, and A is a 2x3 matrix of constant coefficients that represent how much each signal source gets into each detector. So, I have:

$\begin{bmatrix} X_{1 1} & X_{12} \\ \vdots & \vdots \\ X_{N1} & X_{N2} \end{bmatrix} * \begin{bmatrix} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23}\end{bmatrix} = \begin{bmatrix} C_{11} & C_{12} & C_{13} \\ \vdots & \vdots & \vdots \\ C_{N1} & C_{N2} & C_{N3}\end{bmatrix}$

This is a system of linear equations with 3N equations and 2N+6 unknowns. When N = 6, this system should be determined, and in real world experimentation with noise contributions, when N > 6, I should begin to compensate for noise. In practice, I will usually take 100's of observations, so N will usually be 200 to 1000. Without noise contributions, of course this would be an overconstrained problem. One column of C does have to be a linear combination of the other two, however, with every C element having a Gaussian noise contribution of mean zero, it is underconstrained and one can only find the best estimates.

Intuitively, this should be solvable for the 6 elements of A and the 2N elements of X, but I cannot find a treatment for this construction of a problem. I have been searching for linear algebra solution approaches that address having two matrices of unknowns as I have described, but I haven't found anything appropriate yet. I can re-arrange the matrix equation to $ X = C * A^{-1} $, but I haven't seen a treatment for that construction either.

Any suggestions or insights into solving this? Thanks in advance.

share|improve this question
1  
"This is a system of linear equations" Linear on what? If both $X$ and $A$ are unknown, then it's not linear (on those variables), but quadratic. –  leonbloy Apr 20 '12 at 19:33
    
As a quick comment on this, I don't see how you can rearrange this equation to $X = C*A^{-1}$ without further explanation of what you mean by $A^{-1}$ (as $A$ is not square). If $A$ was such that $A*A^{T} = I$, then perhaps you could rewrite this as $X = C*A^{T}$, whence this is linear. But real world data rarely has such nice behavior. –  Nicholas Stull Apr 20 '12 at 21:24
    
A non-square matrix can sometimes have a one-sided inverse based on e.g. $AA^T(AA^T)^{-1} = I$ –  Edward Apr 23 '12 at 18:39

1 Answer 1

If the $N \times 3$ matrix $C$ can be represented as $XA$ where $X$ is $N \times 2$ and $A$ is $2 \times 3$, $C$ must have rank at most $2$. Thus one column of $C$ must be a linear combination of the other two. Do row-reduction on $C^T$ to find what linear combination it is. If, say, $C_{i3} = a C_{i1} + b C_{i2}$, that means that $C = X A$ where $X$ consists of the first two columns of $C$ and $A = \pmatrix{1 & 0 & a\cr 0 & 1 & b\cr}$.

The solution is never unique: given one solution, you can always replace $X$ by $XU$ and $A$ by $U^{-1} A$ where $U$ is any invertible $2 \times 2$ matrix.

share|improve this answer
    
I understand that the solution can never be unique in the way you mention (similar to other matrix factorization problems). The issue of the third column being a linear combination of the other two is mostly true, but doesn't the addition of Gaussian noise to each C element change the situation? Now shouldn't it be usefull additional information to better estimate the relative values of A? –  Edward Apr 23 '12 at 18:28
    
You can use a Singular Value Decomposition to get a good rank-$2$ approximation to $C$. –  Robert Israel Apr 23 '12 at 18:47

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.