Separating a signal into its constituent components First off, I apologize for possibly incorrect use of jargon, as I am new in this area. 

I have a type of signal that I am assuming can be described by an arbitrary number (let's say $n$) of basis vectors, which are linearly independent. I am measuring a signal through one sensor, which gives me a vector of amplitudes. I am able to measure an arbitrary number (let's say $m$) of mixtures of the basis signals, which gives me $m$ measured signals.

The length of my vectors is 150000, sampled at a frequency of 2.5 MHz.

So here is what I think my system of equations will look like.

$$ Y = A X $$

where $Y$ is the matrix containing the mixtures (measurements) and has dimensions of $ m \times 150000 $, $X$ is the matrix containing the basis signals, and has dimensions of $ n \times 150000 $, and $A$ is a coefficient matrix that has dimensions of $ m \times n $.

I need to solve for $A$. However, since $X$ is non-square ($ n \ll 150000$), it does not have an inverse. I guess I could calculate the pseudo-inverse. Does this make sense?

I have also looked into Principal Component Analysis and how it is used in various types of spectroscopy. It seems to me that PCA assumes a set of basis signals, and I don't know whether that has a physical meaning in my case, since my signals are stochastic (but stationary). I used the PCA routine on matrix $Y$ in MATLAB, and it gave me some signals very similar to the input ones, but I did not know what to do with them.

Any ideas?

I think I do have a valid physical model for my signals. Should I perhaps use model-based blind source separation (BSS) ?

Thanks in advance for taking the time.
You do a QR decomposition of $X^T$. Then the system reads as $$Y=AR^TQ^T.$$ Now multiply with $Q$ from the right, $$YQ=AR^T,$$ which already should give you hints on the reliability of the eventual solution, since ideally, the block of $YQ$ right of the leading $m\times n$ block should be zero. If that is true, then $$A=[YQ]_{[1:m,1:n]}[R^T]_{[1:n,1:n]}^{-1}$$.

Note that in the QR decomposition using Householder reflections, you can apply the reflections computed for $X$ simultaneously to $X$ and $Y$, if $$S=I-2 vv^T,$$ then $$X_{next}=XS=X-2(Xv)v^T\text{ and }Y_{next}=YS=Y-2(Yv)v^T$$ and $$X_{final}=R^T,\; Y_{final}=YQ$$.