TITLE: How to transform one nonsquare matrix into another
QUESTION [5 upvotes]: I am modeling the effect of neural activity on synaptic strength. My question, though, is mathematical. 
I have the following differential equation:
$ \tau_{W}  \frac{d\mathbf{W}}{dt}=\mathbf{KWQ}-\alpha\mathbf{R_{\infty}}\mathbf{W}   \quad\mathbf{Q}=\left\langle \vec{u}\vec{u}\right\rangle ,\,\mathbf{R_{\infty}}=\left\langle \vec{v_{\infty}}\vec{v_{\infty}}\right\rangle
$
$\tau_{W}$ and $\alpha$ are constants. $\mathbf{W}$ is $n \times m$.  $\mathbf{R_{\infty}} $ is  $n \times n$. $\mathbf{Q}$ is $m \times m$. $\mathbf{K}$ is $ n \times n$. 

If $\mathbf{Q}$ and $\mathbf{R_{\infty}}$ come from the outer products of vectors, how do I know that they are invertible? 
How would I find the value for $\mathbf{W}$ for which the derivative vanishes? Especially if neither $\mathbf{Q}$ nor $\mathbf{R_{\infty}}$ are invertible.
How would I find $\mathbf{X}$ such that $ \mathbf{X}\mathbf{W} = \mathbf{Q}$ if neither $\mathbf{Q}$ nor $\mathbf{W}$ are invertible?

REPLY [5 votes]: The outer product of two vectors is a rank 1 matrix, and so it's never invertible (as long as the matrices are more than 1 dimensional.)
Your differential equation matrix isn't square, so it doesn't have a determinant. The matrices formed with outer products have determinant $0$.
Your third question needs some clarification. You mean you need a matrix $X$ such that $XW=Q$? This is possible but it really depends on a lot more information than you have given.


"Neither is antisymmetric so I know that they may be invertible."

Not sure what this means either, because invertibility does not bear much relationship to antisymmetry (for example $\begin{bmatrix}0&1\\-1&0\end{bmatrix}$ has inverse $\begin{bmatrix}0&-1\\1&0\end{bmatrix}$, and of course there are noninvertible antisymmetric matrices).