I'm stuck when it comes to solving equations with matrices on both sides when the matrices contain no inverses (so they cannot be cancelled out).
For example, $$\mathbf A = \begin{bmatrix}1&3\\0&0\end{bmatrix}$$ (i.e., no inverse) and $\mathbf y = [a,b]$
I'm looking for all solutions to $\mathbf A\mathbf y = \mathbf A\mathbf u$ in the form $\mathbf u=c\mathbf v+\mathbf w$, but I'm confused on where to start if the $\mathbf A$'s can't be cancelled.
$$\begin{bmatrix}1&3\\0&0\end{bmatrix}\begin{bmatrix}a\\b\end{bmatrix}=\begin{bmatrix}a+3b\\0\end{bmatrix}$$ We have the same for $\mathbf u=\begin{bmatrix}u_1\\u_2\end{bmatrix}$:$$\mathbf A\mathbf u=\begin{bmatrix}u_1+3u_2\\0\end{bmatrix}$$ The $\mathbf y$ is a solution if $a+3b=u_1+3u_2$. Then we can express $a$ as $$a=u_1+3u_2-3b$$ Therefore $$\mathbf y=\begin{bmatrix}a\\ b\end{bmatrix}=\begin{bmatrix}u_1+3u_2-3b\\ b\end{bmatrix}=\begin{bmatrix}u_1+3u_2\\ 0\end{bmatrix}+\begin{bmatrix}-3b\\ b\end{bmatrix}=\mathbf A\mathbf u+\begin{bmatrix}-3\\ 1\end{bmatrix}b$$ To identify with your notation $c=b$, $\mathbf v=\begin{bmatrix}-3\\ 1\end{bmatrix}$, and $w=\mathbf A\mathbf u$