Write $M=[Px\space\space Qy]$ as a product of matrices

Question

Let $M=[Px\space\space Qy]\in\mathbb R^{n\times 2}$, where $P,Q\in\mathbb R^{n\times n},x,y\in\mathbb R^{n}$.

Is there a matrix $B$ such that $M=B[x\space y]$, $B\in\mathbb R^{n\times n}$?

Answer 1

You are looking for a matrix $B$ such that $Bx=Px$ and $By=Qy$. This there exists if and only if $$\operatorname{rank}[Px,Qy]\leq\operatorname{rank}[x,y]\leq n$$

For let $U$ be the sub-space of $\Bbb R^n$ generated by $x,y$ and $V$ its complement in $\Bbb R^n$ so that $\Bbb R^n=U\oplus V$.

There exists a unique linear mapping $f:U\to\Bbb R^n$ such that $f(x)=Px$ and $f(y)=Qy$. Let $g:V\to\Bbb R^n$ be an arbitrary linear mapping. By direct sum decomposition, $f,g$ extends to a unique linear map $h:\Bbb R^n\to\Bbb R^n$, hence $h(x)=Px$ and $h(y)=Qy$. The matrix $B$ of $h$ is the required matrix.

Note that the matrix $B$ is unique if and only if $V=\{0\}$ that's $\operatorname{rank}[x,y]=n$.

For a possible closed form for $B$ in terms of $P$ and $Q$, assume for simplicity $\operatorname{rank}[x,y]=2$ with rank given by the rows $i,j$. Let $I_{ij}$ be the identity matrix without columns $i,j$ and $N\in\Bbb R^{n\times(n-2)}$ be arbitrary. Then each $B$ can be computed as: $$B=[Px,Qy,N][x,y,I_{ij}]^{-1}$$