Separating vectors from linear combination Suppose I have a linear combination of vectors as follows
$
\mathbf{s} = \alpha_1\mathbf{x}_1 + \dots + \alpha_m\mathbf{x}_m + \beta_1\mathbf{y}_1 + \dots + \beta_n\mathbf{y}_n
$
where $\alpha_i, \beta_i \forall i$ are constants, and $\mathbf{x}_i, \forall i$ lies in the null space of a matrix $\mathbf{Z}$.
Is there a way to recover the sum :
$\sum_{i=1}^m\alpha_i\mathbf{x}_i$ from above. I'm aware that 
$\mathbf{Z}\mathbf{s} = \sum_{i=1}^n\beta_i\mathbf{Z}\mathbf{y}_i$.
 A: We are given vectors $x_1,\dotsc,x_m,y_1,\dotsc,y_n\in\Bbb R^N$ so that $\{y_1,\dotsc,y_n\}$ is linearly independent and 
$$
\DeclareMathOperator{Span}{Span}
\Span\{x_1,\dotsc,x_m\}\cap\Span\{y_1,\dotsc,y_n\}=\{\vec 0\}
$$
Let $\Phi$ be the $N\times(m+n)$ matrix
$$
\Phi=
\begin{bmatrix}
x_1 & \dotsb & x_m & y_1 & \dotsb & y_n
\end{bmatrix}
$$
Now, suppose the following condition is met.
Assumption. $\DeclareMathOperator{rank}{rank}N<m+n$ and $\rank(\Phi)=N$
This assumption guarentees that $\Phi$ has a right inverse defined by
$$
\Phi_{\text{right}}^{-1}=\Phi^\top(\Phi\Phi^\top)^{-1}
$$
satisfying $\Phi\Phi_{\text{right}}^{-1}=I_N$.
Now, let 
$$
\Psi=\begin{bmatrix}x_1&\dotsb&x_m&\vec 0_{N\times n}\end{bmatrix}
$$
and consider the equation
$$
T\Phi=\Psi\tag{1}
$$
Applying $\Phi_{\text{right}}^{-1}$ on the right of (1) gives
$$
T=\Psi\Phi_{\text{right}}^{-1}
$$
Here $T$ satisfies $T(x_k)=x_k$ and $T(y_k)=\vec 0$. Hence
\begin{align*}
T\vec s
&= T(\alpha_1 x_1+\dotsb+\alpha_m x_m+\beta_1y_1+\dotsb+\beta_ny_n) \\
&= \alpha_1 T(x_1)+\dotsb+\alpha_m T(x_m)+\beta_1 T(y_1)+\dotsb+\beta_n T(y_n) \\
&= \alpha_1 x_1+\dotsb+\alpha_m T(x_m)+\beta_1\vec 0+\dotsb+\beta_n\vec 0 \\
&= \alpha_1 x_1+\dotsb+\alpha_m x_m
\end{align*}
