What do we know about the solution of this set of linear equations? Let $C \in \mathbb{R}^n$, $A \in \mathbb{R}^{n \times n}$, $Y \in \mathcal{Y}$, and $B : \mathcal{Y} \to \mathbb{R}^n$ be linear, where the linear space $\mathcal{Y} \subset \mathbb{R}^m$ may be proper. Here $n \geq 1$ and $m \geq 1$, so everything is linear and finite-dimensional and fixed up to this point. 
Assume then that for every $P \in \mathbb{R}$ there exists a unique $X \in \mathbb{R}^n$ such that:
\begin{align}
C^TX & = P \\
AX & = BY 
\end{align}
My questions: When does such $X$ exist, and what does this imply of $X$ or $BY$? 
Combining the two equations, on the LHS, there are $n+1$ equations and $n$ unknowns, so it is an over-determined system. However, on the RHS, only the first equation changes. Further, clearly $\ker(C^T) \cap \ker(A) = \{0\}$, for otherwise the solution would not be unique. Apart from that I seem to be stuck.  
As a side note, this system of equations arises as a consequence of properties of certain dynamical systems. In a less general setting I can prove that necessarily $BY = 0$, and (in a slightly more general setting) that $h^T BY = 0$ for some $h \in \mathbb{R}^n$. Something like that would be ideal here, too, but I may need additional assumptions.
Thanks for your help in advance. 
 A: The first thing to notice is that you need $C\neq0$, because you need $C^T X=P$ for each $P$. If you have $C\neq 0$, for each $P$ there will be a whole hyperplane (perpendicular to $C$) of solutions $X$ to the equation $C^T X$, and you will have to change $X$ every time you change $P$.
Now, the equation $AX=BY$ is an $n\times n$ system. If $A$ has rank $n$, you're in trouble, because, there will be only one $X$ that is a solution. But if $A$ is singular (and $BY$ is in the image of $A$), there will be lots of solutions for $X$. For each $P$, the solution set of the first equation is a hyperplane perpendicular to $C$. The solution set of the second equation is an affine subspace, parallel to the kernel of $A$. There are two possibilities:


*

*The kernel of $A$ is perpendicular to $C$.
In this situation, you're in trouble again, because as you change $P$, the hyperplanes of allowed values for $X$ (according to the first equation) move in the direction of $C$. This means that the intersection of both solution sets is null, because they are parallel (because they are both perpendicular to $C$). Of course, there will  be a specific value of $P$ that will make the solution set of the second equation contained inside the hyperplane of the first equation, obut for all other values of $P$, there is no solution.

*The kernel of $A$ is not perpendicular to $C$.
This is what you need. This way, the kernel of $A$ will intersect all hyperplanes. For the solution to be unique, you need the kernel of $A$ to be a line (in other words, you need $A$ to have rank $n-1$).
So, a solution $X$ exists to every $P$, if and only if these three conditions are met:


*

*$C\neq 0$ (actually, this is implied by the next condition)

*The kernel of $A$ not be perpendicular to $C$.
[This means that $\exists v\in kerA$, $C^T v\neq 0$]

*$BY$ is in the image of $A$.
[This means that $\exists u\in\mathbb{R}^n,Au=BY$]
The solution is unique if and only if the above conditions are met, and:


*

*$A$ has rank $n-1$

