Suppose I have the following linear system of equation
$$X = AC$$ with $X$ of dimensions $N \times T$, $A$ of dimension $N \times D$ and $C$ of dimension $D \times T$.
Does it hold in general that $\mathcal{R}(X)= \mathcal{R}(A)$?
My line of thought is the following, but I would like to have some comments and further explanation on it.
Clearly $\mathcal{R}(X) \subseteq \mathcal{R}(A)$, since each column of $X$ is a linear combination of the columns of $A$. Now, if $C$ is invertible, then I can write $XC^{-1}=A$, and hence each column of $A$ is a linear combination of the columns of $X$. Thus $\mathcal{R}(A) \subseteq \mathcal{R}(X)$. In this particular case ($C$ invertible), then $\mathcal{R}(X)= \mathcal{R}(A)$. But does it hold in general?
Also, I did not talk about rank and so on. Is the rank going to influence the above conditions?
SVD view
If we take the SVD of $X$, and we consider its economy size $X= U\Sigma V^\top$ with $\Sigma$ a diagonal $D \times D$ matrix, the matrix $U$ forms an orthonormal basis for $\mathcal{R}(X)$. How is this orthonormal basis $U$ related to $A$? What are the relationship between the SVD of $X$ and the matrix $AS$ as in $ U\Sigma V^\top$ with $\Sigma= AS$?
Thank you.
As you note, we have $\mathcal{R}(X) \subseteq \mathcal{R}(A)$ (and correspondingly, $\operatorname{rank}(X) \leq \operatorname{rank}(A)$). It follows that we have $\mathcal{R}(X) = \mathcal{R}(A)$ if and only if $\operatorname{rank}(X) = \operatorname{rank}(A)$.
With that said, it is easy to see that equality will not hold in general. For example, if $T < N < D$ and $A$ has rank $N$ (i.e. "full row rank"), then the rank of $X$ is at most equal to $T<N$, which means that the ranks cannot be equal.
One sufficient condition for equality of rank (and hence of ranges) is that $C$ has linearly independent rows (i.e. "full row rank"); notably, this can only occur if $D \leq T$. In this case, $C^+$ (the Moore-Penrose inverse of $C$) satisfies $CC^+ = I$, which means that we have $$ \operatorname{rank}(X) \geq \operatorname{rank}(XC^+) = \operatorname{rank}(ACC^+) = \operatorname{rank}(A), $$ which together with the other inequality implies that $\operatorname{rank}(X) = \operatorname{rank}(A)$.