For a full rank overdetermined system $A \in \mathbb{R}^{m\times n}$, is the range always in $n$-dimensional subspace? When the solution $x$ to $Ax =b $ exists for an overdetermined ($m > n$) system $A \in \mathbb{R}^{m\times n}$ with rank $n$, does $b$ always live in $n$-dimensional subspace of $\mathbb{R}^{m}$?
Let $m > n$. If $\rho{(A)}$ = n, then $\mathcal{N}(A)=\{0\}$.
The solution may not exist for every $b \in \mathbb{R}^{m}$. But if it does, is it always in $n$-dimensional subspace?
My thought process was that if there is $x$ that solves $Ax = b$, then $A$ will always have $m-n$ rows of zeros and thus always have $m-n$ number of $0$'s in the column $b$. Thus, $b$ will be a vector that spans only $m - (m-n) = n$ dimension.
For example, in the case where the system $Ax = b$ has a solution for $A$ being $3$-by-$2$, with rank 2, is the right-hand side always in the form of $b = [c_1, c_2, 0]^{\top}$?
Is this correct? If not, what would be an easy counter-example?
I am confused, is this obvious because rank $n$ means that the dimension of the basis vectors that form $b$ is $n$?
 A: The answer to the first question is Yes. Note that the rank of a linear transform T is the dimension of the image/range of the transformation. That means the image is an  $n-$dimensional subspace of $\mathbb{R^m}.$
This also answers your second question. I mean $b$ must be inside the image, which is $\mathbb{R^n}.$ That means you're correct.
The answer to your last question: You are almost correct. By almost I mean the $0$ in $b$ could be any of the three coordinates. You do have the right idea.
Finally, if you consider an  $m \times n$ matrix as a linear map $T: \mathbb{R^n} \rightarrow \mathbb{R^m}$, everything can be explained more geometrically.
Extra comments:
I wrote the following answer a couple of months ago. Think about the dimension theorem.
The dimension theorem (the rank-nullity theorem) can be explained in many ways. I consider it as a consequence of the first isomorphism theorem/splitting lemma. When I teach undergrad matrix-theoretic linear algebra, I start with the equation $Ax=b,$ and I tell my students that the dimension theorem basically says that the number of total variables equals the sum of the number of free variables and the number of "non-free" variables. They find this statement very easy. If I teach a "formal/proof-based" undergrad mathematics class, I tell my students that the dimension theorem basically tells us how much "stuff" we need to put inside the nullspaces to extend it to the given vector space. This last sentence has some geometric appeal. That's what I mean when I said to consider a matrix geometrically. You can consider rank-nullity/the dimension theorem as the linear algebraic analog of the Pigeonhole Principle. Note that for any finite set $A,$ the function $f: A \rightarrow A$ is injective iff surjective iff bijective. It's a consequence of the Pigeonhole Principle. Thank rank-nullity/the dimension theorem gives a similar kind of conclusion for a finite-dimensional vector space $V,$ and any linear map $T: V \rightarrow V.$ Bottom line: You consider matrix as an array of numbers, which is okay. If you think of a matrix as a map between two finite-dimensional vector spaces, a matrix is more than simply arrays of numbers.  I hope this helps.
