Which subspace (column space, row space, null space, left null space) of A is x a part of? Which of the four subspaces contains $x$, for the given equation:
$$A^T Ax=A^Tb$$
where $b$ is a column vector NOT in the column space of $A_{m\times n}$. (This equation is to find the best solution for $Ax=b$). Also, $A$ is assumed to have independent columns.
(Question is from one of the MIT OCW problem sets.)
My attempt: Null space of $A$ is just the zero vector. So $\mathbb R^n$ is entirely the row space. Since $x$ has to be a $n\times 1$ vector, it belongs in the row space.
Is there any other more intuitive way to visualize this? Can I generalize this to any $Ax=b$ also, if rank $=n$ for $A$?
 A: Well, to me your attempt is as intuitive as it gets. Just a remark: you assumed that $m\neq n$ to conclude that $x$ can't be in the column space or left null space, which a priori is not given (at least, it wasn't stated in the question). But this is ok, since, if $m=n$ and $A$ has independent columns, $A$ would be an invertible square matrix, and $A^TAx=A^Tb\implies Ax=b$, but, since $b$ is assumed not to be in the column space of $A$, this case doesn't happen.
Maybe what you're asking for is a more explicit explanation? If so, Let $A=\begin{bmatrix}L_1 \\ \vdots\\ L_m\end{bmatrix}$, where $L_i$ is the $i$-th line, a $1\times n$ vector. If $b=\begin{bmatrix} b_1 \\ \vdots\\ b_m\end{bmatrix}$, we have that
\begin{align*}
A^TAx=A^Tb&\implies \begin{bmatrix} L_1 & \cdots & L_m\end{bmatrix}\begin{bmatrix}L_1 \\ \vdots\\ L_m\end{bmatrix}\textbf{x}=\begin{bmatrix} L_1 & \cdots & L_m\end{bmatrix}\begin{bmatrix} b_1 \\ \vdots\\ b_m\end{bmatrix}\\
&\implies(L_1\cdot L_1+\dots+L_n\cdot L_n)\textbf{x}=b_1L_1+\dots+b_mL_m\\
&\implies\textbf{x}=\dfrac{1}{L_1\cdot L_1+\dots+L_n\cdot L_n}(b_1L_1+\dots+b_mL_m)
\end{align*}
So $\textbf{x}$ is indeed in the row space of $A$ (notice that the denominator of the fraction is clearly non-zero).
A: Your solution is already perfectly simple. By the four subspaces theorem, $x$ can be written as $x = x_1 + x_2$, where $x_1 \in R(A^T)$ and $x_2 \in N(A)$. But the null space of $A$ is trivial, so $x_2 = 0$ and $x \in R(A^T)$.
