In a rectangular matrix it is easy to see that every vector $x$ in $Ax = b$ comes from the row space because $C(A^T)$ and $N(A)$ are orthogonal to each other and span the whole space..
$$
\mathbb{R}^n = C(A^T) \; \oplus \; N(A) \\
A(x_n + x_r) = Ax_n + Ax_r = 0 + Ax_r = Ax_r \\
$$
In a rectangular $m\;x\;n$ matrix the vector $x$ will have $n$ entries, so I can see that something in the rowspace must definitely come from $n$-dimensional space.
In a square matrix with $rank < n$ however…
$$
\mathbb{R}^n = C(A^T) \; \oplus \; N(A) = C(A^T) \; \oplus \; N(A)
$$
the sum of the spaces still spans the whole space $\mathbb{R}^n$, but there seems to be an arbitrary option to view a particular $x$ that is in the column space already as part of the column space $Ax_c = b$ or a combination of the rowspace and the nullspace $A(x_n + x_r) = b$
If something $x$ is in the column space in this situation does that mean it is also in the rowspace? the column space and the rowspace are different subspaces when the rank is less than n, so this seems impossible…
Best Answer
You are correct, if a certain vector is in the column space and the rank of the matrix is less then $n$ in general the vector need not be in the row space. Consider $$\begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix}.$$ The column space of this matrix is $\text{span}(1,0)$ and the row space of this matrix is $\text{span}(0,1)$. Clearly every vector in the column space is not a vector of the row space.