TITLE: Given the solution of $Ax = b$, do we know whether $A$ spans $\mathbb{R}^3$?
QUESTION [1 upvotes]: Let $A$ be a $3 \times 4$ matrix such that all solutions to the equation $A\mathbf{x} = \mathbf{b}$ may be written as:
$$
\left[ \begin{matrix} 1 \\ 2 \\ 3 \\ 4 \end{matrix} \right] + x_3 \left[ \begin{matrix} 3 \\ 2 \\ 1 \\ 0 \end{matrix} \right]  $$
with $x_3 \in \mathbb{R}$. Does $A$ span $\mathbb{R}^3$, or can't we tell?

REPLY [1 votes]: I assume that by "does $A$ span $\Bbb R^3$", you mean whether the columns of $A$ span $\Bbb R^3$, that is, whether the image of (the linear transformation induced by) $A$ is $\Bbb R^3$.
(And that $A$ is a $3 \times 4$ matrix.)

Recall the following:
If $x_0$ is a particular solution to the system $Ax = b$, then the complete set of solutions is given by $$x_0 + \cal N(A),$$
where $\cal N(A)$ denotes the null space of $A$.

In this case, we see that $\cal N (A)$ is spanned by one-vector, that is, it is one dimensional.
Recalling the rank-nullity theorem, we see that $$\operatorname{rank} A = 4 - \operatorname{nullity} A = 3.$$
Since the image of $A$ is a subspace of $\Bbb R^3$, we see that $\operatorname{rank} A = 3$ gives us that $A$ does span $\Bbb R^3$.
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