So the question is to find the span of $v_{1} = \begin{bmatrix} 1 \\ 0 \\ 2 \\ \end{bmatrix}$, $v_{2} = \begin{bmatrix} 3 \\ 1 \\ 1 \\ \end{bmatrix}$, $v_{3} = \begin{bmatrix} 9 \\ 4 \\ -2 \\ \end{bmatrix}$, and $v_{4} = \begin{bmatrix} -7 \\ 3 \\ 1 \\ \end{bmatrix}$.
Before trying to solve this problem, it is important to know what span means. The first thing you need to know is where the vectors live. To figure this out, count the number of components in one of the vectors. In this case, there are 3 components in each vector, so these vectors live in $\mathbb{R}^{3}$. The first component corresponds to the $x$-axis, the second component corresponds to the $y$-axis, and the third component corresponds to the $z$-axis.
Now, if we take two linearly independent vectors, say $v_{1} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$ and $v_{2} = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}$, then these vectors clearly live in $\mathbb{R}^{3}$, since they each have 3 components. Since they are linearly independent, and there are 2 of them, they span a plane in $\mathbb{R}^{3}$. This particular plane is isomorphic to $\mathbb{R}^{2}$, but it is not $\mathbb{R}^{2}$, because vectors in $\mathbb{R}^{2}$ all look like $\begin{bmatrix} x \\ y \\ \end{bmatrix}$ (i.e., they have 2 components only). A plane in $\mathbb{R}^{3}$ is clearly 2-dimensional since that is how we define a plane, but we do not describe it as $\mathbb{R}^{2}$. Instead, we simply say it is a plane in $\mathbb{R}^{3}$.
Here is an example of what I mean: span$\left \{\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \right \}$ is the $XY$-plane in $\mathbb{R}^{3}$, while span$\left \{\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \right \}$ is the $XZ$-plane in $\mathbb{R}^{3}$, and span$\left \{\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \right \}$ is the $YZ$-pane in $\mathbb{R}^{3}$. All three of the spans I just mentioned are isomorphic to $\mathbb{R}^{2}$, but they are distinct planes in $\mathbb{R}^{3}$, which is why we must describe them as planes in $\mathbb{R}^{3}$, rather than just as $\mathbb{R}^{2}$. When described as planes, they can be differentiated from each other, which is good because they are different planes.
Now, on to your question, as you correctly stated, to determine which vectors are linearly independent (in order to determine the dimension of the span), we put the vectors as columns in a matrix and reduce to RREF. So after doing that, we get that
$\begin{bmatrix} 1 & 3 & 9 & -7 \\ 0 & 1 & 4 & 3 \\ 2 & 1 & -2 & 1 \\ \end{bmatrix}$ reduces to $\begin{bmatrix} 1 & 0 & -3 & 0 \\ 0 & 1 & 4 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}$
We see that there are three pivots! The number of pivots is the number of linearly independent columns, and the pivots correspond to the linearly independent columns, so the vectors $ \left \{ \begin{bmatrix} 1 \\ 0 \\ 2 \\ \end{bmatrix}, \begin{bmatrix} 3 \\ 1 \\ 1 \\ \end{bmatrix}, \begin{bmatrix} -7 \\ 3 \\ 1 \\ \end{bmatrix} \right \}$ (which are $v_{1}$, $v_{2}$, and $v_{4}$) are the linearly independent ones, and $v_{3}$ is linearly dependent on them. Since there are three linearly independent vectors, the span of all four vectors is equal to the span of the three linearly independent ones. Three linearly independent vectors span a subspace that is 3-dimensional. But these vectors live in $\mathbb{R}^{3}$, which is 3-dimensional itself, so their span must be equal to $\mathbb{R}^{3}$. If these vectors happened to live in $\mathbb{R}^{4}$, then their span would be a 3-dimensional subspace of $\mathbb{R}^{4}$.
So to answer your question, these four vectors could have spanned a 2-dimensional subspace of $\mathbb{R}^{3}$ if only two of the four were linearly independent. But we had three pivots in our matrix, so three linearly independent vectors, which meant the span of the four was equal to the span of the three linearly independent vectors (i.e., span$\{ v_{1}, v_{2}, v_{3}, v_{4} \} = $ span$\{ v_{1}, v_{2}, v_{4} \}$), and they spanned all of $\mathbb{R}^{3}$ in this case because they live in the 3-dimensional space and their span is 3-dimensional.
The following intuitive explanation can be given.
Each row of the matrix $A$ of a linear map $T:V\to W$ describes one coordinate, with respect to the chosen basis of$~W$, of the images $T(v)$ of vectors in$~V$. (This is a linear form on $V$, hence an element of $V^*$, but there is no need to stress that point.) If one row is a linear combination of other rows, then this information allows reconstructing the corresponding coordinate from those other coordinates, given only the knowledge that $T(v)$ belongs to the image subspace $\def\Im{\operatorname{Im}}\Im(T)\subseteq W$. Therefore if one selects a maximal independent subset of $r$ rows of $A$, the selection defines a subset of $r$ coordinates, such that for $w\in\Im(T)$ all other coordinates of$~w$ can be recovered from those coordinates of$~w$ (by forming appropriate linear combinations).
It should be fairly intuitive that for a subspace of dimension $d$, it requires knowing $d$ (properly chosen) coordinates of a vector of the subspace to know exactly which vector it is. Formally this can be shown as follows. Our $r$ coordinates define a projection $p:\def\R{\Bbb R}W\to\R^r$, and our reconstruction of coordinates defines a linear map $s:\R^r\to W$ (keeping the given coordinates and reconstructing the remaining ones) such that $s(p(w))=w$ for all $w\in\Im(T)$ (we don't care what happens to elements outside $\Im(T)$). Thus the restriction of $p$ to $\Im(T)$ is certainly injective, and if we can show it to be surjective then $r=\dim(\Im(T))$, which is $\operatorname{rk}(T)$, will follow. But if it were not surjective, then the would be at least one nontrivial equation satisfied by all elements of $p(\Im(T))\subset\R^r$, a relation contradicting the independence of our set of $r$ chosen coordinates.
Best Answer
The rank of a matrix $A \in M(m \times n, \mathbb{R})$ is defined as the dimension of its range, i.e. $\dim(R(A))$. We will show that $\dim(R(A)) = \dim(R(A^T))$.
Using the identity $Ax \cdot y = x \cdot A^T y$, it is not difficult to show that $R(A^T)^{\perp} = N(A)$. By taking the orthogonal complement of both sides we get $R(A^T) = N(A)^{\perp}$. We have $\mathbb{R}^n = N(A) \oplus N(A)^{\perp}$ (this holds for any subspace $U$ of $\mathbb{R}^n$, not just $N(A)$). Therefore $\dim(N(A)^{\perp}) = n - \dim(N(A)) = \dim(R(A))$, where the last equality is by the rank-nullity theorem. Alternatively, you can prove that $A \colon N(A)^{\perp} \to R(A)$ is bijective, and hence $N(A)^{\perp} \simeq R(A)$, so $N(A)^{\perp}$ and $R(A)$ must have the same dimension.