Dimension and its consequences Find a basis for $S=[\mathrm{(1,2,3),(3,4,7),(5,-2,3)}]\subseteq \Bbb R^3$ and give the dimension.
Then, putting all the vectors as the columns of a new matrix:
$A=\begin{bmatrix}
  1 & 3 & 5  \\
  2 & 4 & -2 \\
  3 & 7 & 3
\end{bmatrix}$
By row reducing:
$A_R=\begin{bmatrix}
  1 & 0 & -13 \\
  0 & 1 & 0   \\
  0 & 0 & 0
\end{bmatrix}$
Let $C=(c_1,c_2, c_3) \in \Bbb R^3$
$A_R\times C =0 \implies \begin{cases} c_3 = {1\over{13}}c_1 \\ c_2 = {-6\over{13}}c_1 \end{cases}$
Then $S$ is not linearly independent, so I remove the vector $(5,-2,3)$ from $S$
Then now $S=\mathrm{gen}[(1,2,3),(3,4,7)]$
Now $(1,2,3)$ and $(3,4,7)$ are linearly independent and $\dim(S)=2$
So, if $S$ is a basis for $W\subseteq \Bbb R^3$ and $\dim(S)=2 \implies W= \Bbb R^2$
Then, I want to check if $gen(S)=W$.
Let $B=\begin{bmatrix}
  1 & 3 \\
  2 & 4 \\
  3 & 7
\end{bmatrix}
$ which reduced form is 
$ \begin{bmatrix}
  1 & 0 \\
  0 & 1 \\
  0 & 0
\end{bmatrix}
$, $k_1,k_2 \in \Bbb R$
$k_1 \times \begin{bmatrix}1\\0\\0 \end{bmatrix} + k_2 \times \begin{bmatrix} 0 \\ 1 \\0 \end{bmatrix} = \begin{bmatrix} x_1 \\ x_ 2 \\ x_3 \end{bmatrix} \implies \begin{cases} k_1=x_1 \\ k_2 = x_2 \\ x_3 = 0 \end{cases}$
So, finally, my question:
This not generates $\Bbb R^3$, but $\mathrm{gen}(S)=\Bbb R^2$?
 A: By putting the vectors as the columns of the matrix, the leading entries in the row echelon form indicate columns of the original matrix that form a basis of its column space.  In this case, we have:
$$
\overbrace{\begin{bmatrix} 1 & 3 & 5 \\ 2 & 4 & -2 \\ 3 & 7 & 3 \\ \end{bmatrix}}^A
\xrightarrow{\text{row operations}}
\overbrace{\begin{bmatrix} 1 & 0 & -13 \\ 0 & 1 & 6 \\ 0 & 0 & 0 \\ \end{bmatrix}}^{A_R}
$$
(there's a bug in the $A_R$ in the question).  This implies the linear combination $$\color{blue}{13}C_1\color{red}{-6}C_2+\color{purple}{1}C_3=\mathbf{0}$$ of the columns of either $A_R$.  The same linear combination applies for the columns of $A$:
$$\color{blue}{13}\begin{bmatrix} 1 \\ 2 \\ 3 \\ \end{bmatrix}\color{red}{-6}\begin{bmatrix} 3 \\ 4 \\ 7 \\ \end{bmatrix}+\color{purple}{1}\begin{bmatrix} 5 \\ -2 \\ 3 \\ \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix}.$$
We see that $A_R$ has rank $2$, so its column space will be $2$-dimensional, and will be $$\mathrm{span}\{(1,2,3),(3,4,7),(5,-2,3)\}=\mathrm{span}\{(1,2,3),(3,4,7)\}.$$  So $$\{(1,2,3),(3,4,7)\}$$ is a basis (as is $\{(1,2,3),(5,-2,3)\}$ and $\{(3,4,7),(5,-2,3)\}$).

Note: $\{(1,0,0),(0,1,0)\}$ is not a basis: $\mathrm{span}\{(1,0,0),(0,1,0)\}$ does not even contain $(1,2,3)$.

We may also find a basis by putting the vectors as the rows of the matrix:
$$\begin{bmatrix} 1 & 2 & 3 \\ 3 & 4 & 7 \\ 5 & -2 & 3 \\ \end{bmatrix}
\xrightarrow{\text{row operations}}
\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \\ \end{bmatrix}$$
which gives the basis $\{(1,2,3),(0,1,1)\}$ formed by the non-zero rows in the row echelon form.
