<s>[INST] Finding a basis for a column space Let A be the matrix: $$\begin{pmatrix} 1&2&3&2&1&0\\2&4&5&3&3&1\\1&2&2&1&2&1 \end{pmatrix}$$.
Show that {$\bigl( \begin{smallmatrix} 1 \\ 4\\3\end{smallmatrix} \bigr)$, $\bigl( \begin{smallmatrix} 3\\4\\1 \end{smallmatrix} \bigr)$} is a basis for the column space of A. Find a "nice basis for the column space of A. 
So far, I have row reduced A to $$\begin{pmatrix} 1&2&0&-1&4&3\\0&0&1&1&-1&-1\\0&0&0&0&0&0 \end{pmatrix}$$ where the pivots occur in column 1 and column 3, so {(1,2,1),(3,5,2)} should be a "nice" column space? I do not see where {$\bigl( \begin{smallmatrix} 1 \\ 4\\3\end{smallmatrix} \bigr)$, $\bigl( \begin{smallmatrix} 3\\4\\1 \end{smallmatrix} \bigr)$} come from though. 
 [/INST] Building on the insights of @puhsu, you know you have 2 vectors that span the image, columns 1 and 3. Can you construct the target vectors in this basis?
$$
\left[ \begin{array}{r}
  1 \\ 4 \\3 
\end{array} \right]
=
\alpha
\left[ \begin{array}{r}
  1 \\ 2 \\ 1 
\end{array} \right]
   +
\beta
\left[ \begin{array}{r}
  3 \\ 5 \\ 2 
\end{array} \right], \quad 
\alpha = 7, \ \beta = -2.
$$
$$
\left[ \begin{array}{r}
  3 \\ 4 \\ 1 
\end{array} \right]
=
\alpha
\left[ \begin{array}{r}
  1 \\ 2 \\ 1 
\end{array} \right]
   +
\beta
\left[ \begin{array}{r}
  3 \\ 5 \\ 2 
\end{array} \right], \quad 
\alpha = -3, \ \beta = 2.
$$
</s>