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.
Best Answer
You can just show, that rank of matrix $A$ is equal to $2$ (you actually did it by reducing $A$). So if rank is $2$ the dimension of a column space is also two. It means that there are two elements in the basis of $A$ column space. And you only need to notice that vectors $(1, 4, 3)^T$ and $(3, 4, 3)^T$ are linearly independent, so they form a basis of a column space.
Note that you can easily get this two vectors via simple matrix transformations (working with columns). For example $(3, 4, 3)^T = (3, 5, 2)^T - (0, 1, 1)^T$ and $(1, 4, 3)^T = (1, 2, 1)^T + 2(0, 1, 1)^T$ and so on you can show that after all this matrix transformations, you will reduce it to $$ \begin{pmatrix} 0 & 0 & 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 0 & 4 & 4 \\ 0 & 0 & 0 & 0 & 3 & 1 \end{pmatrix} $$