Given an $m\times n$ matrix $A$ with $m\leq n$, with the rank of $A$ being less than $n$, is it necessarily true that the columns in $A$ representing the free variables are linear combinations of the pivot columns? If I am to figure out the column space of $A$, without having to calculate which of the columns are redundant (i.e. linear combinations of other columns), can I reliably say that $C(A)$ is the span of all (and only) the pivot columns in $A$? I was watching a video by Khan Academy where it seemed that this was the case, at least for the example given… but I don't know if it generalizes for all matrices $A$ where the null space does not equal $\{\vec{0}\}$
Example:
$$A=\left[\begin{array}{rrrr}1 & 1 & 1 & 1\\ 2 & 1 & 4 & 3 \\ 3 & 4 & 1 & 2 \end{array}\right]$$
Its column space is the span of the two vectors $\left[\begin{array}{r}1\\2\\3 \end{array}\right]$ and $\left[\begin{array}{r}1\\1\\4\end{array}\right]$, which just so happen to be the only two pivot columns. The other two are free variable columns.
Best Answer
This is true for all matrices. Elementary row operations preserve linear relationships between the columns of a matrix. Suppose we have a matrix $A$ with columns $\mathbf{a}_i$ along with the Reduced Row Echelon Form $R$ with columns $\mathbf{r}_i$. Then for any set of coefficients $c_i$, we have $$\sum_{i=1}^nc_i\mathbf{a}_i = \mathbf{0}\iff \sum_{i=1}^nc_i\mathbf{r}_i = \mathbf{0}$$ The pivot columns in $R$ correspond to a basis for the columnspace of $R$, it follows that the same columns in $A$ form a basis for the columnspace of $A$.