Relationship between the column space of a matrix $A$ and its non-free (pivot) columns 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.
 A: 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 $A$.
A: It is not necesary true. We have a result which says

If the matrices $A$ and $B$ are related by a elementary row operation, then their row spaces are equal. Hence, row-equivalent matrices have the same row space, and hence also, the same row rank.

But with column space is different, another related result is

Row operations do not change the column rank.

which do not say nothing about column space, only column rank. For example, consider the effect on the column space of this row reduction
$$\left( {\begin{array}{*{20}{c}}
   1 & 2  \\
   3 & 4  \\
\end{array}} \right)\mathop  \to \limits^{ - 2{\rho _1} + {\rho _2}} \left( {\begin{array}{*{20}{c}}
   1 & 2  \\
   0 & 0  \\
\end{array}} \right).
$$
The column space of the left-hand matrix contains vectors with second component that is nonzero. But the column space of the right-hand matriz is different because it contains only vectors whose second component is zero.
Above comments says that we can not, always, express the column space of a initial matrix as span of the pivot columns of the echelon form of this matrix because, as above example, it can change with row operations. The only thing that do not change is the column rank.
