Column rank, linear independence and RRE form

if I have a matrix A like this:

A = $\displaystyle \left[\begin{matrix}1 & 1 & 0 & 0 & 1 & 0 & 0\\1 & 1 & 0 & 0 & 0 & 1 & 0\\1 & 1 & 0 & 0 & 0 & 0 & 1\\1 & 0 & 1 & 0 & 1 & 0 & 0\\1 & 0 & 1 & 0 & 0 & 1 & 0\\1 & 0 & 1 & 0 & 0 & 0 & 1\\1 & 0 & 0 & 1 & 1 & 0 & 0\\1 & 0 & 0 & 1 & 0 & 1 & 0\\1 & 0 & 0 & 1 & 0 & 0 & 1\end{matrix}\right]$

when I do a reduced row echelon form rref in matlab it comes out like this:

$\displaystyle \left[\begin{matrix}1 & 0 & 0 & 1 & 0 & 0 & 1\\0 & 1 & 0 & -1 & 0 & 0 & 0\\0 & 0 & 1 & -1 & 0 & 0 & 0\\0 & 0 & 0 & 0 & 1 & 0 & -1\\0 & 0 & 0 & 0 & 0 & 1 & -1\\0 & 0 & 0 & 0 & 0 & 0 & 0\\0 & 0 & 0 & 0 & 0 & 0 & 0\\0 & 0 & 0 & 0 & 0 & 0 & 0\\0 & 0 & 0 & 0 & 0 & 0 & 0\end{matrix}\right]$

Well I thought that when there is a pivot in a column, then that column is linearly independent, but I can see simply that column one in matrix A, $\displaystyle c_1$ is a combination of $\displaystyle c_2 $ and $\displaystyle c_3 $ and $\displaystyle c_4 $.

So why is $\displaystyle c_1$ linearly independent?