This page on theorem 8.2 states that, Neither of the operations of the gaussian elimination changes the row space of an $m \times n$ matrix after applying the operation. It says later that this is only true about the row space and not the column space.
I can clearly see how multiplying and adding two vectors does not change the row space. Let's assume any pair of two dimensional non parallel non zero vectors. These vectors span $R^2$. Thus it does not matter how we combine them linearly, they will still span $R^2$.
The column space for any two non zero non parallel vectors can be thought of as being two dimensional vectors, spanning another two dimensional space again. Lets call this one $R^2_c$.
Now here is my question, doing gaussian elimination on the columns of a matrix, will firstly, do nothing to the span of the column space, $R^2_c$, because that space still can be spanned with the two new column vectors, and secondly, it will result in two new row vectors in $R^2$ that can still span $R^2$. So it seems to me by doing linear combinations on the column space, neither the row nor the column space change.
What am I doing wrong?
Consider the matrix $$\begin{pmatrix}1&5\\2&10\end{pmatrix}.$$ Its column space is one-dimensional, spanned by $\binom12$ (or equivalently by $\binom5{10}$ since these vectors are proportional). After a row operation, subtracting twice the first row from the second, the matrix becomes $$\begin{pmatrix}1&5\\0&0\end{pmatrix}.$$ The column space of this is still one-dimensional, but it's a quite different one-dimensional space from before; the new one is spanned by $\binom10$. (Edit to correct an error in a comment: The column space of the original matrix was a line of slope $2$ (not $5$); the column space of the new matrix has slope $0$.)