Why Gaussian elimination on the columns changes the column space? 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? 
 A: 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 $0$.)
A: First of all, you should clarify two defination: Vector space $R^n$ and Subspace .
Vector space $R^n$ : The space $R^n$ consists of all column vectors v with n components.
Subspace : A subspace of a vector space is a set of vectors (including 0) that satisfies
two requirements: If v and ware vectors in the subspace and c is any scalar, then 
(1)v + w is in the subspace
(2)cv is in the subspace
consider a matrix by 5*2.
The combination of first column and second column forms the column space. The column space is a subspace of $R^5$, not $R^n$ itself！
When doing gaussian elimination, you do change the iner structure of each column, then you do change the column space. the column space is a subspace of $R^n$, not $R^n$.
