Why column spaces are different in one matrix and its echelon form? The text doesn't show me why just tells me if I want to find out the column space of a given matrix, I should use a reduced row-echelon form, then find leading one's column in RREF, and finally find the corresponding columns in the original one.
It tells me that we should do so cause they usually have different column spaces. But why that happens?
 A: Here is a simple example. Let 
$$A=\begin{pmatrix}1&1&1\\1&1&1\\1&1&1\end{pmatrix}$$
Its reduced row echelon form is
$$\begin{pmatrix}1&1&1\\0&0&0\\0&0&0\end{pmatrix}$$
Do you think the column space of the reduced echelon form, i.e., span of $\begin{pmatrix}1\\0\\0\end{pmatrix}$, can span the original column space?
Notice the original column space is spanned by $\begin{pmatrix}1\\1\\1\end{pmatrix}$.
A: The transformation of a given matrix A into its echelon form U (or reduced echelon form R) is made through a series of elementary row operations. Even if A has no zero element, these operations may reset (to zero) some rows in case A is rectangular or contains linear dependencies.
Now, suppose for the sake of the explanation that A has no zero elements and that some elimination steps zeroed the k-th row, which means the k-th component of all columns of U (or R) is zero. Given that the column space of a matrix is, of course, a linear combination of its columns, it is clear that there is no way of yielding some vector c=Ux in which the ck (the k-th component of the vector c) is different of zero. However, it is perfectly possible to yield a vector b=Ax with non-zero components since we assumed for clarity that A had no zeroes. Consequently, there are vectors that can be spanned by A that cannot be spanned by U - as KittyL put beautifully in his/her answer.
Note, however, that the dimensions of the column spaces of A and U (or R) are the same and, interesting enough, that their null spaces are identical (see here).
