The pivot columns form the basis of the column space Let A be our matrice and B be our matrixes after row reduction.
I know the columns with pivots form the basis of column space of B. But after row reduction, we have added constants to random spots in the column vectors. This completely destroys the column space so I know this can be countered by using the original columns before they were row reduced and form the column space of A. 
But why does this follow? Why does it make sense that after completely destroying the column space, we can just say that the columns with pivots are the basis if we take their originals.
I know this question been asked before but I still don’t get it
 A: If $A$ and $B$ are row equivalent, then their null spaces are the same. That is to say that:
$$ A \mathbf x = \mathbf 0 \Leftrightarrow B\mathbf x = \mathbf 0. $$
Because matrix-vector multiplication can be thought of as creating a linear combination of the columns of the matrix, this implies that
$$ x_1 \mathbf a_1 + x_2 \mathbf a_2 + \cdots x_n\mathbf a_n = \mathbf 0 \Leftrightarrow x_1 \mathbf b_1 + x_2 \mathbf b_2 + \cdots x_n\mathbf b_n = \mathbf 0$$
where $\mathbf a_i$ is the $i$th column of $A$ and $\mathbf b_i$ is the $i$th column of $B$. The above equivalency tells us that the columns of $A$ are related in the same way as the columns of $B$, i.e., if a set of column vectors of $A$ are linearly independent, then the corresponding set of columns of $B$ will also be linearly independent (and vice versa). If $B$ is in reduced echelon form, then it is obvious that the pivot columns of $B$ are linearly independent (they should be distinct standard basis vectors). Thus, the pivot columns of $A$ must also be linearly independent.
A: Another answer: because when you perform row reduction on a matrix you're NOT "destroying" the column space. Think about the coefficients in a column as the coordinates of this column-vector in the standard basis:
$$
\overrightarrow{\mathbf{a}} 
=
\begin{pmatrix}
a^1 \\
a^2 \\
\vdots \\
a^n
\end{pmatrix}
= 
a^1 \mathbf{e_1} + a^2\mathbf{e_2} +  \dots + a^n \mathbf{e_n}
$$
Here $\mathbf{e_1}, \dots , \mathbf{e_n} $ is the standard basis of $\mathbb{R}^n$.
When you perform a row reduction what you're actually doing is changing the basis of this column-vector.
For instance, a legitimate operation in a row reduction process is to multiply a row by a non zero real number: $\lambda \neq 0$.
If we do this to the first row of our column-vector up there, you could think that you have obtained a new vector
$$
\overrightarrow{\mathbf{b}} 
=
\begin{pmatrix}
\lambda \cdot a^1 \\
a^2 \\
\vdots \\
a^n
\end{pmatrix} \ .
$$
But in fact, this $\overrightarrow{\mathbf{b}}$ is the same vector $\overrightarrow{\mathbf{a}}$: just the basis (and the coordinates, accordingly) has changed: 
$$
\overrightarrow{\mathbf{b}} = \lambda a^1 (\frac{1}{\lambda} \mathbf{e_1}) + a^2\mathbf{e_2} +  \dots + a^n \mathbf{e_n} = \overrightarrow{\mathbf{a}}
$$
(Now the basis is $(\frac{1}{\lambda} \mathbf{e_1}), \mathbf{e_2},   \dots ,\mathbf{e_n} $.)
Exercise. Convince yourself that the rest of the allowed operations in a row reduction procedure can also be seen as changing the basis of the column-vectors. 
Hence, in the end, you're NOT changing the column-vectors. Therefore, if the pivot columns in the matrix $B$ are linearly independent, so are the corresponding same columns in matrix $A$. Just because they are the same vectors: only their coordinates have changed.
