# Pivot Columns and Linear Indepenence

If I have a matrix $(A)$ and I want to find the basis for $C(A)$ (column space)

I first find the $rref(A)$. Is it true that the columns of $(A)$ corresponding to the pivot columns of $rref(A)$ form the basis of $C(A)$?

And if so, please explain why this is true. Why, are the columns of $(A)$ corresponding the the pivot columns of $rref(A)$ linearly independent just because the pivot columns are independent?

• Are you OK with thinking about linear transformations? If so, think about what the column space means for $A$ as a linear transformation, and then think about how you are changing (or not changing) this linear transformation when you apply elementary row operations to obtain the reduced row echelon form. – Clinton Boys Oct 10 '12 at 22:30
Suppose that we have a matrix $A$ with reduced row echelon form $R$. Then there is a series of elementary matrices which bring $A$ to $R$. Let $E$ be their product so that $$EA=R$$ Now if we write the matrix column-wise, we have $$E\begin{pmatrix}\mathbf{a_1} & \cdots & \mathbf{a_n}\end{pmatrix} = \begin{pmatrix}E\mathbf{a_1} & \cdots & E\mathbf{a_n}\end{pmatrix} =\begin{pmatrix}\mathbf{r_1} & \cdots & \mathbf{r_n}\end{pmatrix}$$ where $\mathbf{a_i}$ and $\mathbf{r_i}$ are the columns of $A$ and $R$ respectively. From this we get the equality $$E\mathbf{a_i} = \mathbf{r_i}$$ for $1\le i \le n$. Now consider the equation $$c_1\mathbf{r_1} + \cdots + c_n\mathbf{r_n} = \mathbf{0}$$ for scalars $c_i$. We rewrite this as $$c_1\left(E\mathbf{a_1}\right) + \cdots + c_n\left(E\mathbf{a_n}\right) = E\left(c_1\mathbf{a_1} + \cdots + c_n\mathbf{a_n}\right)=\mathbf{0}$$ since $E$ is a product of elementary matrices, it is invertible. In particular this implies $$c_1\mathbf{a_1} + \cdots + c_n\mathbf{a_n}\iff c_1\mathbf{r_1} + \cdots + c_n\mathbf{r_n} = \mathbf{0}$$ And it's easy to see that this holds for any subset of the columns. What this means is that elementary row operations preserves linear relations between columns. If a set of columns vectors are linearly independent in $R$ then they will remain linearly independent after transformation $R$ to $A$ via elementary row operations. Likewise, if a subset of the columns of $R$, say $\left\{\mathbf{r_{k_1}},\ \cdots,\ \mathbf{r_{k_\ell}}\right\}$ span the columnspace of $R$, then each $\mathbf{r_i}$ can be written as a linear combination of the set, say $$c_1\mathbf{r_{k_1}} + \cdots + c_{\ell}\mathbf{r_{k_\ell}} = \mathbf{r_i}$$ then correspondingly, we will have $$c_1\mathbf{a_{k_1}} + \cdots + c_{\ell}\mathbf{a_{k_\ell}} = \mathbf{a_i}$$ so that the corresponding columns of $A$ remain a spanning set.