# basis of a matrix that is transposed.

So lets say the transposed matrix is this:

Span{(0,-3,6,6,4,-5),(2,1,4,2,8,-9),(3,7,-5,-8,8,9),(3,9,-9,-12,6,15)} The RREF for the original matrix is

$$\begin{pmatrix} 1 & 0 & 3 & 2 & 0 & -24 \\ 0 & 1 & -2 & -2 & 0 & 7 \\ 0 & 0 & 0 & 0 & 1 & 4 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix}$$

The RREf for the transposed matrix would then be different Im assuming.

If I knew the basis for the original matrix is there a faster way to find the basis for the transposed matrix or do I have to find the reef of the this transposed matrix and find the basis using the long way?

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The non-zero rows of the RREF forms a basis for the rowspace of the matrix. The pivot columns in turn serve as an "index" for a basis of the columnspace. For example, suppose we are given a matrix $A$ with RREF $R$, $$A = \begin{pmatrix}\mathbf{a_1} & \mathbf{a_2} & \cdots & \mathbf{a_n}\end{pmatrix} \ \sim \ R=\begin{pmatrix}\mathbf{r_1} & \mathbf{r_2} & \cdots & \mathbf{r_n}\end{pmatrix}$$ where $\mathbf{a_i}$ and $\mathbf{r_i}$ are the respective column vectors of $A$ and $R$. Now further suppose that $$\left\{\mathbf{r_{i_1}},\ \mathbf{r_{i_2}},\ \cdots,\ \mathbf{r_{i_k}}\right\}$$ is the set of pivot columns of $R$. Then the corresponding columns of $A$ $$\left\{\mathbf{a_{i_1}},\ \mathbf{a_{i_2}},\ \cdots,\ \mathbf{a_{i_k}}\right\}$$ forms a basis for the columnspace of $A$. Of course this will in general not be the same basis as the one produced by row reducing $A^\mathrm{T}$, but there is rarely a need to prioritize one basis over the other.