The question from my test book asks to find the basis for the span of the set:
$$\{\begin{pmatrix}1\\2\\1\end{pmatrix},\begin{pmatrix}3\\1\\-1\end{pmatrix}, \begin{pmatrix}1\\-3\\-3\end{pmatrix}\}\subseteq \mathbb{R}^3
$$
The solution shows that the professor transposed the matrix and then the row reduced to find the 2 bases $\left\langle\begin{pmatrix}1\\2\\1\end{pmatrix},\begin{pmatrix}0\\-5\\-4\end{pmatrix} \right\rangle
$. However, I understand that if I just row reduced without transposing, I would be finding the basis for the row space. Can someone please explain why I have to write the vectors as rows (i.e., transpose) to find the basis for the span?
Thanks!
Best Answer
Without transposing, you can use the row echelon form to recover a basis of the column space. It has the advantage of being expressed as a subset of the original vectors. See, for example, lectures 37 and 38 of my course. The punchline is this: Take the columns of the original matrix in which you find pivots in the row echelon form.