5. Basis within a spanning set

Problem. Among the list of vectors

$\left[\begin{array}{r}1\\ 2\\ 1\end{array}\right]$, $\left[\begin{array}{r}2\\ 4\\ 2\end{array}\right]$, $\left[\begin{array}{r}1\\ 3\\ 3\end{array}\right]$, $\left[\begin{array}{r}2\\ 7\\ 8\end{array}\right]$, $\left[\begin{array}{r}3\\ 8\\ 8\end{array}\right]$

choose some that form a basis for the span of all five.

Solution. Form the matrix $M$ with these as columns and row reduce; let's say $E$ is the row-reduced matrix. In $E$, notice that the non-pivot columns are in the span of the pivot columns. Since $M$ has the same relations between columns as $E$, the same pivot columns in $M$ are the ones you want for your basis.

Here $E = \left[\begin{array}{ccccc}1&2&0&-1&0\\ 0&0&1&3&0\\ 0&0&0&0&1\end{array}\right]$. The pivot columns are columns 1, 3, and 5. So in the original list of vectors, the first, third, and fifth make a basis.

Note. If you are asked simply to find a basis for the span of a list of vectors, with no method specified, you have a choice:

(1) Make these the rows of a matrix and row-reduce to get a basis for the row space, as in Section , or (2) rewrite these vectors as column vectors and select some to be a basis, as in this section.