I just need someone to check if I'm doing this properly.
Find a basis for: $span\{(1,1,1),(1,0,1),(0,0,0),(0,1,0)\}$
I formed matrix A using these 4 vectors as columns;
$A=\begin{bmatrix}1&1&0&0\\1&0&0&1\\1&1&0&0\end{bmatrix}^T$
$rref = \begin{bmatrix}1&0&0&-1\\0&1&0&1\\0&0&0&0\end{bmatrix}^T$
I'm a bit uncertain about the next step. I have pivots in the first two columns, meaning they are linearly independent, so does that mean my basis is just the first two columns of $A$?
$basis=\{ (1,1,1), (1,0,1) \}$
$\endgroup$ 72 Answers
$\begingroup$The span is all linear combinations of the basis vectors.
$(1,1,1) = (1,0,1) + (0,1,0)$ and $(0,0,0) = 0*(1,0,1)$ so we can remove these two vectors from a potential basis.
$\alpha(1,0,1) + \beta(0,1,0) = (\alpha,\beta,\alpha) = (0,0,0)$ only when $\alpha = \beta = 0$.
Thus $(1,0,1)$ and $(0,1,0)$ are linearly independent and thus a basis of $\text{span}\{(1,1,1),(1,0,1),(0,0,0),(0,1,0)\}$.
$\endgroup$ $\begingroup$In general, if we take the vectors as columns and operate row reduced form, we will receive to pivot
In pivot matrix the columns which have leading 1, are not directly linear independent, by help of that we choose linear independent vector from main span vectors.
$\endgroup$
