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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 $A$?

$basis=\{ (1,1,1), (1,0,1) \}$

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    I think you should operate row reduction on the *transposed* matrix to see that – or operate column-reduction. – 2017-02-04
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    I'm not sure if I understand... I should write the vectors as rows rather than as columns? If so - how do I get a set of linearly independent vectors from this? – 2017-02-04
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    It's to interpret the results more easily: when you have a linear combination of rows, it exactly corresponds to a linear combination of the vectors. This is not the case without transposing. – 2017-02-04
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    So the math is right, but I haven't written it very well? Or the math is wrong and I've written it nonsensically? – 2017-02-04
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    Row reduction as you wrote it leads to determining the *linear relations* between the vectors. Here you obtain the relation $v_1-v_2+v_4=0$. So, as we're in dimension $3$, you know the system of vectors has rank $2$. – 2017-02-04
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    I'm really sorry, I'm so bad at this that I'm having trouble following how this ties back to transposing/column-reduction. Is it better with my (minor) edits? – 2017-02-04
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/53030/discussion-between-bernard-and-kookiemonster). – 2017-02-04

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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 $\text{span}\{(1,1,1),(1,0,1),(0,0,0),(0,1,0)\}$.