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Show that $u \in \text{span}(v + w, w + u, u + v)$ My first approach is to set the matrix as follows and placing a 1 where a variable exists.

\begin{bmatrix}0&1&1&1\\1&0&1&0\\1&1&0&0\end{bmatrix}

reduces down to

\begin{bmatrix}1&0&0&-1/2\\0&1&0&1/2\\0&0&1&1/2\end{bmatrix}

$a = -\frac{1}{2}, b = \frac{1}{2}, c = \frac{1}{2}$. Now I am lost on justifying on why it spans, because it consists of a trivial solution that is consisent.

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    If you don't explain, in particular to yourself, how you form that matrix perhaps you won't be able to understand what is going on...what are the meaning of those numbers there? – 2017-02-23
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    This looks like half a conversation, most of which is going on in your head. How can anyone know what you are talking about? – 2017-02-23

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For example

$$\frac12(u+w)+\frac12(u+v)-\frac12(v+w)\;\ldots$$

Another way, with matrices, assuming $\;u,v,w\;$ are linearly independent: denote

$$\begin{cases}&u\leftrightarrow(1,0,0)\\&v\leftrightarrow (0,1,0)\\&w\leftrightarrow (0,0,1)\end{cases}\implies u+v=(1,1,0)\,,\,\,u+w=(1,0,1)\,,\,\,v+w=(0,1,1)$$

Now just check the above last three vectors are linearly independent (for example, using a matrix...) and thus obviously $\;u\;$, and any other three dimensional vector, belongs to their span...

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    Thanks, didn't think about showing it as linear independent. – 2017-02-23