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I need to show that these vectors span the $\mathbb{R}$-vectorspace in $\mathbb{R}^3$.

$\left[ \begin {array}{c} 1\\ 1\\ -2\end {array} \right]$, $\left[ \begin {array}{c} -1\\ -2\\ 3\end {array} \right]$, $\left[ \begin {array}{c} 2\\ 0\\ -2\end {array} \right]$, $\left[ \begin {array}{c} 3\\ 1\\ -3\end {array} \right]$,$\left[ \begin {array}{c} 1\\ 1\\ 1\end {array} \right]$

I don't really know how to do this, im thinking about putting into a Matrix and put on RREF. But I cant seem to get anything good out of it. This is what I get: $\left[ \begin {array}{cccccc} 1&0&4&0&-4&-5\,c-3\,a-6\,b \\ 0&1&2&0&-2&-2\,c-a-3\,b\\ 0&0&0 &1&1&c+a+b\end {array} \right] $

Can i use this, to solve my problem?

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    A collection of vectors span $\mathbb{R}^n$ if the vectors are in $\mathbb{R}^n$ and the matrix of these vectors has a pivot in every row. – 2017-02-12

2 Answers 2

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Hints:

With the vectors as rows, form a $\;5\times3\;$ matrix. Reduce it with Gauss elementary operations. At the end, you have to remain with at least three different rows that are not all-zeros. The vectors corresponding to those rows are a basis ofr $\;\Bbb R^3\;$ .

BTW, what you did also works for the specific question you ask, as you got the matrix you formed has rank three...

Why does the above work?

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Yes, that's good. Now you can conclude that the linear system $$ \alpha_1v_1+\alpha_2v_2+\alpha_3v_3+\alpha_4v_4+\alpha_5v_5=y $$ has a solution for every vector $y$, hence every vector is in the span.

More easily the matrix $$ A=\begin{bmatrix} 1 & -1 & 2 & 3 & 1 \\ 1 & -2 & 0 & 1 & 1 \\ -2 & 3 & -2 & -3 & 1 \end{bmatrix} $$ by the very same reduction has rank $3$, so it has a right inverse $R$. Thus the linear system $Ax=y$ has at least the solution $x=Ry$.

Without linear systems: the span of those vectors is the column space of $A$; since $A$ has rank $3$, the span is the whole $\mathbb{R}^3$.