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The given vector is $(6,1,-6,2)$ The vector set is $\{(1,-1,-1,0), (-1,0,1,1), (1,1,-1,1)\}$.

How does one prove that the vector is in the span of the set?

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The vector $(6,1,-6,2)$ is in the span of $(1,-1,-1,0)$, $(-1,0,1,1)$, and $(1,1,-1,1)$ if and only if there exist real numbers $\alpha,\beta,\gamma$ such that $$(6,1,-6,2) = \alpha(1,-1,-1,0) + \beta(-1,0,1,1) +\gamma(1,1,-1,1).$$

This becomes a system of four equations in three unknowns, by looking at each coordinate in turn: $$\begin{array}{rcrcrcl} \alpha & - & \beta & + & \gamma & = & 6\\ -\alpha & & &+&\gamma &=& 1\\ -\alpha &+&\beta &-&\gamma &=&-6\\ &&\beta &+ &\gamma&=&2. \end{array}$$

If the system has a solution, this solution witnesses the fact that $(6,1,-6,2)$ lies in the span of the other three vectors. If the system has no solutions, then this shows the vector does not lie in the span of the other three.

P.S. Curly brackets, { and }, are usually used to denote sets, not vectors.

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  • $\begingroup$ thank you Arturo, and thank you more for {} notation. $\endgroup$ – Maysam Jul 2 '11 at 6:29

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