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I have a set of vectors:

$$V=\lbrace x\in\Bbb{R}^5:x_1+x_2+x_3+x_4+x_5=0\rbrace$$

I need to find basis for this set of vectors. What is the "algorithm" for solving such problems? Where should I start?

EDIT:

Is this a valid basis: $\lbrace[1,0,0,0,-1]^T,[0,1,0,0,-1]^T,[0,0,1,0,-1]^T,[0,0,0,1,-1]^T\rbrace$?

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    $\begingroup$ What's the dimension of this subspace? That's how many linearly indep vectors you need to find. And what are some straight-forward, non-zero vectors in this space? $\endgroup$ – Simon S Nov 8 '14 at 18:35
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    $\begingroup$ (1,-1,0,0,0), (0, 1, -1, 0, 0), ... $\endgroup$ – user135520 Nov 8 '14 at 18:36
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Express one of the variables in terms of the others, for example $x_5 = - \sum_{i=1}^4 x_i$, to which you assign arbitrary values. Then you can write a general vector in $V$ in terms of a linear combination of four vectors which form a basis.

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  • $\begingroup$ Is my answer correct? $\endgroup$ – qiubit Nov 8 '14 at 18:42
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    $\begingroup$ @qiubit It is indeed, note that it may be worth to show that your set of vectors is also linearly independent to show that it is a basis (you have shown spanning already) but this is trivial. $\endgroup$ – suarko Nov 8 '14 at 18:44
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You have a homogeneous system in a 5 dimensional space. Your vector space is described by only one linear equation. So the cardinality of your basis is undoubtedly 4, which is also the dimesion of your vector space (R-C's Theorem). You get a basis giving your 4 free parameters a random value. For example, let $x_5=t, x_4=s, x_3=u, x_2=v, x_1=-t-s-u-v$. Then, you give the values $1$ for one parameter and $0$ for the others.

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