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I'm stuck on this one question that was on my math work sheet. Say that a certain vector space $V$ consists of triples of real numbers $u=(u_1,u_2,u_3)\in \mathbb{R}^3$ with vector addition and scalar multiplication defined by $$(u_1,u_2,u_3)+(v_1,v_2,v_3)=(u_1+v_1,u_2+v_2,u_3+v_3-2u_1v_1)$$$$k(u_1,u_2,u_3)=(ku_1,ku_2,ku_3+(k-k^2)u_1^2)$$ For what value of $a$ is $(2,0,a)$ in the span Of $(1,1,1),(-1,1,1)\subset V$ I might have a method to do this but I'm not sure what it means by triples of real numbers

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Do those operations actually give a vector space? Don't remember seeing anything like that before. –  Gerry Myerson Mar 13 '13 at 2:15
    
Oh, "triples of real numbers" just means things like $(\sqrt2,\pi,-49.873)$. –  Gerry Myerson Mar 13 '13 at 2:17
    
Reminds me of the heisenberg group structure almost. –  BaronVT Mar 13 '13 at 2:53
    
Pair: $(x,y)$, triple: $(x,y,z)$, quadruple: $(x,y,z,w)$, $n$-tuple: $(x_1,x_2,\ldots,x_n)$. See the Wikipedia entry on tuple. –  user1551 Mar 13 '13 at 6:12
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up vote 4 down vote accepted

If $(2,0,a)$ is in $\langle (1,1,1),(-1,1,1)\rangle$ then there exists scalars $z_1,z_2 $ such that $$z_1(1,1,1)+z_2(-1,1,1)=(2,0,a)$$

Can you show when this can be true?

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+1 Peter. and quick, too! –  amWhy Mar 13 '13 at 2:16
    
Thanks for the quick response Peter, I'm pretty sure I can handle it from there. –  Chance Mar 13 '13 at 2:37
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