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Given a set of vectors:

$M = \{(a, b, c)\ |\ a^2= b^2 = c^2 \vee 3a = b+c \}$

Is M a subspace of $\mathbb{Q^3}$?

I've already done the multiplication with a scalar, but I'm stuck at the addition of two vectors:

So if we have two vectors $(a_1,b_1,c_1)$ and $(a_2,b_2,c_2)$ which fulfill these conditions, for their sum we have:

$a_1^2 + 2a_1a_2 + a_2^2 = b_1^2 + 2b_1b_2 + b_2^2$, and thus

$a_1a_2 = b_1b_2$

For both $a_1$ and $a_2$ we have $a^2-b^2=0 \Rightarrow a = b \vee a = -b $

If $a_1=b_1$ then $a_2 = b_2 \vee a_2 = -b_2$, so I guess $M$ is not a subspace, because the equality doesn't hold always. Am I right?

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    $\begingroup$ $\Bbb Q_3=\Bbb Q^3$? $x=a$? $y=b$? $z=c$? $\endgroup$ – Lord Shark the Unknown Dec 21 '18 at 21:04
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Well, the triples $(1,1,1)$ and $(-1,1,1)$ are elements of $M$, however their sum $(1,1,1) + (-1,1,1) = (0,2,2)$ is not, so $M$ cannot be a linear subspace.

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