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I am having trouble wrapping my head around proving the axiom: "vector plus another vector within a set exists in the set". Mainly, I am not sure how to go about proving it with the given polynomial.

V = {[x,y,z] such that 3x-2y+z=0}

My initial thought was that since you are proving a polynomial with the trivial solution, then V must be a vector space. What can I do to prove whether this is a vector space? Thank you.

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    A vector space is closed under addition. The sum of two polynomials of equal degree form a polynomial of the same degree. This would suggest that $p_n(x)$ (polynomials of degree n) meets criteria defining a vector space. – 2017-02-15

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The simpler proof:

The equation $3x-2y+z=0$ is the equation of a plane that passes through the origin, that is a subspace of dimension $2$ of $\mathbb{R}^3$ .

So the set of vectors $[x,y,z]$ that satisfies such equation are the vectors of the plane, and are elements of a vector space.

If you want to prove directly the addition closure take two vectors $$ \vec x=[x,y,z] \quad \mbox{with} \quad 3x-2y+z=0 $$ and

$$ \vec v= [u,v,w] \quad \mbox{with} \quad 3u-2v+w=0 $$

and note that: $$ \vec x +\vec v=[x+u,y+v,z+w] $$ so that: $$ 3(x+u)-2(y+v)+(z+w)=3x-2y+z+3u-2v+w=0+0=0 $$

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    Thanks! Very straightforward – 2017-02-15