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.
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: $$ 3(x+u)-2(y+v)+(z+w)=3x-2y+z+3u-2v+w=0+0=0 $$