prove if the function is vector space Let $\Bbb R^3$ be 3-dimensional euclidean space.
Is the set of all vectors in $\Bbb R^3$ that have the form $v=(v_1,v_2,v_3)$ where $5v_1-3v_2+2v_3=0$ itself a vector space?
so my take:
$v+w = (v_1+w_1,v_2+w_2,v_3+w_3)$ satisfies $5(v_1+w_1)-3(v_2+w_2)+2(v_3+w_3) = 0 + 0 = 0$
and now I know I have to use similar method to use $tv$ has this form but I'm not quite sure how to write this out. 
 A: If $v:=(v_1,v_2,v_3)$ is in your set, then
$$
5v_1-3v_2+2v_3=0
$$
hence $tv=(tv_1,tv_2,tv_3)$ is such that
$$
5(tv_1)-3(tv_2)+2(tv_3)=t(5v_1-3v_2+2v_3)=t\cdot0=0
$$
So, $tv$ is also in your set.
A: It's a bit cleaner if you write the original equation as $[5,-2,2] \cdot x =0$. Let $v,w$ be two such vectors. Obviously its a nonempty set as the zero vector satisfies this.
Then, you have $[5,-2,2] \cdot (a v + w) = [-5,-2,2] \cdot (a v) + [5,-2,2] \cdot w = a ([-5, -2, 2] \cdot v) + [5,-2,2] \cdot w = a (0) + 0 = 0$.
Thus, the set is a subspace because its closed under addition and scalar multiplication.
A: Let $A$ be the $3\times 1$ matrix 
$$
A=\begin{bmatrix}5 & -3 & 2\end{bmatrix}
$$
and let $T:\Bbb R^3\to\Bbb R$ be the linear map $T(v)=Av$. Then
\begin{align*}
\ker T
&= \left\{v\in\Bbb R^3:T(v)= 0\right\} \\
&= \left\{v\in\Bbb R^3:Av= 0\right\} \\
&= \left\{(v_1,v_2,v_3)\in\Bbb R^3:5\,v_1-3\,v_2+2\,v_3=0\right\}
\end{align*}
That is, your collection is precisely $\ker T$. Now, since the kernel of a linear map is a subspace of $\Bbb R^3$ we get your result for free!
