All vectors in R3 with v1,v2 2v3 0

Consider all vectors in $R_3$ with $3v_1-v_3=0$ , $2v_1+3v_2-4v_3=0$ . Is this given set of vectors a vector space? Give Reasons. ( $v_1$, $v_2$, ... denote three components. Give Reasons. ( $v_1$, $v_2$, ... denote three components.

let $v_1$ = [ a b c ] $v_2$ = [ d e f ] $v_3$ = [ g h i ]

I have the feeling you're mixing components with vectors. I'm assuming they mean to consider vectors of the form $(v_1,v_2,v_3) \in \mathbb{R^3}$ , satisfying the given two conditions. Suppose $\vec v = (v_1,v_2,v_3)$ and $\vec w = (w_1,w_2,w_3)$ are two such vectors; you then want to check if:

• $\vec v + \vec w = (v_1+w_1,v_2+w_2,v_3+w_3)$ also satisfies the given conditions; i.e. is in the set;
• for any $k \in \mathbb{R}$, $k\vec v = (kv_1,kv_2,kv_3)$ also satisfies the given conditions; i.e. is in the set.

If the set is non-empty (check whether the zero vector is in the set; it clearly is) and if the answer to the two questions above is yes, then the set is indeed a subspace of $\mathbb{R^3}$.

Depending on what you have seen about linear subspaces, there may be simpler or more elegant arguments to avoid the straightforward verifications in the method above.

Alternative 1

The given equations determine planes through the origin. Vectors satisfying both lie on the intersection of both planes, which is a line through the origin. Lines through the origin are (one-dimensional) subspaces of $\mathbb{R^3}$, so the answer is yes.

Alternative 2

The two conditions come down to vectors $(v_1,v_2,v_3)$ satisfying: $$\color{blue}{\begin{pmatrix} 3 & 0 & -1 \\ 2 & 3 & -4 \end{pmatrix}}\begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$$ so it is exactly the null space (or kernel) of the blue matrix. The null space of a matrix (or linear transformation) is always a subspace.