How to show that the given 2 sets is a vector space?

Question

I am having a problem with this weird vector space problem

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.

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

$3v_1-v_3=0$ $+$ $2v_1+3v_2-4v_3=0$ $=$ $ 5v_1+3v_2-5v_3=0$

the zero vector exists because $5(0)+3(0)-5(0)=0=0$

$v_1 + v_2$ = [ a b c ] + [ d e f ] = [ a+d b+e c+f ]

$v_1 + v_2$ = [ a+d b+e c+f ] = [ d+a e+b f+c ] = v2 + v1

$(v_1 + v_2) + v_3$ = [ a+d b+e c+f ] + [ g h i ] = [ a+d+g b+e+h c+f+i ] = [ a b c } + [ d+g e+h f+i ] = $v_1 + (v_2 + v_3)$

$v_1$ + 0 = [ a b c ] + [ 0 0 0 ] = [ a b c ] = $v_1$

v1 + (-v1) = [ a b c ] + { -a -b -c ] = [ 0 0 0 ]

$kv_1$ = k [ a b c ] = [ ka kb kc ]

$k (v_1+v_2)$ = k [ a+d b+e c+f ] = [ ka+kd) kb+ke kc+kf ] = [ ka kb kc ] + [ kd ke kf ] = $kv_1 + kv_2$

(k+L) v1 = (k+L) [ a b c ] = [ ka+La kb+Lb kc+Lc ] = [ ka kb kc ] + [ La Lb Lc ] = kv1 + Lv1

$k(Lv_1)$ = k [ La Lb Lc ] = (kL) [ a b c ] = (kL) v1

$1(v_1)$ = 1 [ a b c ] = [ a b c ] = $v_1$

Is this a legit prove that it is a vector space?

Also, if i pick [ 0 1 0 ] , [ 1 1 0 ] , and [ 0 1 1 ] and i put into the equation $5v_1+3v_2-5v_3$= [ 3 3 -5 ] which is not equal to 0 , so this vector does not belong to the set so it is not a vector space ?

Answer 1

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.