$W = { (x_1, x_2, x_3)^T: 3x_1 +1/4x_2=0 }. $Is W a subspace? I'm learning about proving whether a subset of a vector space is a subspace.
It is my understanding that to be a subspace this subset must:
Have the 0 vector.
Be closed under addition (add two elements and you get another element in the subset).
Be closed under multiplication by a scalar (one element by any scalar yields another element in the subset).
But still I don't know how to use that on this example..
If someone could help me I would be really thankful.
 A: Essentially, we have:
$$W \begin{array}[t]{l}=\{ (x_1, -12 x_1, x_3)^T: x_1, x_3 \in \mathbb R\}\\
= \{(x_1, -12x_1,0)^T + (0,0,x_3)^T: x_1, x_3 \in \mathbb R\}\\
=\{x_1 \cdot \underbrace{(1,-12,0)^T}_{\vec{e}_1} + x_3\cdot \underbrace{(0,0,1)^T}_{\vec{e}_2}: x_1, x_3 \in \mathbb R\}
.
\end{array}$$
Thus, every element in $W$ can be written as a linear combination of the vectors $\vec{e}_1 = (1,-12,0)^T$ and  $\vec{e}_2 = (0,0,1)^T$.
Let's take $2$ vectors $\vec v, \vec u \in W$. Then, it holds:
$\vec v = a \cdot \vec{e}_1 + b\cdot \vec{e}_2$ and $\vec u = c\cdot \vec{e}_1 + d\cdot \vec{e}_2$, with $a,b,c,d \in \mathbb R$.
I think now you can prove the $2$ properties which must hold.

Additional comment:
$\begin{array}{l}
3x_1 + \frac{1}{4}x_2 = 0\\
\frac{1}{4}x_2 = -3x_1\\
x_2 = -12 x_1
\end{array}
$ 
and also:
$ (a,-12a,0)^T + (0,0,b)^T = (a, -12a,b)^T$. It is just addition of $2$ vectors.
A: I presume that you have to or want to prove that $W$ is a vector space straight from the definition of vector space. What you have to show is exactly as you say:


*

*Show that the vector $(x_1, x_2, x_3) = (0, 0, 0)$ satisfies the equality $3 x_1 + 1/4 x_2 = 0$.

*Assuming that $(x_1,x_2,x_3)$ and $(x_1',x_2',x_3')$ both satisfy that equality, show that the sum $(x_1+x_1',x_2+x_2',x_3+x_3')$ also satisfies that equality.

*Assuming that $(x_1,x_2,x_3)$ satisfies that equality and taking an arbitrary $a$, show that $(ax_1,ax_2,ax_3)$ satisfies that equality.

