Prove that it's a subspace of $\mathbb{R}^3$ I have the following definition of $V_a$:
$$V_a := \{(x, y, z)^T \in \mathbb{R}^3 : y = 3x - az\}, \quad \text{for $a \in \mathbb{R}$}.$$
My first problem: I don't understand this definition. Which role does $a$ play in this formula? How do the vectors look like? Why there's only $y$ given, what about $x$ and $z$?
Second problem (which is no wonder taking into account the first one): How can I find a basis of $V_a$ for all $a \in \mathbb{R}$
And one more: How can I prove that $V_a$ is a subspace of $\mathbb{R}^3$?
Please, help me to understand and to solve these problems.
 A: For every $a \in \mathbb{R}$ you take, you get a different $V_a$. For example,
$$V_0 = \{ (x,y,z)^T \in \mathbb{R}^3 \colon x,y,z \in \mathbb{R}, \ y = 3x \} = \{ (x,3x,z)^T \colon x,z \in \mathbb{R} \}.$$
For a general $a$, you have
$$V_a = \{ (x,y,z)^T \in \mathbb{R}^3 \colon x,y,z \in \mathbb{R}, \ y = 3x-az \} = \{ (x,3x-az,z)^T \colon x,z \in \mathbb{R} \}.$$
Note that
$$(x,3x-az,z)^T = x(1,3,0)^T + z(0,-a,1)^T$$
for all $x,z \in \mathbb{R}$. So, the set $\{(1,3,0)^T, (0,-a,1)^T\}$ spans $V_a$ for every $a \in \mathbb{R}$. To prove that it's basis, you need to show that these two vectors are linearly independent.
I'll let you take over now. Feel free to ask if you get stuck.
A: Firstly, $a\in \mathbb{R}$ is just some fixed constant. You want to show that $V_a$ is a subspace for all $a$.
Second, it's not that $y$ is given, none of $x,y,z$ is "given". You are told that $V_a$ is the set of all $(x,y,z)$ that satisfy $y=3x-az=0$, i.e. all points $(x,y,z)$ that lie on the plane given by $3x-y-az=0$. Namely $V_a$ is the plane described by $3x-y-az=0$.
Now what do you know about planes? They are subspaces iff they contain the origin. It is easy to see that $30-0-a0=0$ so indeed $V_a$ is a subspace.
Finally, to find a basis for $V_a$ all you need is any two nonparallel vectors in the plane. Try to find two such vectors by looking for two of the form $(0,1,z_1)$ and $(1,0,z_2)$ which lie on the plane. Any two nonparallel vectors will work, but these seem the obvious choices. (If $a=0$ be careful though, maybe other choices would be best.)
