Proving vector subspace is closed on multiplication 
I'm trying to prove if F1 is a subspace for R^4. I've already proved that the 0 vector is in F1 and also F1 is closed under addition. I'm confused on how to be able to prove it is closed under multiplication?
I have:
Take some scalar of alpha. alpha * F1 = (alphax, alphay, alphaz, alphat) = alpha3x + alpha2y + alphaz = -alphat. But what do I do from here?
 A: Maybe it would be easier to understand if you took some $(x, y, z, t) \in \mathcal{F}_1$ and checked whether $\alpha (x, y, z, t) = (\alpha x, \alpha y, \alpha z, \alpha t)$ also belongs to $\mathcal{F}_1$.
A: $\mathcal{F}_1 = \{ (x, y, z, t) \in \mathbf{R}^4 : 3 x + 2 y + z + t = 0 \}$.
S1) $\mathbf{0} \in \mathcal{F}_1$.
This holds obviously.
S2) $\mathcal{F}_1 $ is closed under vector addition.
Suppose $\mathbf{v}_1, \mathbf{v}_2 \in \mathcal{F}_1$.
Then we take
$$
\mathbf{v}_1 = (x_1, y_1, z_1, t_1), \ \
\mathbf{v}_2 = (x_2, y_2, z_2, t_2).
$$
Then they satisfy
$$
3 x_1 + 2 y_1 + z_1 + t_1 = 0, \ \ 
3 x_2 + 2 y_2 + z_2 + t_2 = 0 \tag{1}
$$
Adding the two equations in (1), we get
$$
3 (x_1 + x_2) + 2 (y_1 + y_2) + (z_1 + z_2) + (t_1 + t_2) = 0.
$$
Thus,
$$
\mathbf{v}_1 + \mathbf{v}_2 \in \mathcal{F}_1
$$
Hence, S1) has been proved.
S2) $\mathcal{F}_1$ is closed under scalar multiplication.
Let $\alpha \in \mathbf{R}$ and $\mathbf{v} \in \mathcal{F}_1$.
We take $\mathbf{v} = (x, y, z, t)$. Then
$$
3 x + 2 y + z + t = 0 \tag{2}
$$
Multiplying both sides of (2) by $\alpha$, we get
$$
3 (\alpha x) + 2 (\alpha y) + (\alpha z) + (\alpha t) = 0
$$
This shows that $\alpha \mathbf{v} \in \mathcal{F}_1$.
Hence, S3) has been proved.
Hence, $\mathcal{F}_1$ is a subspace of $\mathbf{R}^4$. $ \ \ \ \ \blacksquare$
