Proving a subset of $\mathbb{R}^4$ is a vector space. Define $U = \left\{\begin{bmatrix} a + b + c \\ c \\ a+c \\ b \end{bmatrix} : a,b,c \in \mathbb{R}\right\} \subseteq \mathbb{R}^4$.
Show that $U$ is a real vector space with the usual vector addition and scalar multiplication.
I am new to the concept of vector spaces and I have no idea how to prove this. I would really appreciate if someone could help me understand the process of proving this.
 A: I will assume that you know that $\mathbb{R}^4$ is a real vector space with the usual notions of vector addition and scalar multiplication.
There are $3$ things we need to verify in order to see that $U$ is a subspace of $\mathbb{R}^4$:


*

*$0 \in U$. (Where by $0$ I mean the $0$ vector.)

*If $u,v \in U$, then $u+v \in U$.

*If $u \in U$ and $r \in \mathbb{R}$, then $ru \in U$.





*

*By setting $a = b = c = 0$, we see that $0 \in U$.





*Suppose $u = \begin{bmatrix} a_u + b_u + c_u \\ c_u \\ a_u + c_u \\ b_u \end{bmatrix}$ and $v = \begin{bmatrix} a_v + b_v + c_v \\ c_v \\ a_v + c_v \\ b_v \end{bmatrix}$.


Then $$u + v = \begin{bmatrix} (a_u + b_u + c_u) + (a_v + b_v + c_v) \\ c_u + c_v \\ (a_u + c_u) + (a_v + c_v) \\ b_u + b_v \end{bmatrix} = \begin{bmatrix} (a_u + a_v) + (b_u + b_v) + (c_u + c_v) \\ c_u + c_v \\ (a_u + a_v) + (c_u + c_v) \\ b_u + b_v \end{bmatrix} \in U.$$



*I'll leave this to you.

A: Write
$$
\begin{bmatrix} a + b + c \\ c \\ a+c \\ b \end{bmatrix}=
\begin{bmatrix} a \\ 0 \\ a \\ 0 \end{bmatrix}+
\begin{bmatrix} b \\ 0 \\ 0 \\ b \end{bmatrix}+
\begin{bmatrix} c \\ c \\ c \\ 0 \end{bmatrix}=
a\begin{bmatrix} 1 \\ 0 \\ 1 \\ 0 \end{bmatrix}+
b\begin{bmatrix} 1 \\ 0 \\ 0 \\ 1 \end{bmatrix}+
c\begin{bmatrix} 1 \\ 1 \\ 1 \\ 0 \end{bmatrix}
$$
so you can recognize that $U$ is the span of the three vectors
$$
\begin{bmatrix} 1 \\ 0 \\ 1 \\ 0 \end{bmatrix}\qquad
\begin{bmatrix} 1 \\ 0 \\ 0 \\ 1 \end{bmatrix}\qquad
\begin{bmatrix} 1 \\ 1 \\ 1 \\ 0 \end{bmatrix}
$$
hence a subspace. It is also the image of the linear map
$$
f\colon\mathbb{R}^3\to\mathbb{R}^4
\qquad
f\left(\begin{bmatrix}a\\b\\c\end{bmatrix}\right)=
\begin{bmatrix}
1 & 1 & 1 \\
0 & 0 & 1 \\
1 & 0 & 1 \\
0 & 1 & 0
\end{bmatrix}
\begin{bmatrix}a\\b\\c\end{bmatrix}
$$
(this is just a reformulation of being the span of the above three vectors).
