Determine is $U$ and $V$ subspace and find basis and dimension? $V=R^{3}$.
$U=\left\{ (x_{1},1,x_{3} )^{T}\in \mathbb R ^{3}: x_{1},x_{3}\in \mathbb R  \right\} $
$W=\left\{ (x_{1},x_{2},x_{3} )^{T}\in \mathbb R ^{3}: x_{1}+2x_{2}+x_{3}=0  \right\} $


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*Determine are sets $U,W\subset V$, subspace of $V$?

*Find basis and dimension of $U$ and $W$ if it is possible?


I am not sure that I completely know how this should work, so I hope someone can say is it good.
What I have done:
-U is not subspace of V because it is not closed under the addition(because of this 2).
$\begin{pmatrix} x_{1}  \\ 1 \\ x_{3} \end{pmatrix} +\begin{pmatrix} y_{1} \\ 1 \\ y_{3} \end{pmatrix}=\begin{pmatrix} x_{1}+y_{1} \\ 2 \\ x_{3}+y_{3} \end{pmatrix}  $
-W is subspace because it is closed under the addition and under the scalar multiplication.
I can not find basis and dimension for U, right?
Basis for W: 
$B=s\begin{pmatrix} -2 \\ 1 \\ 0 \end{pmatrix}+t\begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}$  
$Dim(B)=2$
Can someone say me is this good, because I am not sure I completely understand rules like closed "under the scalar multiplication" and "under the addion"?
 A: It is true, but for the last question a basis is a set of vectors, so you should write :
$$B=\left\{\begin{pmatrix} -2 \\ 1 \\ 0 \end{pmatrix},\begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}\right\}$$
And $W$ is closed under the scalar multiplication means that : $\forall u \in W, \forall  \lambda \in \mathbb{R},\lambda u\in W$.
$W$ is closed under the addition means that : $\forall u,v \in W, u+ v\in W$.
Let's verify these properties for $W$, let $\begin{pmatrix} a \\ b \\ c \end{pmatrix}$ and $\begin{pmatrix} a' \\ b' \\ c' \end{pmatrix}$ elemnts of $W$ and $\lambda \in \mathbb{R}$. 
$\begin{pmatrix} a \\ b \\ c \end{pmatrix}+\begin{pmatrix} a' \\ b' \\ c' \end{pmatrix}=\begin{pmatrix} a+a' \\ b+b' \\ c+c' \end{pmatrix}$
and $a+a'+2(b+b')+c+c'=(a+2b+c)+(a'+2b'+c')=0$. So $\begin{pmatrix} a \\ b \\ c \end{pmatrix}+\begin{pmatrix} a' \\ b' \\ c' \end{pmatrix} \in W$. So $W$ is closed under addition.
$\lambda \begin{pmatrix} a \\ b \\ c \end{pmatrix}=\begin{pmatrix} \lambda a \\ \lambda b \\ \lambda c \end{pmatrix}$ and $\lambda a+2\lambda b+\lambda c=\lambda(a+2b+c)=0$. So $\lambda \begin{pmatrix} a \\ b \\ c \end{pmatrix} \in W$. So $W$ is closed under scalar multiplication.
