Finding a basis for a subspace given an equation Consider the vector space $\mathbb{R}^4$ over $\mathbb{R}$ with its subspaces defined to be
$U = \{(x_{1}, x_{2}, x_{3},x_{4}) : 2x_{2} = x_{3} = x_{4} \}$
$W= \{(x_{1}, x_{2}, x_{3},x_{4}) : x_{1} = -x_{ 2}= x_{3} \}$
Find basis for $U, W, U\cap W$
Now the thing throwing me off is the I'm use to having a subspace say $2x+4y-3z = 0$ If we solve for can $ z$ easily find a basis.
However I never dealt with multiple equals signs nor intersections. How would I go about finding the basis for these any hints and help appreciated sorry if I misspell some things as I am in a phone typing.
 A: $\bullet$  $U: 2x_2=x_3=x_4$ gives $x=(x_1,x_2,x_3,x_4)=(x_1,x_2,2x_2,2x_2)=x_1(1,0,0,0)+x_2(0,1,2,2)$ and a basis is $${\cal B}_U=((1,0,0,0),(0,1,2,2))$$
$\bullet$ $W: x=(x_1,-x_1,x_1,x_4)=x_1(1,-1,1,0)+x_4(0,0,0,1)$;
$${\cal B}_W=((1,-1,1,0),(0,0,0,1))$$
$\bullet$  $U \cap W: \left\{\begin{array}{l}  2x_2=x_3=x_4 \\x_1=-x_2=x_3  \end{array}  \right. $ gives:$x_3=2x_2=-x_2$  then $x_2=x_3=0$ then $x_1=x_2=x_3=x_4=0$  and there is no basis since $$U \cap W =\{(0,0,0,0)\}$$
A: To get you started, notice that $U$ is the collection of vectors of the form
$$
(x_1,\frac{x_4}{2},x_4,x_4)
$$
where $x_1$ and $x_4$ are arbitrary, since $x_3 = x_4$ and $x_2 = \frac{x_4}{2}$. In other words, $U$ is the collection of vectors of the form 
$$
x_1(1,0,0,0)+x_4(0,1/2,1,1).
$$
From here you should be able to find a basis for $U$. A similar approach will work for $W$, and to find a basis for $U \cap W$, notice that a vector $(x_1,x_2,x_3,x_4)$ in $U\cap W$ satisfies the equations $2x_2=x_3=x_4$ and $x_1=-x_2=x_3$.
A: Notice that $U$ and $W$ are really null spaces -- the solution sets for a set of homogeneous equations.
For example: $U$ is the solution set for the equations: $2x_2=x_3=x_4$ which can be broken into $2x_2=x_3$ and $x_3=x_4$ so that...
$$ \begin{array}{cccccc}  & 2x_2 & -x_3 & & = & 0 \\
                          &      &  x_3 & -x_4 & = & 0 \end{array}$$
which leads to the matrix...
$$ \begin{bmatrix} 0 & 2 & -1 & 0 & : & 0 \\ 0 & 0 & 1 & -1 & : & 0 \end{bmatrix} \sim 
   \begin{bmatrix} 0 & 1 & 0 & -1/2 & : & 0 \\ 0 & 0 & 1 & -1 & : & 0 \end{bmatrix}$$
Notice that columns 2 and 3 are pivot columns and the other columns aren't. This means the variables that go with columns $1$ and $4$ (i.e. $x_1$ and $x_4$) are free variables. 
We get: $x_4=s$, $x_3-x_4=0$ so $x_3=s$, $x_2-0.5x_4=0$ so $x_2=0.5s$, and $x_1=t$. Thus 
if $(x_1,x_2,x_3,x_4) \in U$, we must have $(x_1,x_2,x_3,x_4)=s(0,0.5,1,1)+t(1,0,0,0)$ for some $s,t \in \mathbb{R}$.
Therefore, $\beta = \{(0,1/2,1,1), (1,0,0,0)\}$ is a basis for $U$.
A similar calculation works for $W$. 
Finally, how do we deal with $U \cap W$? Well, by definition a vector belongs to $U \cap W$ only if it belongs to both $U$ and $W$. This means that the components of $(x_1,x_2,x_3,x_4) \in U \cap W$ must satisfy both systems of equations. In other words $U \cap W$ is the solution set of the homogeneous system...
 $$ \begin{array}{cccccc}  & 2x_2 & -x_3 & & = & 0 \\
                          &      &  x_3 & -x_4 & = & 0\\
                          x_1 & + x_2 & & & = & 0 \\
                          x_1 & & -x_3 & & = & 0
 \end{array}$$
