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.
Best Answer
$\bullet$ $U: 2x_2=x_3=x_4$ 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 $$U \cap W =\{(0,0,0,0)\}$$