# Linear algebra - Basis and dimension of subspaces

Let $U$ and $W$ be subspaces of $\Bbb R^4$.

$U=\{ (x_1,x_2,x_3,x_4)| x_2+x_3+x_4=0\}$

$W=\{ (x_1,x_2,x_3,x_4)| x_1+x_2=0 \ and \ x_3-2x_4=0\}$

Find the basis and dimension of the subspaces $U\cap W$ and $U+W$.

Is $W\oplus U=\Bbb R^4$ ?

I'm not really sure on how to start this question, I don't see how to do the intersection or the addition of these two subspaces so any advice would be appreciated.

$(x_1,x_2,x_3,x_4) \in U\cap V$ iff $$x_2+x_3+x_4 = 0 \Rightarrow x_2 = - x_3 -x_4$$ $$x_1 + x_2 = 0 \Rightarrow x_1 = -x_2$$ $$x_3 - 2x_4 = 0 \Rightarrow x_3 = 2x_4$$ and so $$x_3 = 2x_4, x_2 = -3x_4, x_1 = 3x_4$$ The only "variable" is $x_4$, so $\text{dim}(U\cap V) = 1$ (Can you find a basis for $U\cap V$ now?)

For $U+W$, start by trying to find a spanning set for $U$ and $W$, and add them together component-wise. This should help you find a spanning set for $U+W$. After that, check to see if that set is linearly independent, and if not, eliminate redundant vectors.

• I see, so the basis for the intersection should be $(0,0,0,1)$ ? The spanning sets for U and W respectively should be $(1,0,0,0),(0,0,0,1)$ so the span of the addition is the same and it's dimension is 2. – GinKin Dec 11 '13 at 10:51
• $(0,0,0,1)$ is not an element of the intersection (in fact, it is not an element of $W$, because $x_3 - 2x_4 = 0 - 2\cdot 1 \neq 0$ – Prahlad Vaidyanathan Dec 11 '13 at 10:52
• We can represent everything in the intersection using $x_4$ so continuing from your calculation above: $(3x_4,-3x_4,2x_4,x_4)$. The basis should have 4 vectors so it should be 3000, 0-300, 0020, 0001. I'm pretty sure this time I got it right. Is it? – GinKin Dec 11 '13 at 11:54