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.

  • $\begingroup$ 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. $\endgroup$ – GinKin Dec 11 '13 at 10:51
  • 1
    $\begingroup$ $(0,0,0,1)$ is not an element of the intersection (in fact, it is not an element of $W$, because $x_3 - 2x_4 = 0 - 2\cdot 1 \neq 0$ $\endgroup$ – Prahlad Vaidyanathan Dec 11 '13 at 10:52
  • $\begingroup$ 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? $\endgroup$ – GinKin Dec 11 '13 at 11:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.