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W1:= {(x,y,z) : x,y,z E R, x+y -5z = 0} ⊆R^3 W2:= ⊆R^3 , e1 := (1,0,0) and e2:= (0,1,0) I need to give a description of the intersection of W1 and W2, and find the basis and dimension of the intersection.

I calculated the dimension of W1 =2 and the dimension of W2 is also 2, however I’m unsure how to find the intersection of W1 and W2 and how to find the basis and dimension of the intersection.

I know the basis is a subset that is linearly independent and spans, however as I can’t work out what intersection of W1 and W2 actually is I don’t know how to do any more of the question, thank you.

  • $W_2$ is the set of vectors in $\Bbb R^3$ such that the third entry is always zero. Are there any vectors in $W_1$ whose third entry is zero? Yes. For example $(3,-3,0)$ satisfies $x+y-5z=0$. Having noticed this, can you continue? – JMoravitz Nov 20 '19 at 13:13

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Let $(x,y,z)\in W_1\cap W_2$. If $\dim(W_1\cap W_2)=2$, then that means that $W_2=W_1\cap W_2$. But clearly neither $e_1$ nor $e_2$ satisfy the equation for $W_1$. So this is false. So that means that $\dim(W_1\cap W_2)$ is either $1$ or they intersect trivially. Now, $(1,-1,0)=e_1-e_2\in W_1\cap W_2$, so this tells us that $W_1\cap W_2$ is in fact one dimensional and hence any non-zero vector in it gives us a basis. E.g. $(1,-1,0)$.

Also for a more general approach see this: How to find basis for intersection of two vector spaces in $\mathbb{R}^n$

take008
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  • That makes sense thank you, I’m just checking what does it mean for something to intersect trivially? –  Nov 20 '19 at 13:25
  • $W_1$ and $W_2$ intersect trivially if the only common vector is $(0,0,0)$. – take008 Nov 20 '19 at 13:26