Let $W=sp \{e_1,e_2,e_3,e_4\}, U= sp\{(1,-2,1,0),(0,3,-1,1)\}$ be vector spaces both are linearly independent.

Show that $U\cap W = sp\{(3,0,1,2)\}$.

I know that $\dim U\cap W =1$.

Now since every vector $v$ that is in the intersection is in both $U,W$ so when I do: $ae_1+be_2+ce_3+de_4=xu_1+yu_2$

I get: $a=x, \\b= -2x+3y, \\c = x-y,\\ d=y$

But this looks like it has a dimension of two. What am I missing here?

  • 1
    $\begingroup$ Since you are working with 4-triples, $W$ is all of 4-space. The intersection will be $U$, which has two dimensions. $\endgroup$ – Rory Daulton Jun 23 '15 at 11:23

The intersection is the entire space $U$ since $W$ is the entire $4-$dimensional vector space. You can see it using the Grassman formulae

$$dim(W+U)=dimW+dimU-dim(W\cap U); $$

now: $dimW=4,dimU=2$; since $U\subset W$ then $dim(W+U)=4$, then $$ 4=4+2-dim(W\cap U),$$ so $dim(W\cap U)=2$.

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  • $\begingroup$ Thanks, I noticed now what I was doing wrong, looking at two different problems simultaneously. $\endgroup$ – shinzou Jun 23 '15 at 11:35

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