# About finding intersection between two vector spaces

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?

• Since you are working with 4-triples, $W$ is all of 4-space. The intersection will be $U$, which has two dimensions. – 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 $dim(W\cap U)=2$.