Dimension of intersection of subspaces proof

[Dansuer]

Homework Statement

$V$ is a vector space with dimension n, $U$ and $W$ are two subspaces with dimension k and l.
prove that if k+l > n then $U \cap W$ has dimension > 0

Homework Equations

Grassmann's formula

$dim(U+W) = dim(U) + dim(W) - dim(U \cap W)$

The Attempt at a Solution

Suppose k+l >n.
Suppose that $dim(U \cap W) \leq 0$

since the dimension can't be negative $dim(U \cap W) = 0$

then Grassman formula reduces to

$dim(U+W) = dim(U) + dim(W)$
$dim(U+W) = k +l > n$

this is a contraddiction because $U+W$ has dimension grater than the dimension of his enclosing space.

is this a valid proof?

 

[HallsofIvy]

What, exactly, are you trying to prove?

 

[Dansuer]

that $dim(U \cap W) > 0$
since you are asking me that question do i have to assume that my proof is wrong?

EDIT: ops

I have to prove that if k+l > n then $dim(U \cap W) > 0$