Dimension of two subspaces Good evening. I'm trying to show that if the sum of the dimensions of two subspaces of a vector space exceeds the dimension of space then these subspaces have a vector in common. I have trouble to build the proof, because no how is this possible. 

Let $V$ be the vector space, and $W$, $U$ be the two subspaces. Choose bases $\{w_1,\dots,w_m\}$ and $\{u_1,\dots,u_n\}$ for $W$ and $U$, respectively.
If $m+n>\dim V$, then the set $\{w_1,\dots,w_m,u_1,\dots,u_n\}$ is linearly dependent, so there are scalars $c_1,\dots,c_{m+n}$ not all zero such that
$$ c_1w_1+\dots+c_mw_m+c_{m+1}u_1+\dots+c_{m+n}u_n=0 $$
Therefore
$$ c_1w_1+\dots+c_mw_m=-(c_{m+1}u_1+\dots+c_{m+n}u_n) $$
and neither side can be zero because the original sets $\{w_1,\dots,w_m\}$ and $\{u_1,\dots,u_n\}$ are linearly independent.
Thus $c_1w_1+\dots+c_mw_m$ is a non-zero vector in $U\cap W$.
By the way, a refinement of this argument can show that
$$ \dim(W+U)=\dim(W)+\dim(U)-\dim(W\cap U)$$
which will immediately solve the problem.
