Given two subspaces $U,W$ of vector space $V$, how to show that $\dim(U)+\dim(W)=\dim(U+W)+\dim(U\cap W)$ 
Let $U,W$ be subspaces of a vector space $V$. Show that
  $$\dim(U)+\dim(W)=\dim(U+W)+\dim(U\cap W)$$
  Hint: Show that the map given by $L:U×W\to V$ given by $L(u,w)=u-w$ is linear.

I can show that $L:U×W\to V$ given by $L(u,w)=u-w$ is a linear map. I also know that the dimension of $U×W$ is $\dim(U)+\dim(W)$. What do I do next? Any hints?
 A: The range of the map $L$ is clearly $U+W$.
Now the nullspace is:
$$
\mbox{Ker} \;L=\{(u,w)\;;\; u=w\in U\cap W\}
$$
so it is isomorphic to $U\cap W$ via the map $v\longmapsto (v,v)$.
By the rank-nullity theorem applied to $L$, we find:
$$
\mbox{rank}\;L+\mbox{null}\;L=\mbox{dim}\;(U\times W)
$$
which yields the desired formula, which is sometimes called Grassmann formula..
A: Hint: assume  $\dim  ( U \cap W)=k$ and let  $B_{ ( U \cap W)}={[x_1,...x_k]}$ then expand this base for W and U  let these base are  $$B_U={[x_1,...x_k,y_1,..,y_n]}$$  $$B_w={[x_1,...x_k,z_1,..,z_m]}$$ then prove C={$x_1,...x_k$,$y_1$,..,$y_n$,$z_1$,..,$z_m$} is base for W+U you can show C generate W+U and C is independent.
A: Does the following suffice to prove it?
If we proceed to show the symmetric differences:
$dim(U \times W) = [ dim(U \setminus W) + dim(U \cap W) ] + [ dim(W \setminus U) + dim(U \cap W) ] $,
And we have:
$dim(U \oplus W) = dim(U \setminus W) + dim(W \setminus U) + dim(U \cap W)$
Thus,
$dim(U \times W) = dim(U \oplus W) + dim(U \cap W)$
A: By the second isomorphism theorem for vector spaces, $U/(U\cap W)\cong U+W/W$. For any finite dimensional vector space U with a subspace W, $\dim U/W =\dim U-\dim W$, which is clear if one takes a basis for U: $\{u_1,...,u_r,u_{r+1},...,u_n\}$ where $u_1,...,u_r$ form a basis of W. Then $\dim U-\dim (U\cap W)=\dim(U+W)-\dim W,$ which implies $\dim(U)+\dim(W)=\dim(U+W)+\dim(U\cap W)$.
