Given two subspaces $ U, W $ of vector space $ V $, how to show that $ \text{Dim}( U) + \text{Dim} (W) = \text{Dim}(U+W) + \text{Dim} ( U \cap W) $

Question

Let $ U, W $ be subspaces of vector space $ V $. Show that $$ \text{Dim}( U) + \text{Dim} (W) = \text{Dim}(U+W) + \text{Dim} ( U \cap W) $$

Given Hint: Show that map given by $ L:U \times W \to V $ given by $ L(u,w) = u-w $ is a linear map.

I can show that $ L:U \times W \to V $ given by $ L(u,w) = u-w $ is a linear map. Also I know that dimention of $U\times W$ is $ \text{Dim}( U) + \text{Dim} (W) $. What do I do next? Any hints?

Answer 1

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..

Answer 2

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.