# 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?

• Show that $\text{Dim}(U+W)= \text{Dim}( U) + \text{Dim} (W) -\text{Dim} ( U \cap W)$.
– A.D
Jan 29, 2013 at 19:12
• Isn't that same? I think Dimension of $u-w$(image of map), will be $\text{Dim}(U+W)$, will the dimension of kernel of transformation will have $\text{Dim}(U\cap W)$ ?? Jan 29, 2013 at 19:13
• You're on the right track. What are the nullspace and the range of $L$? Then conclude with the rank-nullity theorem: en.wikipedia.org/wiki/Rank%E2%80%93nullity_theorem Jan 29, 2013 at 19:13
• @julien am I right?? will they share the same null space? Jan 29, 2013 at 19:14
• Share? There's only one map here. Its range is $L+W$. Its nullspace is isomorphic to $L\cap W$. Jan 29, 2013 at 19:16

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..

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.

• i thought about it too, but this is a bit awkward (since i was working on linear transformation than vector spaces). thanks anyway!! Jan 29, 2013 at 19:23
• @testuser How exactly do you claim that this is awkward? This is the most straightforward way to solve it. Jan 29, 2013 at 19:24
• @DoctorBatmanGod context :P Jan 29, 2013 at 19:25

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,

$$dim(U \times W) = dim(U \oplus W) + dim(U \cap W)$$

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