# Dimensions

• Aug 31st 2008, 11:13 AM
mivanova
Dimensions
Hi,
I have this problem to solve.

Prove or give a countterexample: If U_1, U_2, U_3 are subspaces of a finite dimensional vector space V, then
dim(U_1+U_2+U_3)=dim(U_1)+dim(U_2)+dim(U_3)-dim(U_1(interset)U_2)-dim(U_2(interset)U_3)-dim(U_3(interset)U_1)+dim(U_1(interset)U_2(interse t)U_3)

• Aug 31st 2008, 06:40 PM
NonCommAlg
Quote:

Originally Posted by mivanova
Hi,
I have this problem to solve.

Prove or give a countterexample: If U_1, U_2, U_3 are subspaces of a finite dimensional vector space V, then
dim(U_1+U_2+U_3)=dim(U_1)+dim(U_2)+dim(U_3)-dim(U_1(interset)U_2)-dim(U_2(interset)U_3)-dim(U_3(interset)U_1)+dim(U_1(interset)U_2(interse t)U_3)

this is a good question! here's a counter-example: let $V=\mathbb{R}^2, \ e_1=(1,0), \ e_2 = (0,1).$ so $\{e_1,e_2\}$ is a basis for $V$ as a vector space over $\mathbb{R}.$ let $U_1=\mathbb{R}e_1, \ U_2=\mathbb{R}e_2, \ U_3=\mathbb{R}(e_1+e_2).$
obviously: $\dim U_1=\dim U_2 = \dim U_3 = 1.$ see that: $U_1+U_2+U_3=V$ and: $U_1 \cap U_2=U_2 \cap U_3=U_1 \cap U_3 = U_1 \cap U_2 \cap U_3 = \{(0,0)\}.$ thus: $\dim(U_1+U_2+U_3)=\dim V = 2,$ but:
$\sum_{j=1}^3 \dim U_j - \sum_{1 \leq i < j \leq 3}\dim(U_i \cap U_j) + \dim(U_1 \cap U_2 \cap U_3) = 3. \ \ \ \square$