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