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Math Help - Dimensions

  1. #1
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    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)

    Can you help me, please!
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  2. #2
    MHF Contributor

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    Quote Originally Posted by mivanova View Post
    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 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
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