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Common supplementary subspaces

  1. Jun 5, 2015 #1
    1. The problem statement, all variables and given/known data
    Let ##E## be a finite dimensional vector space, ##A## and ##B## two subspaces with the same dimension.
    Show there is a subspace ##S## of ##E## such that ##E = A \bigoplus S = B \bigoplus S ##

    2. Relevant equations

    ##\text{dim}(E) = n##
    ##\text{dim}(A) = \text{dim}(B) = m \le n ##

    ## {\cal B} = (e_1,...,e_n) ## is a basis of ##E##
    ## A = \text{span}(e_{i_1},...,e_{i_m}) ##
    ## B = \text{span}(e_{j_1},...,e_{j_m}) ##

    ## S_A = \text{span}((e_i)_{i \neq i_1,...,i_m } )##
    ## S_B = \text{span}((e_i)_{i \neq j_1,...,j_m } )##

    ## E = A \bigoplus S_A = B \bigoplus S_B ##

    3. The attempt at a solution

    I find this exercise hard and I can't finish it. Could you help please ?


    The case ##A = B## is easy, ## S = S_A = S_B ##

    I assume now that ##A\neq B ##. I want to show the result by induction based on the decreasing dimensions of ##A## and ##B##.
    1 - If ## m = n ## then ## A = B = E ## and ## S = \{0\} ## works
    2 - If A and B are hyperplanes ## m = n-1 ##, then ## S_A = \text{span}(e_k) ##, and ## S_B = \text{span}(e_\ell) ##, with ##k\neq \ell##.
    Put ## S = \text{span}(e_k + e_\ell) ##.
    - Then for any ##x \in A\cap S##, there are scalars ##(\lambda_i)_{i\neq k}## and ##\mu ## such that ## x = \sum_{i\neq k } \lambda_i e_{i} = \mu (e_k + e_\ell ) \Rightarrow 0 = \mu (e_k + e_\ell ) - \sum_{i\neq k } \lambda_i e_{i} = \mu e_k + ( \mu - \lambda_l) e_l - \sum_{i\neq k,l } \lambda_i e_{i} ##. Since ##{\cal B}## is a basis of ##E##, then ##\mu = 0## and ##x = 0##. So ##A \cap S =\{0\}##. We can do the same for ##B## be so that ##B\cap S=\{0\}##
    - For any ##x \in E##, there are scalars ##(\lambda_i)_{i = 1...n}## such that ## x = \sum_{i= 1}^n \lambda_i e_i ##. Reordering the terms,
    ## x = \lambda_k (e_k + e_\ell) + ((\lambda_\ell - \lambda_k) e_\ell + \sum_{i\neq k,\ell}^n \lambda_i e_i) \Rightarrow x \in S + A \Rightarrow S + A = E ##
    ## x = \lambda_l (e_k + e_\ell) +( (\lambda_k - \lambda_\ell) e_k + \sum_{i\neq k,\ell}^n \lambda_i e_i) \Rightarrow x \in S + B \Rightarrow S + B = E ##​
    The two points above show that ##E = A \bigoplus S = B \bigoplus S ##
    3 - Assume that it works for ## m = n,n-1,...,r+1 ##. I want to show it works for ##m = r ##
    - If ##S_A \cap S_B \neq \emptyset ##, then there is a vector ##e_k \in S_A \cap S_B ## such that ## S_A = \text{span}(e_k)\bigoplus S_A'##, and ##S_B = \text{span}(e_k)\bigoplus S_B'##, where ## S'_A = \text{span}((e_i)_{i \neq k,i_1,...,i_m } )##
    and ## S'_B = \text{span}((e_i)_{i \neq k,j_1,...,j_m } )##
    So ## E = (A \bigoplus \text{span}(e_k)) \bigoplus S_A' = (B \bigoplus \text{span}(e_k)) \bigoplus S_B' ##. By induction hypothesis, there is a subspace ##S'## such that :
    ## E = (A \bigoplus \text{span}(e_k)) \bigoplus S' = (B \bigoplus \text{span}(e_k)) \bigoplus S' ##. So ## S = \text{span}(e_k) \bigoplus S' ## works.
    - If ##S_A \cap S_B = \emptyset ##, I don't know how to finish that part !!!
     
    Last edited: Jun 5, 2015
  2. jcsd
  3. Jun 5, 2015 #2
    Point 3.1 does not convince me anymore, I need to rework that ! Sorry.
    And ##S_A \cap S_B \neq \emptyset ##, always, because the intersection of subspaces is a subspace, so it contains 0.
     
    Last edited: Jun 5, 2015
  4. Jun 5, 2015 #3
    I think I might have more. Point 2 was more important than I thought.

    Any finite dimensional vector space can be broken into the direct sum of an hyperplane and a vectorial line : ## E = H \bigoplus D ##

    Point 2 on OP shows that if ##E = H_1 \bigoplus D_1 = H_2 \bigoplus D_2##, then there exist a vectorial line ##D## such that ## E = H_1 \bigoplus D = H_2 \bigoplus D ##.

    So now assume that the dimension of ##S_A## and ##S_B## is 2. They can both be broken into a direct sum of an hyperplane and a vectorial line, so :
    ## E = A \bigoplus S_A = A \bigoplus ( H_1 \bigoplus D_1) = (A \bigoplus H_1) \bigoplus D_1 ##,
    ## E = B \bigoplus S_B = B \bigoplus ( H_2 \bigoplus D_2) =(B \bigoplus H_2) \bigoplus D_2 ##.
    We have that ##A \bigoplus H_1## and ##B \bigoplus H_2## are hyperplanes of ##E##, so there is a line ##D## such that
    ##E = (A \bigoplus H_1) \bigoplus D = (B \bigoplus H_2)\bigoplus D ##. Commute ##H_1## and ##D##, as well as ##H_2## and ##D##, and do the same. At the end, ## E = A \bigoplus (D' \bigoplus D) = B \bigoplus (D' \bigoplus D) ##

    Repeat this operation by induction. Is this idea good ?
     
  5. Jun 5, 2015 #4

    HallsofIvy

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    If A is a subspace of E then, given any basis for A, b= {v1, v2, ..., vm}, there exist a basis for E that contains b.
     
  6. Jun 6, 2015 #5
    Thank you for your reply. I really need some feedback on this exercise, which is horrible !
    I understand your comment, in the general case, I should have completed a basis of A and B with vectors of the basis ##{\cal B}## of E, but to tell you the truth, I was completely helpless at first, so I took a less general case in order to see clearer. Please allow me to restate it taking into account your comment.

    Let ##{\cal B} = \{ e_1, ..., e_n\}## be a basis of ##E##

    0 - If ##A = B ##, since every subspace of a finite dimensional vector space has at least one supplementary space, ##A## and ##B## share this supplementary space.

    1 - If ##A## and ##B## have dimension ##n##, then ##A = B = E##, and with remark 0, they have a common supplementary space which is ##S=\{0\}##

    2 - In general, a finite dimensional vector space ##W## can be expressed as the direct sum of an hyperplane ##H## and a vectorial line ##D##. If you consider a basis ##(f_1,...,f_p)## of that vector space, put ##H = \text{span}(f_1,...,f_{p-1})##, and ## D = \text{span}(f_p)##. It is easy to see that ## W = H + D## and ## H \cap D = \{0\} ##, so that ##W = H \bigoplus D##

    3 - If ##A## and ##B## have dimension ##n-1##, they are hyperplanes of E. One can complete a basis of ##A## (respectively ##B##) with one vector ##e_A \in {\cal B}## (respectively ##e_B##) in order to form a new basis of ##E##. Setting ##D_A = \text{span}(e_A)##, ##D_B = \text{span}(e_B)##, and following remark 2, then ## E = A \bigoplus D_A = B \bigoplus D_B##.

    4 - If ## D_A = D_B## then ##D_A## is a common supplementary space of ##A## and ##B##

    5 - Assume ##D_A \neq D_B ##, which amounts to saying that ##e_A \in B## and ##e_B \in A ##. Set ##D = \text{span}(e_A + e_B) ##. I want to show that ##E = A \bigoplus D## (respectively ##E = B \bigoplus D##).
    • We already know that ##A + D \subset E##. For any ##x\in E = A \bigoplus D_A##, it can be expressed uniquely as ##x = a + \lambda e_A##, ##a\in A## and ##\lambda## scalar. So ## x = (a - \lambda e_B) + \lambda (e_A + e_B) \Rightarrow x \in A + D \Rightarrow E \subset A + D ##. That shows ## E = A + D ##.
    • Now we need to show that ##A\cap D = \{0\}##. If ##x \in A \cap D##, then it has the form ## x = \sum_{i = 1}^{n-1} \lambda_i a_i = \mu (e_A + e_B) ## when ##(a_1,...,a_{n-1})## is a basis of A. Furthermore ##e_B \in A ## so ## e_B = \sum_{i = 1}^{n-1} \alpha_i a_i##. It follows that ## \mu e_A + \sum_{i=1}^{n-1} (\mu \alpha_i - \lambda_i) a_i = 0##. Since ##(a_1,...,a_{n-1}, e_A)## is a basis of ##E##, then ##\mu = 0## and ## x = 0##. That shows ##A\cap D = \{0\}##
    • Doing the same thing for ##B##, I get what I want.

    6 - Assume that if the dimension of ##A## and ## B## is greater or equal to ##r + 1##, then they have a common supplementary.
    Assume that now ## A ## and ## B ## have dimension r. Then,
    ## E = A \bigoplus S_A = A \bigoplus (D_A \bigoplus H_A) = (A \bigoplus D_A) \bigoplus H_A ##
    ## E = B \bigoplus S_B = B \bigoplus (D_B \bigoplus H_B) = (B \bigoplus D_B) \bigoplus H_B ##
    Both ##A \bigoplus D_A## and ##B \bigoplus D_B## have dimension ##r+1##, so they have a common supplementary ##H## in ##E##.
    Commuting the terms, ## E = ( A \bigoplus H ) \bigoplus D_A = (B\bigoplus H) \bigoplus D_B ##. Now ##A \bigoplus H## and ##B \bigoplus H## are hyperplanes of ##E## so they have a common supplementary ##D##.
    It follows that ## E = A \bigoplus (H \bigoplus D) = B \bigoplus (H \bigoplus D)##. So by induction, it can be shown.
     
  7. Jun 6, 2015 #6
    I apologize for putting my thread on top of the list like that, I know it's not very polite :sorry:, but I would like to know if post #5 answers the question.
     
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