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Math Help - Linear Algebra Preliminaries in "Finite Reflection Groups" by Grove and Benson

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    Super Member Bernhard's Avatar
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    Linear Algebra Preliminaries in "Finite Reflection Groups" by Grove and Benson

    I think, that I have understood the folowing argument in Grove & Benson. Can someone please check my argument?

    Basically, to repeat the text I am trying to understand: (see attachement for Page 1 of Grove and Benson)In the Preliminaries to Grove and Benson "Finite Reflection Groups"

    On page 1 (see attachment) we find the following:

    "If  \{ x_1 , x_2, ... x_n \}  is a basis for V, let  V_i be the subspace spanned by  \{ x_1, ... , x_{i-1} , x_{i+1}, ... x_n \} , excluding  x_i .

    If  0 \ne y_i \in {V_i}^{\perp} , then  < x_j ,   y_i > \ = 0 for all  j \ne i , but  < x_i , y_i > \ne 0 , for otherwise  y_i \in V^{\perp} = 0 "

    ================================================== ===================

    Thus given "  \{ x_1 , x_2, ... x_n \}  is a basis for V, let  V_i be the subspace spanned by  \{ x_1, ... , x_{i-1} , x_{i+1}, ... x_n \} , excluding  x_i "

    I am trying to show that:"If  0 \ne y_i \in {V_i}^{\perp} , then  < x_j ,   y_i > \ = 0 for all  j \ne i , but  < x_i , y_i > \ne 0 , for otherwise  y_i \in V^{\perp} = 0 "

    ================================================== ===================

    First show the following:

    If  < x_j , y_i > = 0 for all j then  y_i  must equal zero ........ (1)

    -------------------------------------------------------------------------------------------------------------------------------------------

    Proof of (1)

     y_i  must belong to V [since it belongs to a subspace  {V_i}^{\perp}  of V]

    Thus we can express  y_i  as follows:[

    tex] y_i = a_1 , x_1 + a_2 , x_2 + .... + a_n , x_n [/tex]

    =   < y_i , x_1 > x_1 + < y_i , x_2 > x_2 + ..... + < y_i , x_n > x_n .... (2)

    =  < x_1, y_i > x_1 + < x_2 , y_i > x_2 + .... + < x_n , y_i > x_n  .... (3)

    If every inner product in (2) or (3) is zero then clearly [itex] y_i [/itex] = 0

    {Problem! Expansion (2) or (3) requires  \{ x_1, x_2, .... , x_n \} to be orthonormal! But of course it could be made orthonormal!}

    -----------------------------------------------------------------------------------------

    But we know (1) does not hold because we have assumed  y_i \ne  \ 0

    But we also know that  < x_j , y_i > = 0 for all  j \ne i  since the  x_j with  j \ne i belong to  V_i  which is orthogonal to  {V_i}^{\perp} .Thus  < x_i , y_i >  \ne \ 0, for otherwise  y_i \in  {V_i}^{\perp} = \ 0

    Is this reasoning correct?

    I would appreciate it very much if someone can confirm the correctness of my reasoning.[Problem: Grove and Benson actually conclude the above argument by saying the following: (see attachement)

    But  < x_i , y_i >  \ne \ 0, for otherwise  y_i \in  {V}^{\perp} = \ 0 but I think this is a typo as they mean  y_i \in  {V_i}^{\perp} = \ 0?? Am I correct? ]

    Hope someone can help.

    Peter
    Last edited by Bernhard; March 4th 2012 at 12:56 AM.
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