# Thread: Linear Algebra Preliminaries in "Finite Reflection Groups" by Grove and Benson

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

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

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

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

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First show the following:

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

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Proof of (1)

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

Thus we can express $\displaystyle y_i$ as follows:[

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

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

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

If every inner product in (2) or (3) is zero then clearly $y_i$ = 0

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

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

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

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

Hope someone can help.

Peter