# Linear Algebra Preliminaries in Finite Reflection Groups

1. Mar 3, 2012

### Math Amateur

Linear Algebra Preliminaries in "Finite Reflection Groups

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 $y_i \in V^{\perp} = 0$"

I do not completely follow the argument as to why $< x_i , y_i > \ \ne 0$ despite Grove and Bensons attempt to explain it. Can someone please (very explicitly) show why this is true?

Why would $y_i \in V^{\perp}$ necessarily be equal to 0 if $< x_i , y_i > \ = \ 0$?

Another issue I have is the folowing:

$y_i$ is defined as a non-zero vector belonging to ${V_i}^{\perp}$.

How do we know that $y_i \in V^{\perp}$ ?

Peter

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Last edited: Mar 3, 2012
2. Mar 3, 2012

### Office_Shredder

Staff Emeritus
Re: Linear Algebra Preliminaries in "Finite Reflection Groups

If $x_1,...,x_n$ is a basis of V, and $\left< y, x_j \right> = 0$ for all j, then y=0 necessarily.

Proof: Suppose that $y = a_1 x_1 +... + a_n x_n$ for some numbers a1,...,an. Then if y is non-zero, $\left< y, y \right> \neq 0$. But
$$\left< y, y \right> = \left< y, \sum a_j x_j \right> = \sum a_j \left< y, x_j \right> = 0$$ since every inner product in that summation is zero. So y must have been zero to begin with

I'm not where your confusion is with yi being in the orthogonal space. They haven't defined yi to be anything in particular, they're just saying, suppose they happened to pick such a vector

It may help to do an example. Let's do it in 2 dimensions: x1 = (1,0) and x2 = (1,1). Then y1 is perpendicular to everything EXCEPT for x1. In particular, y1 is perpendicular to x2 which means y1 must be of the form (a,-a) for some number a. Then <x1,y1> = a = 1 implies that a=1. So y1 = (1,-1).

<y2,x1> = 0 implies that y2 is of the form (0,c) for some number c. <y2,x2> = c, so for the inner product to be 1 it must be c=1, So y2 = (0,1)

Last edited: Mar 3, 2012
3. Mar 3, 2012

### Math Amateur

Re: Linear Algebra Preliminaries in "Finite Reflection Groups

Thanks!

Will just work through that now!

Peter

4. Mar 4, 2012

### Math Amateur

Re: Linear Algebra Preliminaries in "Finite Reflection Groups

I think, that (with the help received in the last post) I have understood the folowing argument in Grove & Benson. Can someone please check my argument.

Basically, to repeat the text I am trying to underrstand:

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 $y_i \in V^{\perp} = 0$"

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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 $y_i \in V^{\perp} = 0$"

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First show the following:

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

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Proof of (1) [following post by Office_Shredder]

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

Thus we can express $y_i$ as follows:

$y_i = a_1 , x_1 + a_2 , x_2 + .... + a_n , x_n$

= $< y_i , x_1 > x_1 + < y_i , x_2 > x_2 + ..... + < y_i , x_n > x_n$ .... (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 $y_i$ = 0

{Problem! Expansion (2) or (3) requires $\{ x_1, x_2, .... , x_n \}$ 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 $y_i \ne \ 0$

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: Mar 4, 2012
5. Mar 16, 2012

### Math Amateur

Re: Linear Algebra Preliminaries in "Finite Reflection Groups

Thanks

Peter