Linear Algebra Preliminaries in Finite Reflection Groups

[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

 

[Office_Shredder]

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,...,a_n. 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 y_i being in the orthogonal space. They haven't defined y_i to be anything in particular, they're just saying, suppose they happened to pick such a vectorIt 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)

 

[Math Amateur]

Thanks!

Will just work through that now!

Appreciate your help!

Peter

 

[Math Amateur]

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

 

[Math Amateur]

Thanks

Peter