Linear Algebra Preliminaries in Finite Reflection Groups

In summary, Grove and Benson's "Finite Reflection Groups" discusses the preliminaries of linear algebra in regards to finite reflection groups. They define the subspace V_i as the span of all basis vectors except for x_i. If a non-zero vector y_i belongs to the orthogonal space of V_i, then it is perpendicular to all basis vectors except for x_i. This means that <x_j, y_i> = 0 for all j except i. However, <x_i, y_i> must be non-zero in order for y_i to not be equal to 0. This is because if <x_i, y_i> was equal to 0, then y_i would also be in the orthogonal space of V,
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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 [itex] \{ x_1 , x_2, ... x_n \} [/itex] is a basis for V, let [itex] V_i [/itex] be the subspace spanned by [itex] \{ x_1, ... , x_{i-1} , x_{i+1}, ... x_n \} [/itex], excluding [itex] x_i [/itex].

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

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

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

Another issue I have is the folowing:

[itex] y_i [/itex] is defined as a non-zero vector belonging to [itex] {V_i}^{\perp} [/itex].

How do we know that [itex] y_i \in V^{\perp} [/itex] ?

Peter
 

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

Proof: Suppose that [itex] y = a_1 x_1 +... + a_n x_n [/itex] for some numbers a1,...,an. Then if y is non-zero, [itex] \left< y, y \right> \neq 0[/itex]. But
[tex] \left< y, y \right> = \left< y, \sum a_j x_j \right> = \sum a_j \left< y, x_j \right> = 0[/tex] 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 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)
 
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  • #3


Thanks!

Will just work through that now!

Appreciate your help!

Peter
 
  • #4


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

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


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

Thus given "[itex] \{ x_1 , x_2, ... x_n \} [/itex] is a basis for V, let [itex] V_i [/itex] be the subspace spanned by [itex] \{ x_1, ... , x_{i-1} , x_{i+1}, ... x_n \} [/itex], excluding [itex] x_i [/itex]" I am trying to show that:

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

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

First show the following:

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

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

Proof of (1) [following post by Office_Shredder]

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

Thus we can express [itex] y_i [/itex] as follows:

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

= [itex] < y_i , x_1 > x_1 + < y_i , x_2 > x_2 + ... + < y_i , x_n > x_n [/itex] ... (2)

= [itex] < x_1, y_i > x_1 + < x_2, y_i > x_2 + ... + < x_n , y_i > x_n [/itex] ... (3)

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

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

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

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

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

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

Hope someone can help.

Peter
 
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  • #5


Thanks

Peter
 

1. What is linear algebra?

Linear algebra is a branch of mathematics that deals with linear equations, vector spaces, and linear transformations. It is used to solve problems related to systems of equations, geometry, and data analysis.

2. What are finite reflection groups?

Finite reflection groups are groups of transformations that can be represented as a composition of reflections in a finite-dimensional vector space. These groups have a finite number of elements and are important in the study of symmetry in geometry and physics.

3. What are the applications of linear algebra in finite reflection groups?

Linear algebra is used to study the properties and representations of finite reflection groups. It is also used to determine the symmetry and geometric structure of these groups and their associated objects such as polytopes, crystals, and lattices.

4. What are some common examples of finite reflection groups?

Some common examples of finite reflection groups include the symmetric group, dihedral group, and Coxeter groups. The symmetric group is the most basic example and consists of all permutations of a set of elements. The dihedral group is a reflection group that represents the symmetry of a regular polygon, while Coxeter groups are used to study symmetries in higher dimensions.

5. Why are finite reflection groups important?

Finite reflection groups are important in many areas of mathematics and physics. They help us understand the concept of symmetry and its applications, such as in crystallography and group theory. They also have practical applications in computer graphics, coding theory, and data compression.

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