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 &lt; x_j , y_i &gt; \ = 0 for all j \ne i, but &lt; x_i , y_i &gt; \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 &lt; x_j , y_i &gt; \ = 0 for all j \ne i, but &lt; x_i , y_i &gt; \ne 0 , for otherwise y_i \in V^{\perp} = 0"

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First show the following:

If &lt; x_j , y_i &gt; = 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

= &lt; y_i , x_1 &gt; x_1 + &lt; y_i , x_2 &gt; x_2 + ... + &lt; y_i , x_n &gt; x_n ... (2)

= &lt; x_1, y_i &gt; x_1 + &lt; x_2, y_i &gt; x_2 + ... + &lt; x_n , y_i &gt; 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 &lt; x_j , y_i &gt; = 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 &lt; x_i , y_i &gt; \ne \ 0, for otherwise y_i \in {V_i}^{\perp} = \ 0Is 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 &lt; x_i , y_i &gt; \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