Proof of combination of reflections

In summary, V is a n-dimensional euclidean space, U and W are n-1 dimensional subspaces of V, and s_U \circ s_W = s_W \circ s_U if and only if s_W^{\perp}, U^{\perp} are perpendicular.
  • #1

gop

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0

Homework Statement



V is a n-dimensional euclidean space. U and W are n-1 dimensional subspaces of V.
U and W define a reflection (because of their property as n-1 dimensional subspaces).

Show that
[tex]s_U \circ s_W = s_W \circ s_U[/tex]

if and only if

[tex]W^{\perp}, U^{\perp}[/tex]

are perpendicular.


Homework Equations



[tex]W^{\perp}[/tex] is the subspace of V such that every vector in [tex]W^{\perp}[/tex] is perpendicular to W.
 
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  • #2
This is really a two dimensional problem. The only vectors that are affected by the reflections are the normals to W and U. The n-2 other independent vectors that are common to the W and U subspaces are not. So tackle it in two dimensions first and then think how to extend it.
 
  • #3
Well I tried to do it in 2D and then to generalize it to n then I get

[tex]s_{w}&=&2\cdot\sum_{i=1}^{n}\langle w_{i},v\rangle\cdot w_{i}-v[/tex]
[tex]s_{u}&=&2\cdot\sum_{i=1}^{n}\langle u_{i},v\rangle\cdot w_{i}-v[/tex]
Then with [tex]\langle u_{i},v_{i}\rangle=0[/tex]

[tex]s_{w}\circ s_{u}&=&2\cdot\sum_{i=1}^{n}\langle w_{i},2\cdot\sum_{i=1}^{n}\langle u_{i},v\rangle\cdot w_{i}-v\rangle\cdot w_{i}-2\cdot\sum_{i=1}^{n}\langle u_{i},v\rangle\cdot w_{i}+v[/tex]
[tex]=&-2\cdot\sum_{i=1}^{n}\langle w_{i},v\rangle\cdot w_{i}-2\cdot\sum_{i=1}^{n}\langle u_{i},v\rangle\cdot w_{i}\ \ +v[/tex]

Now if W and U are perpendicular we have [tex]W=U^{\perp}[/tex]. Thus we can write every vector in V as the sum of the projection with respect to W and to U which gives us

[tex]-2v+v=-v[/tex]

I can repeat the same with [tex]s_{u}\circ s_{w}[/tex] and get at the same result.


However I have still no idea how to proof the if and only if part. Thus how to show that this is the only case where the two functions are equal.
 
  • #4
That's pretty confusing. I have no idea what all of those things are. If you assume u and w are normalized (which you may as well), you can write s_u(x)=x-2*<x,u>u and s_w(x)=x-2*<x,w>w. If you form s_u(s_w(x))-s_w(s_u(x)) you should get <u,w> times a linear combination of u and w. The linear combination can only vanish if u and w are linearly dependent (which is the trivial case where they are parallel), so <u,w> must vanish. Try it again without all the messy subscripts.
 
  • #5
Okay so I did it with just u and w (v is what you called x) and got the desired result

[tex]\langle v,u\rangle\cdot\langle u,w\rangle\cdot w-\langle v,w\rangle\cdot\langle u,w\rangle\cdot u&=&0[/tex]

Now I have to argue why this result holds in n-dimensional space.
I now do

[tex] s_{w}&=&2\cdot\sum_{i=1}^{n-1}\langle w_{i},v\rangle\cdot w_{i}-v\\&=&2\cdot(v-v_{\perp})-v\\&=&v-2\cdot v_{\perp}\\&=&v-2\langle v,w_{n}\rangle\cdot w_{n}[/tex]

where w_1, ..., w_(n-1) is the orthonormal base of the subspace W and w_n is the othornomal extension to a base of V.
Then the n-dimensional case is the two-dimensional case with w=w_n and v=v_n
 
  • #6
gop said:
Okay so I did it with just u and w (v is what you called x) and got the desired result

[tex]\langle v,u\rangle\cdot\langle u,w\rangle\cdot w-\langle v,w\rangle\cdot\langle u,w\rangle\cdot u&=&0[/tex]

Now I have to argue why this result holds in n-dimensional space.
I now do

[tex] s_{w}&=&2\cdot\sum_{i=1}^{n-1}\langle w_{i},v\rangle\cdot w_{i}-v\\&=&2\cdot(v-v_{\perp})-v\\&=&v-2\cdot v_{\perp}\\&=&v-2\langle v,w_{n}\rangle\cdot w_{n}[/tex]

where w_1, ..., w_(n-1) is the orthonormal base of the subspace W and w_n is the othornomal extension to a base of V.
Then the n-dimensional case is the two-dimensional case with w=w_n and v=v_n

That's it! Good job. You are actually done. You didn't assume anything about the dimensionality of the space to get that result. There is nothing else to do. I only mentioned doing it in two dimensions first in case you were working with an explicit basis or something. But you bypassed that step.
 
  • #7
Thank you for your help! Once you see the idea it'r really simple I guess...
( I was also able to solve the second problem you gave me a hint too so thank you again)
 

1. What is "Proof of combination of reflections"?

"Proof of combination of reflections" is a mathematical concept that involves using two or more reflections to determine the properties of a geometric figure. It is commonly used in geometry and can help prove statements about symmetry and congruence.

2. How does "Proof of combination of reflections" work?

The process of "Proof of combination of reflections" involves taking two or more reflections of a figure and analyzing their combined effect. This can help determine the properties of the figure, such as its line of symmetry or angles of reflection.

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While "Proof of combination of reflections" can be a powerful tool in geometry, it does have some limitations. It is only applicable to figures that have reflective symmetry, and it does not take into account other types of transformations, such as translations or rotations.

5. How can "Proof of combination of reflections" be helpful in problem-solving?

"Proof of combination of reflections" can be a useful problem-solving technique, as it allows for the determination of properties of a figure without the need for complex calculations. It can also provide a visual representation of the problem, making it easier to understand and solve.

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