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Proof of combination of reflections

  • Thread starter gop
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  • #1
gop
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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.
 

Answers and Replies

  • #2
Dick
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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
gop
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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
Dick
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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
gop
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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
Dick
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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
gop
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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)
 

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