Proof of combination of reflections

[gop]

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
s_U \circ s_W = s_W \circ s_U

if and only if

W^{\perp}, U^{\perp}

are perpendicular.

Homework Equations

W^{\perp} is the subspace of V such that every vector in W^{\perp} is perpendicular to W.

 

[Dick]

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.

 

[gop]

Well I tried to do it in 2D and then to generalize it to n then I get

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

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
=&-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

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

-2v+v=-v

I can repeat the same with s_{u}\circ s_{w} 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.

 

[Dick]

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.

 

[gop]

Okay so I did it with just u and w (v is what you called x) and got the desired result

\langle v,u\rangle\cdot\langle u,w\rangle\cdot w-\langle v,w\rangle\cdot\langle u,w\rangle\cdot u&amp;=&amp;0

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

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

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

 

[Dick]

Okay so I did it with just u and w (v is what you called x) and got the desired result

\langle v,u\rangle\cdot\langle u,w\rangle\cdot w-\langle v,w\rangle\cdot\langle u,w\rangle\cdot u&amp;=&amp;0

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

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

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.

 

[gop]

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)