Orthogonal operator and reflection

[drawar]

Homework Statement

Let $n$ be a unit vector in $V$ . Define a linear operator $F_n$ on $V$ such that
$$F_n(u) = u-2\langle u, n \rangle n \; \mathrm{for} \; u \in V.$$
$F_n$ is called the reflection on $V$ along the direction of $n$. Let $S$ be an orthogonal linear operator on $V$ and let $W = \left\{u \in V | S(u) = u\right\}$. Suppose $W^\perp$ is not empty and $w$ is a nonzero vector in $W^\perp$.

(a) Find a unit vector $n$ such that $F_n(S(w))=w$.

(b) Prove that $W \subsetneq E_1(F_n \circ S)$, where $n$ is the unit vector obtain in (a) and $E_1(F_n \circ S)$ denotes the eigenspace of $F_n \circ S$ associated with 1.

Homework Equations

$S$ is orthogonal iff $||S(u)||=||u||$ for all $u \in V$

The Attempt at a Solution

Okay I admit I have no idea how to do either of the questions but I would really appreciate some help to do (a) because without getting it done first, I wouldn't be able to proceed to (b). So right now all I can do for (a) is writing down the equation I'm supposed to solve, hoping something useful would pop up and lead me straight to the answer but it doesn't seem that easy:
$$S(w)-2 \langle S(w),n \rangle n = w.$$
Honestly I'm stuck at here, is there any way I can do about it?

 

[bloby]

Perhaps not the best way but: what is $F_n^2$, how can (a) be restated? And is there a way to recover n knowing v and $F_n(v)$?

 

[drawar]

Perhaps not the best way but: what is $F_n^2$, how can (a) be restated? And is there a way to recover n knowing v and $F_n(v)$?

$F_n^2=I$, so the equation in (a) reduces to $S(w)=F_n(w)$ (- I guess this requires $F_n$ to be bijective or at least injective but I don't know how to prove it). From here $n$ can be found by subtracting $F_n(w)$ (or equivalently $S(w)$) from $w$ followed by a normalization. Does that sound ok?

 

[bloby]

(- I guess this requires $F_n$ to be bijective or at least injective but I don't know how to prove it).

Yes. Either verify the solution found or simply use the definition of injective.

From here $n$ can be found by subtracting $F_n(w)$ (or equivalently $S(w)$) from $w$ followed by a normalization. Does that sound ok?

Yes.

Is the part (b) ok?

 

[drawar]

Is the part (b) ok?

No, unfortunately. Would appreciate further hints if possible!

 

[bloby]

I gave perhaps too much hints for (a): it's not magical trick, I just figured out what $F_n$ does on a vector in $\mathbb R^3$, (do a sketch) hence the name and the intuitions.

For (b) there are two things to prove: $W \subset E_1(F_n \circ S)$ and there is a vector $\not\in W$ that belongs to $E_1(F_n \circ S)$.

What does it means for a vector v to be in $E_1(F_n \circ S)$?

 

[drawar]

I gave perhaps too much hints for (a): it's not magical trick, I just figured out what $F_n$ does on a vector in $\mathbb R^3$, (do a sketch) hence the name and the intuitions.

For (b) there are two things to prove: $W \subset E_1(F_n \circ S)$ and there is a vector $\not\in W$ that belongs to $E_1(F_n \circ S)$.

What does it means for a vector v to be in $E_1(F_n \circ S)$?

$E_1(F_n \circ S)=\left\{w \in V|F_n(S(w))=w\right\}$. $W \subset E_1(F_n \circ S)$ is because of the fact that $S$ is orthogonal. It remains to find a vector that $\not\in W$ but belongs to $E_1(F_n \circ S)$, which is nothing but $w$.

 

[bloby]

$E_1(F_n \circ S)=\left\{w \in V|F_n(S(w))=w\right\}$. $W \subset E_1(F_n \circ S)$ is because of the fact that $S$ is orthogonal. It remains to find a vector that $\not\in W$ but belongs to $E_1(F_n \circ S)$, which is nothing but $w$.

Yes, yes, and the definition of W and the fact that $w$ of (a) belongs to $W^\perp$

 

[drawar]

Yeah, that also. Thanks for being so patient with me. I really appreciate your instructive help and guidance.

 

[bloby]

You're welcome