Solving Isometries Proofs: Geometry Revisions & Help

[nlews]

Have been revising geometry today and have came across some proofs that I can't seem to find in books, but I can't get through either. Any help would be great.

Let A be a 3x3 orthogonal matrix and let x and y be vectors in R^3

a) Show that detA = +/- 1

b) Show that the length of Ax is the same as that of x and that x and y are orthogonal iff Ax and Ay are orthogonal
Suppose further that A represents a rotation through angleθ , with axis of rotation along the unit vector n, show that if m is a unit vector orthogonal to n, then n.m^Am = sinθ

attempt at a)

from defn of orthogonal matrix (Atr.A = I)

det(Atr . A) = det(I) = 1
using standard results such as det(A.B) = detA. detB and det(Atr) = det(A)
we have det(Atr).det(A) = det(A)^2
detA^2=1
therefore det A = +\- 1

b) struggling to start.

Thank you in advance

 

[CompuChip]

For b) if you have a vector v with transpose tr(v), how do you calculate the length of v?

You also might want to know that tr(x y) = tr(y) tr(x).

 

[nlews]

The only way I know how to find the length of a vector is by squaring and square-rooting. I have never used the transpose to find the length. Ahh I am very stuck!

 

[CompuChip]

Can you convince yourself that
||v||² = v . v^T
where the dot denotes the inner product?

 

[nlews]

I can sort of see it, but working on a proof right now. Will check back later if I get one out. For some reason I just don't see how this answers the question! Sorry for being a pain and thank you for your advice so far.

 

[wofsy]

Your proof of part a is correct.

For part b.

Try using the formula <x,Ay> = <Atrx,y> which works because A is orthogonal.

In the rotation question, you are looking at the area of the parallelogram spanned by the two unit vectors m and Am.

 

[CompuChip]

Well, if the formula I gave is valid for any vector, then in particular it holds for Ax.

So if you calculate (Ax) . (Ax)^T you should get precisely the length x . x^T back.