Does an Improper Orthogonal Matrix Always Have a Determinant of -1?

[Mindscrape]

The question is (true or false) if Q is an improper 3 x 3 orthogonal matrix then Q^2 = I.

The way I have approached it so far has been a brute force method. I'm not really sure if this will be true or false, and I have a feeling it is false, but I can't construct a good counter-example. So, I have been trying to prove it is true, which is becoming tedious and lengthy as I go through the inner products.

I started on another more algebraic approach too. I know that Q^T Q = I, so if QQ = I, then Q = Q^{-1} = Q^T. Also det(Q) = -1, since it is improper. From here I am not quite sure either.

 

[StatusX]

Well, it's true in even dimensions. I don't know what more to say without giving it away.

Edit: Ok, I'll tell you this: stop trying to prove it.

 

[Mindscrape]

Yes, in an earlier question I found it was true for a 2x2, and I can see how that would extend for any even dimensioned matrix.

My idea in trying to prove the 3x3 case true was to at some point find where the preposition becomes false. Usually it is pretty easy, for me at least, to find counter examples to matrices, but with the orthogonal matrix I am having trouble thinking of a counter example since the orthogonal matrix is so limiting.

 

[StatusX]

You build one up out of lower dimensional matrices.

Edit: Actually, I don't think it's true for any dimension greater than 2. Sorry.

 

[Hurkyl]

How did you prove it for 2 dimensions? That might suggest what you need to find a counterexample in 3 dimensions.

 

[matt grime]

It's obviously not true: think geometrically.

 

[Mindscrape]

It's obviously not true: think geometrically.

What do you mean? If you have a set of 3 orthogonal vectors that form an orthogonal matrix, which must mean that the vectors themselves are orthonormal, and if you find the inner product of all the vectors then they will not necessarily still be orthonormal in lR^3?

 

[matt grime]

Umm. What? Orthogonal matrices are generated by rotations and reflections. In R^2, any improper matrix is a reflection. The question asks is this true in any dimension, and the answer is clearly no. So, prove that the composition of a reflection in a plane and a rotation is or is not in general a reflection in another plane.

(I think we all agree it isn't - so just find a counter example. Status X's hint gives you the answer easily - treat things as block matrices, and remember that proper orthogonal matrices (rotations) are not, in general, self inverse

 

[Mindscrape]

Okay, I think I was making it a lot tougher than it should have been. First counter example I tried worked.

I didn't realize that an improper matrix indicated a reflection though. Your geometrical response makes sense now.