# Improper orthogonal matrix

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.

## Answers and Replies

StatusX
Homework Helper
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.

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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
Homework Helper
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.

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

matt grime
Homework Helper
It's obviously not true: think geometrically.

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