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Prove that an orthogonal transformation T in R^{m}has 1 as an eigenvalue if the determinant of T equals 1 and m is odd. What can you say if m is even?

The attempt at a solution

I know that I can write R^{m}as the direct sum of irreducible invariant subspaces W_{1}, W_{2}, ..., W_{s}which are mutually orthogonal and have dimension 1 or 2. They can't all have dimension 2 for otherwise m would be even. Thus, there is at least one with dimension 1, say W_{1}. For any nonzero w in W_{1}, T(w) = w or T(w) = -w. In the former case, we're done. The latter case is giving me problems though. If T has -1 as an eigenvalue, I imagine that the determinant of T will not be 1 anymore (since I haven't used this fact), but I don't understand how. Any tips?

Oh, and if m is even I imagine that T may not necessarily have eigenvalues since it's minimal polynomial may be multiples of powers of irreducible quadratics.

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# Homework Help: Orthogonal Transformations with Eigenvalue 1

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