# Homework Help: Orthogonal Transformations with Eigenvalue 1

1. Nov 23, 2008

### e(ho0n3

The problem statement, all variables and given/known data
Prove that an orthogonal transformation T in Rm 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 Rm as the direct sum of irreducible invariant subspaces W1, W2, ..., Ws 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 W1. For any nonzero w in W1, 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.

2. Nov 23, 2008

### tiny-tim

Hi e(ho0n3!

Hint: you haven't yet use the fact that the determinant is the product of the individual determinants.

3. Nov 23, 2008

### e(ho0n3

And where would I use that fact exactly? I'm not calculating the determinant of a product as far as I know.

4. Nov 23, 2008

### tiny-tim

Perhaps I'm misunderstanding the problem …

but what can you say about the determinant of one irreducible subspace?

5. Nov 23, 2008

### e(ho0n3

Nothing. Determinants are defined for matrices and linear transformations, not subspaces. Hmm...but now you've made me think of the following:

Since we're told that the determinant of T is 1, does that mean the determinant of the restriction of T to each irreducible subspace is 1 as well? I know that for an irreducible subspace of dimension 2, T has determinant equal to 1. For an irreducible subspace of dimension 1, then it could be 1 or -1 depending on the eigenvalue of T on this subspace. Right?

6. Nov 23, 2008

### tiny-tim

Exactly!

And there's an odd number of dimension-1 subspaces, so … ?

7. Nov 23, 2008

### e(ho0n3

So, if the determinant of T on each of the dimension-1 subspaces is -1, then the determinant of T on the direct sum of these one-dimensional subspaces is -1, right? (And also analogously for the two-dimensional subspaces.) Let's write Rm = U + V where U is the direct sum of the one-dimensional subspaces and V is the direct sum of the two-dimensional subspaces. Since the determinant of T on U is -1 and the determinant of T on V is 1, does that mean the determinant of T on U + V is -1?

8. Nov 23, 2008

### tiny-tim

Yes …

so how many of the individual 1-dimensional subspaces can have negative determinant?

9. Nov 23, 2008

### e(ho0n3

An even number of them. So we have at least one 1-dimensional subspace where T has determinant 1 and so T has eigenvalue 1. Right?

10. Nov 23, 2008

### tiny-tim

Woohoo!

11. Nov 23, 2008

### e(ho0n3

Thanks a lot for your help.