Orthogonal Transformations with Eigenvalue 1

[e(ho0n3]

Homework Statement
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 W₁, W₂, ..., 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 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.

 

[tiny-tim]

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?
…
I know that I can write R^m as the direct sum of irreducible invariant subspaces W₁, W₂, ..., 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 W₁.…. Any tips?

Hi e(ho0n3!

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

 

[e(ho0n3]

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

 

[tiny-tim]

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

Perhaps I'm misunderstanding the problem …

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

 

[e(ho0n3]

Perhaps I'm misunderstanding the problem …

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

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?

 

[tiny-tim]

I know that for an irreducible subspace of dimension 2, T has determinant equal to 1

Exactly!

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

 

[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 R^m = 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?

 

[tiny-tim]

… 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?

Yes …

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

 

[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?

 

[tiny-tim]

Woohoo! ​

 

[e(ho0n3]

Thanks a lot for your help.