# A²=Id => A diagonalizable?

1. Dec 23, 2011

### quasar987

Hello matrix gurus,

Is is true that if A is real with A²=I (eigenvalues ±1), it is diagonalizable over R?

What if I add that A is in O(m,m), where O(m,m) is the split indefinite orthogonal group of 2m x 2m matrices M such that $M^TI_{m,m}M=I_{m,m}$, where $I_{m,m}$ is the block diagonal matrix diag(I_m,-I_m) for I_m the m x m identity matrix?

Thanks

Last edited: Dec 23, 2011
2. Dec 23, 2011

### micromass

Yes, an involution is always diagonalizable over the reals.

We use the following result:

See http://en.wikipedia.org/wiki/Diagonalizable_matrix A proof can be found in Hoffman & Kunze theorem 6 page 204.

The minimal polynomial of an involution is $x^2-1=(x+1)(x-1)$ and satisfies our theorem.

Last edited: Dec 23, 2011
3. Dec 24, 2011

### lavinia

A.v - v and A.v + v are both eigen vectors of A. They can not both be zero.
Use this inductively to find a basis of the vector space made up of eigen vectors of A.

e.g.
If A.v is not a multiple of v then both A.v - v and A.v + v are non-zero eigen vectors. Choose any other vector that is not in the span of v and A.v. and continue.

One certainly does not need all of the reals to diagonalize A. The rationals will work. I think all you really need to be able do is divide by 2.

Over the integers A can not always be diagonalized. The map (1,0) -> (1,-1) (0,1) -> (0,-1) is an example. The problem of inequivalent integer representations of groups is daunting.

Last edited: Dec 24, 2011
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