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[SOLVED] Diagonalizing a 3x3 matrix
I want to show that a real 3x3 matrix, A, whose square is the identity is diagonalizable by a real matrix P and that A has (real) eigenvalues of modulus 1.
None.
Since any matrix is diagonalizable over the complex numbers, I deduced that since there exists a complex matrix P such that PAP^{-1} = diag{x,y,z} (so x,y,z the eigenvalues of A), then diag{x^2,y^2,z^2} = (PAP^{-1})^2 = Id, hence the eigenvalues are square roots of 1 therefore must be real of modulus 1 as required.
I'm not totally sure my reasoning is sound. Even if it is, there is still the problem that P may be complex.
Homework Statement
I want to show that a real 3x3 matrix, A, whose square is the identity is diagonalizable by a real matrix P and that A has (real) eigenvalues of modulus 1.
Homework Equations
None.
The Attempt at a Solution
Since any matrix is diagonalizable over the complex numbers, I deduced that since there exists a complex matrix P such that PAP^{-1} = diag{x,y,z} (so x,y,z the eigenvalues of A), then diag{x^2,y^2,z^2} = (PAP^{-1})^2 = Id, hence the eigenvalues are square roots of 1 therefore must be real of modulus 1 as required.
I'm not totally sure my reasoning is sound. Even if it is, there is still the problem that P may be complex.