Question

absolute value of eigen value of orthogonal matric is 1 proof

Answer 1

(1):

Let Av=λvAv=λv with v≠0v≠0, i.e. λλ is an eigenvalue of AA. Then

0<vtAv=λvtv=λ∥v∥2.0<vtAv=λvtv=λ‖v‖2.

Since ∥v∥2>0‖v‖2>0, we get λ>0λ>0.

(2):

You certainly mean that the determinant of AA is ±1±1, since the statement about the eigenvalues is not true, for consider the orthogonal matrix

(cosθsinθ−sinθcosθ)(cos⁡θ−sin⁡θsin⁡θcos⁡θ)

This represents a rotation and has therefore complex eigenvalues.

But if AA is orthogonal, then AtA=AAt=IAtA=AAt=I, therefore applying the detdet to both sides and using the multiplication law for determinants, we obtain

(detA)2=1(detA)2=1

Therefore detA=±1detA=±1.