absolute value of eigen value of orthogonal matric is 1 proof
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(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)(cosθ−sinθsinθcosθ)
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.
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)(cosθ−sinθsinθcosθ)
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.
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