Let a be a matrix of order n and satisfy AAT = e. the determinant of a is less than zero. It is proved that - 1 is an eigenvalue of A
Proof: | a + e|
= |A+AA^T|
= |A(E+A^T)|
= |A||(E+A)^T|
= |A||A+E|
So | a + e | (1 - | a |) = 0
Because | a|
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