The eigenvalues for such mapping can only be .
Thus
let T:V -> V denote an isometry on V, where V is a real inner product space.
prove that the eigenvalues corresponding to different eigenvectors of T are orthogonal.
here is what i did.
let v1, v2 be eigenvectors corresponding to eigenvalues u1, u2 respectively.
since T is an isometry on V, <T(v1) T(v2)> = <v1 v2>.
and
<T(v1) T(v2)> = <u1v1 u2v2> = u1u2<v1 v2> = <u1 u2>
so u1u2<v1 v2>-<v1 v2> = 0
(u1u2-1)<v1 v2> = 0
how do i show that u1u2-1 does not equal 0 or this is not at all correct?
please help.