Let be an arbitrary inner product on . Suppose is self adjointing with respect to this inner product. Prove that has real eigenvalues.
My instinct would be to do something like this: given an eigenvalue , let be a corresponding eigenvector. Then
The thing about this is that the inner product is defined on , not , and I'm having a hard time changing my thought process away from what I did above. I've been thinking maybe it should be since maybe the question would be trivial otherwise. Any thoughts?
Since the set of real numbers is a subset of the complex numbers, even though <u, v> is defined to be a real number, you can think of it as a function from pairs of vectors to a complex number.
For example, with inner product <(a, b), (c, d)>= ac+ bd has inner product always real but the matrix has characteristic equation which has i and -i as roots. Since we can't have Av= iv in , the corresponding linear transformation does NOT have any eigevalues. But we can, temporarily switch to "over so that any linear transformation has eigenvalues, show, as you did, that, for a self-adjoint linear transformation, they must be real and so assert that any self-adjoint linear transformation on a vector space over the real numbers has eigenvalues.
EDIT: I think I've shown that you can extend any inner product over R^n to one over C^n by taking
ComplexInnerProduct(a + bi, c + di) = <a, c> + <b, d> + i[<b, c> - <a, d>]
so I guess it's irrelevant.