91Ó°ÊÓ

Assume that we have a self-adjoint operator, denoted by A. That means \(A^* = A\), where \(A^*\) is the adjoint of A. Let \(\lambda\) be an eigenvalue of A and \(v\) be the corresponding eigenvector such that \(Av = \lambda v\). Now, let's take the inner product of both sides of the equation with v: \( \langle Av, v \rangle = \langle \lambda v, v \rangle \) Since A is self-adjoint, we can replace A with its adjoint: \( \langle A^*v, v \rangle = \langle \lambda v, v \rangle \) Now, use the property of the adjoint: \(\langle Ax, y \rangle = \langle x, A^*y \rangle \): \( \langle v, A^*v \rangle = \langle \lambda v, v \rangle \) Since \(A^* = A\), we can write: \( \langle v, Av \rangle = \langle \lambda v, v \rangle \) Recall that \( Av = \lambda v \): \( \langle v, \lambda v \rangle = \langle \lambda v, v \rangle \) Factor out the scalar \(\lambda\): \( \lambda \langle v, v \rangle = \overline{\lambda} \langle v, v \rangle \) Divide both sides by \(\langle v, v \rangle\): \(\lambda = \overline{\lambda}\) Since \(\lambda\) is equal to its complex conjugate, it must be real. Thus, all eigenvalues of self-adjoint operators are real.

Now, let's assume that A is a normal operator with all real eigenvalues, and show that it is self-adjoint. Let \(\lambda\) be an eigenvalue of A and \(v\) be the corresponding eigenvector such that \(Av = \lambda v\). Take the inner product of both sides of the equation with v: \( \langle Av, v \rangle = \langle \lambda v, v \rangle \) Since \(\lambda\) is real, we can write: \( \langle Av, v \rangle = \lambda \langle v, v \rangle \) Next, let's compute the inner product \(\langle v, Av \rangle\). Apply the property of the adjoint: \(\langle Ax, y \rangle = \langle x, A^*y \rangle\): \( \langle v, Av \rangle = \langle v, A^*v \rangle \) Since \(Av = \lambda v\), we can write: \( \langle v, \lambda v \rangle = \langle v, A^*v \rangle \) Factor out the scalar \(\lambda\): \( \lambda \langle v, v \rangle = \langle v, A^*v \rangle \) Now, comparing the results from taking the inner product with \(Av\) and \(v\) in the equation \(Av = \lambda v\): \(\lambda \langle v, v \rangle = \langle Av, v \rangle = \langle v, A^*v \rangle\) This equality holds for all eigenvectors of A. Since the inner product is linear, it also holds for linear combinations of eigenvectors. Thus, the equality holds for all vectors in the complex inner-product space. Therefore, A is self-adjoint. We have shown that a self-adjoint operator leads to real eigenvalues, and real eigenvalues lead to a self-adjoint operator. This proves the equivalence of the two statements.