self-adjunct operator on a complex Hilbert space

Question

Let $T$ be a self-adjoint operator on a complex Hilbert space $H$.

Show that, if $Tx=\mu x$ with $x \neq 0$, so $\mu \in \mathbb{R}$.

I've already tried to use this book "Linear Functional Analysis, 2 ed - Bryan Rynne, M.A. Youngson" in order to prove that, but i'm kind a stuck, any help or hint?

Answer 1

It's only one line: $$\mu\langle x,x\rangle=\langle \mu x,x\rangle=\langle Tx,x\rangle=\langle x,T^{\ast} x\rangle=\langle x,Tx\rangle=\langle x,\mu x\rangle=\overline\mu\langle x,x\rangle$$ Since $x\neq 0$ by definition of an eigenvector, we can divide by $\langle x,x\rangle=\lVert x\rVert^2$ to get $\mu=\overline\mu$.

Answer 2

Note that $\langle Tx,x\rangle=\mu\|x\|^2$, but you also have$$\langle Tx,x\rangle=\langle x,Tx\rangle=\overline\mu\|x\|^2.$$So, $\mu=\overline\mu$.