TITLE: For self-adjoint operators, eigenvectors that correspond to distinct eigenvalues are orthogonal
QUESTION [5 upvotes]: So I was looking for a proof for the next theorem.
$V$ is inner product space
$T: V\rightarrow V$ self adjoint linear map.
$ \lambda_{1},\lambda_{2} \in \mathbb{F}$ so that $ \lambda_{1} \neq \lambda_{2}$ 
$ v_{1},v_{2} \in V$ so that   $ 0_{v} \neq v_{1} \neq v_{2} \neq 0_{v}$
$T(v_{1}) = \lambda_{1}v_{1}$
$T(v_{2}) = \lambda_{2}v_{2}$ 
then $\langle v_{1},v_{2}\rangle = 0$
I searched for a proof but I did not find it.
Please show me the proof or give me a link to where the proof is.
Thanks in advanced!

REPLY [9 votes]: Here is how you advance, since $T$ is self-adjoint, then 
$$ \langle  T v_1,v_2 \rangle =\langle   v_1,Tv_2 \rangle \implies  \langle  \lambda_1 v_1,v_2 \rangle =\langle   v_1,\lambda_2v_2 \rangle \implies (\lambda_1 - \bar{\lambda_2} )  \langle  v_1,v_2 \rangle =0  $$
$$\implies \langle  v_1,v_2 \rangle =0,$$
since $\lambda_1 \neq \lambda_2$
Note: Check the inner product properties we used here.