For self-adjoint operators, eigenvectors that correspond to distinct eigenvalues are orthogonal

Question

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!

Answer 1

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.

Answer 2

If $T$ is self-adjoint, $$\langle v_1, Tv_2\rangle = \langle Tv_1, v_2\rangle .$$

What do you get when you simplify each side, and use the properties of inner products?