Show that if $A$ is an $n$ by$ $n symmetric matrix and $\lambda \neq \mu$ are eigenvalues of $A$, then $ E_\mu \subset (E_\lambda )^\perp $.

Question

I started off as: Let $v\in E_\lambda$ . Given $u\in E_\mu$, $\left< u , v \right> = 0 $ because $u$ and $v$ are orthogonal. My question is that since the inner product of $u$ and $v$ is zero, would that suffice to conclude that $u$ is an element of $ (E_\lambda) ^\perp $? If not, what would be a better way to go about proving the statement above?

Answer 1

Do something like $ =\lambda\langle v,w\rangle=\langle \lambda v,w\rangle =\langle Av,w\rangle=\langle v,A^tw\rangle =\langle v,Aw\rangle =\langle v,\mu w\rangle=\mu\langle v,w\rangle $ to get $\langle v,w\rangle =0$, for the respective eigenvectors.

And yes, this means $E_{\mu}\subset E_{\lambda}^{\perp}$.