Problem with 2nd part of exercise on symmetric matrices 
Let $M$ be a symmetric matrix (ie $M = M^T$) of the dimension $n \times n$, and a $\langle\cdot,\cdot\rangle$ be a standard scalar product on $\mathbb{R}^n$. Show that
($1$) $\langle u,Mv \rangle$ = $\langle Mu, v\rangle$  
($2$)Deduce from this that if $\lambda \neq \lambda^{\prime}$ are the different eigenvalues of the symmetric matrix $M$ with the eigenvectors $v$ and $v^{\prime}$, $\langle v, v^{\prime}\rangle=0$, $v$ and $v^{\prime}$ are perpendicular.

I know how to show (1) but I do not have idea how from this deduce (2). 
 A: Suppose that $Mv=\lambda v$ and $Mw=\mu w$ with $\lambda\neq \mu$. Then
$$ \lambda\langle v,w\rangle=\langle Mv,w\rangle=\langle v,Mw\rangle=\mu\langle v,w\rangle$$
which implies that
$$ (\lambda-\mu)\langle v,w\rangle=0$$
Since $\lambda\neq \mu$, this means that $\langle v,w\rangle=0$, i.e. $v$ and $w$ are orthogonal.
A: We have $\langle Mv, v'\rangle = \lambda \langle v,v'\rangle$ on the one hand and $\langle v, Mv'\rangle = \lambda'\langle v,v'\rangle$ on the other. Now since the two things are equal by part (1),$(\lambda-\lambda')\langle v,v'\rangle =0 $, and since $\lambda\neq\lambda'$, $\langle v,v'\rangle =0$.
A: Assuming part (1):
In (1), choose $u = v'$ to get
$$
\left< v',Mv \right> = \left< v',\lambda v \right> = \lambda \left< v', v \right>
\\ \left< Mv',v \right> = \left< \lambda' v', v \right> = \lambda' \left< v', v \right>\\
\lambda \left< v', v \right>=  \lambda' \left< v', v \right> \\
(\lambda-\lambda') \left< v', v \right>= 0
$$
and since $\lambda-\lambda' \neq 0$,$$ \left< v', v \right>= 0$$
