Symmetric matrix with different eigenvalues Let be $ A\in \mathbb{R}^{2,2} $ a symmetric matrix with different eigenvalues $\lambda_1,\lambda_2$ and $ v_1 $ the eigenvector of the eigenvalue $ \lambda_1 $.
Show: If $ v $ is orthogonal to $ v_1 $ then $ v $ is the eigenvector of $ A $ of the eigenvalue $ \lambda_2 $.
My Idea:
Let be $ A\in \mathbb{R}^{2,2} $ a symmetric matrix with different eigenvalues $\lambda_1,\lambda_2$ and $ v_1 $ the eigenvector of the eigenvalue $ \lambda_1 $. For all $ v\in \mathbb{R}^2 $ with $ 
v_1^T\cdot v=0 $ we get
$$ \begin{align}&(\lambda_1-\lambda_2)\cdot v_1^T\cdot v=0\\[15pt]\Leftrightarrow \quad &v_1^T\cdot A\cdot v=\lambda_1\cdot v_1^T\cdot v=\lambda_2\cdot v_1^T\cdot v=v_1^T\cdot \lambda_2\cdot v \\[15pt]\Leftrightarrow \quad &v_1^T\cdot \underbrace{(A-\lambda_2\cdot I_2)\cdot v}_{=(*)}=0\end{align}$$
Now I would like to show that $ (*)=0 $ but I don't have any idea how I could do this.
 A: $v_2$ is the eigenvector of $\lambda_2$, then
$$\lambda_2\langle v_1, v_2\rangle=\langle v_1, Av_2\rangle \overset{(*)}{=}\langle  Av_1,v_2\rangle=\lambda_1\langle v_1, v_2\rangle.$$
At the star, you need $A$ to be symmetric. Hence, we get
$$(\lambda_1-\lambda_2)\cdot \langle v_1,v_2\rangle =0. $$
Since $\lambda_1\neq\lambda_2$ we get $ \langle v_1,v_2\rangle =0$. Observe, that $H:=(\operatorname{span}\{v_1\})^{\bot}:=\{w:\langle v_1,w\rangle=0\}$ is a subspace of $\mathbb R^2$, because from $\langle v_1, u\rangle =  0$ and $\langle v_1,w \rangle =  0$ follows $\langle v_1, u+\mu w \rangle =  0$. Its dimension is at least 1, since $v_2\in H$ and $v_2\neq 0$. Its dimension is less than 2, since $v_1\notin H$, because $\langle v_1,v_1\rangle \neq 0$. Hence, its dimension is exactly 1.
Let $u\neq 0$ such that $H=\operatorname{span}\{u\}$.  Because $\langle v_1,v\rangle =\langle v_1,v_2\rangle=0$, we have $v,v_2\in H$, i.e. there are $\mu_1,\mu_2\in \mathbb R$ with $v=\mu_1 u$ and $v_2=\mu_2 u$. This yields $\mu_2 v = \mu_1\mu_2 u=\mu_1v_2$.
