Show that eigenvectors of a symmetric matrix  are orthogonal 

Since $\lambda_1 \neq \lambda_2$, we have
\[(1)     A \xi_1 =\lambda_1 \xi_1     \Rightarrow     \xi_2^{T} A \xi_1=\lambda_1 \xi_2^{T} \cdot \xi_1\]
and 
\[(2)     A \xi_2 =\lambda_2 \xi_2      \Rightarrow     \xi_1^{T} A \xi_2=\lambda_2 \xi_1^{T} \cdot \xi_2.\]
Since $A$ is symmetric, it follows from (1)  and (2) that 
\[\lambda_1 \xi_1^{T} \cdot \xi_2=(\lambda_1 \xi_2^{T} \cdot \xi_1 )^T= (\xi_2^{T} A \xi_1)^T= \xi_1^{T} A^T \xi_2 =\xi_1^{T} A \xi_2=\lambda_2 \xi_1^{T} \cdot \xi_2\]
\[\Rightarrow  \lambda_1 \xi_1^{T} \cdot \xi_2 =\lambda_2 \xi_1^{T} \cdot \xi_2   \Rightarrow (\lambda_1-\lambda_2) \xi_1^{T} \cdot \xi_2 =0 \]
\[\Rightarrow  \xi_1^{T} \cdot \xi_2 =0,\]
and thus $\xi_1$ and $\xi_2$ are orthogonal.  Here we have used the properties of the matrix transpose (See: https://en.wikipedia.org/wiki/Transpose ).