In the present discussion we assume that our matrix is real and symmetric, that is $\mathbf{A}\in {\mathbb{R}}^{n\times n}$. The matrix $\mathbf{A}$ has $n$ eigenvalues $\lambda_1\dots \lambda_n$ (distinct or not). Let $\mathbf{D}$ be the diagonal matrix with the eigenvalues on the diagonal $$\mathbf{D}= \left( \begin{array}{ccccccc} \lambda_1 & 0 & 0 & 0 & \dots &0 & 0 \\ 0 & \lambda_2 & 0 & 0 & \dots &0 &0 \\ 0 & 0 & \lambda_3 & 0 &0 &\dots & 0\\ \dots & \dots & \dots & \dots &\dots &\dots & \dots\\ 0 & \dots & \dots & \dots &\dots &\lambda_{n-1} & \\ 0 & \dots & \dots & \dots &\dots &0 & \lambda_n \end{array} \right).$$ If $\mathbf{A}$ is real and symmetric then there exists a real orthogonal matrix $\mathbf{S}$ such that $$\mathbf{S}^T \mathbf{A}\mathbf{S}= \mathrm{diag}(\lambda_1,\lambda_2,\dots ,\lambda_n),$$ and for $j=1:n$ we have $\mathbf{A}\mathbf{S}(:,j) = \lambda_j \mathbf{S}(:,j)$.