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 \( \mathbf{A}\mathbf{S}(:,j) = \lambda_j \mathbf{S}(:,j) \).