How to show symmetric matrices are orthogonally diagonalizable

Solution 1:

If you are allowed to use the spectral theorem, the proof of (v) is easy:

The spectral theorem says that for a real symmetric matrix $A$ there is an orthonormal set of eigenvectors $\{ \bf{e}_i\}$ such that $$ \begin{array}{c} A{\bf e}_i = \lambda_i {\bf e}_i \\ {\bf e}_i^T A = \lambda_i {\bf e}_i^T \\ \left< {\bf e}_i^T , {\bf e}_j \right> = \delta_{ij} \end{array} $$ Now consider $U$ such that $$ U_{ij} = [{\bf e}_i]_j $$ Then $AU$ is a square matrix with column $i$ equal to ${\bf e}_i$ for each $i$. From which: $$ [U^T(AU)]_{ki} = \lambda_i \left< {\bf e}_k^T , {\bf e}_i \right> = \lambda_i \delta_{ki} $$ so $U^TAU = U^T(AU)$ is a diagonal matrix with elements $\lambda_i$.