Let $A$ be a real $n \times n$ symmetric matrix. Then:

1. All eigenvalues of $A$ are real
2. Eigenvectors corresponding to distinct eigenvalues are orthogonal
3. $\mathbb{R}^n$ has an orthonormal basis consisting of eigenvectors of $A$
4. $A$ is orthogonally diagonalizable: $A = Q\Lambda Q^T$ where:
   • $Q$ is orthogonal with $Q^TQ = I$
   • $\Lambda = \text{diag}(\lambda_1, \ldots, \lambda_n)$ contains the eigenvalues
   • Columns of $Q$ are orthonormal eigenvectors

This decomposition is unique up to reordering eigenvalues and sign choices for eigenvectors.

Let $A$ be an $n \times n$ Hermitian matrix (i.e., $A^* = A$). Then:

1. All eigenvalues of $A$ are real
2. $\mathbb{C}^n$ has an orthonormal basis of eigenvectors
3. $A$ is unitarily diagonalizable: $A = U\Lambda U^*$ where $U$ is unitary and $\Lambda$ is real diagonal

Two symmetric matrices $A$ and $B$ are simultaneously diagonalizable by the same orthogonal matrix if and only if $AB = BA$ (they commute).

If $AB = BA$, there exists orthogonal $Q$ such that both $Q^TAQ$ and $Q^TBQ$ are diagonal.

For symmetric matrix $A$ and nonzero $\mathbf{x} \in \mathbb{R}^n$, the Rayleigh quotient is: $R(\mathbf{x}) = \frac{\mathbf{x}^TA\mathbf{x}}{\mathbf{x}^T\mathbf{x}}$

Let $\lambda_{\min}$ and $\lambda_{\max}$ be the smallest and largest eigenvalues. Then: $\lambda_{\min} \leq R(\mathbf{x}) \leq \lambda_{\max}$

Moreover, $R(\mathbf{x})$ achieves its maximum $\lambda_{\max}$ when $\mathbf{x}$ is the corresponding eigenvector.