T1: Eigenvectors of a Symmetric Matrix

July 18, 2024

• Diagonalizability
• Spectral Theorem
  • Example

$\textbf{Theorem 1}$. Eigenvectors of a symmetric matrix with distinct eigenvalues are orthogonal.

$\textbf{Proof}$.

(To do later).

Diagonalizability

$\textbf{Theorem}$. A matrix $A_{n\times n}$ is diagonalizable if and only if $A$ has a complete set of eigenvectors, i.e, $n$ linearly independent eigenvectors for $A$.

By Gram-Schmidt method, we can always transform these $n$ linearly independent eigenvectors into $n$ orthonormal basis.

Spectral Theorem

Example

Let $A$ be an $n\times n$ symmetric real matrix. We can rewrite $A$ as follows.

$$\left( \begin{matrix} \vert && \cdots && \vert \\ \textbf{v}_1 && \cdots && \textbf{v}_n\\ \vert && \cdots && \vert \end{matrix} \right) \left( \begin{matrix} \lambda_1 \\ && \dots\\ &&&& \lambda_n \end{matrix} \right) \left( \begin{matrix} -&\textbf{v}_1 &- \\ -& \vdots &-\\ -& \textbf{v}_n &- \end{matrix} \right)$$

where $\textbf{v}_i$ are orthonormal vectors.

$\textbf{Theorem}$[Spectral Decomposition]. Given a matrix $A_{n\times n}$ with spectrum $\sigma(A) = \{\lambda_1,\ldots, \lambda_k\}$ is diagonalizable iff there exists matrices $\{G_1,\ldots,G_k\}$ such that

$$A = \lambda_1 G_1 + \ldots + \lambda_i G_k$$

where $G_i$ satisfies the following properties:

1. $G_i G_j = 0$ for any $i\neq j$.

2. $\sum G_i = \textbf{I}$.