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· by Xianbin
\(\textbf{Theorem 1}\). 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:
-
\(G_i G_j = 0\) for any \(i\neq j\).
-
\(\sum G_i = \textbf{I}\).