Spectral Theorem

Definition

$A=A^{†}\iff A=Q\mathbf{\Lambda }{Q}^{†}$ where $Q$ is a Unitary Matrix, and $\mathbf{\Lambda }=\mathrm{diag}({\lambda }_1,{\lambda }_2,\dots ,{\lambda }_n)$

A matrix $A$ is a Hermitian Matrix if and only if $A$ is Unitary Diagonalizable Matrix.

Facts

Every Hermitian Matrix is diagonalizable, and has real-valued Eigenvalues and orthonormal eigenvector matrix.

For the Hermitian Matrix ${\mathbf{A}}^{†}=\mathbf{A}$, the every eigenvalue is real.

Proof For the eigendecomposition $\mathbf{Ax}=\lambda \mathbf{x}$, ${\mathbf{x}}^{†}\mathbf{Ax}={\mathbf{x}}^{†}\lambda \mathbf{x}=\lambda {\mathbf{x}}^{†}\mathbf{x}=\lambda ||\mathbf{x}||\Rightarrow \lambda =\frac{{\mathbf{x}}^{†}\mathbf{Ax}}{||\mathbf{x}||}$. Since $\mathbf{A}$ is hermitian, ${\mathbf{x}}^{†}\mathbf{Ax}\in \mathbb{R}$ and norm is always real. Therefore, every eigenvalue is real.

For the Hermitian Matrix ${\mathbf{A}}^{†}=\mathbf{A}$, the eigenvectors from different eigenvalues are orthogonal.

Proof Let the eigendecomposition $\mathbf{A}{\mathbf{x}}_1={\lambda }_1{\mathbf{x}}_1,\mathbf{A}{\mathbf{x}}_2={\lambda }_2{\mathbf{x}}_2$ where ${\lambda }_1\ne {\lambda }_2,{\mathbf{x}}_1\ne {\mathbf{x}}_2$. By the property of hermitian matrix, $(\mathbf{A}{\mathbf{x}}_1{)}^{†}{\mathbf{x}}_2={\mathbf{x}}_1^{†}{\mathbf{A}}^{†}{\mathbf{x}}_2={\mathbf{x}}_1^{†}{\lambda }_2{\mathbf{x}}_2={\lambda }_2{\mathbf{x}}_1^{†}{\mathbf{x}}_2$ and $(\mathbf{A}{\mathbf{x}}_1{)}^{†}{\mathbf{x}}_2=({\lambda }_1{\mathbf{x}}_1{)}^{†}{\mathbf{x}}_2={\bar{\lambda }}_1{\mathbf{x}}_1^{†}{\mathbf{x}}_2={\lambda }_1{\mathbf{x}}_1^{†}{\mathbf{x}}_2$ So, ${\lambda }_2{\mathbf{x}}_1^{†}{\mathbf{x}}_2={\lambda }_1{\mathbf{x}}_1^{†}{\mathbf{x}}_2\Rightarrow ({\lambda }_1-{\lambda }_2){\mathbf{x}}_1^{†}{\mathbf{x}}_2=0$. Since ${\lambda }_1\ne {\lambda }_2$, ${\mathbf{x}}_1^{†}{\mathbf{x}}_2=0$. Therefore, ${\mathbf{x}}_1^{†},{\mathbf{x}}_2$ are orthogonal.