ThmDex – An index of mathematical definitions, results, and conjectures.
Eigenvectors for a symmetric complex matrix are orthogonal
Formulation 0
Let $A \in \mathbb{C}^{N \times N}$ be a D6159: Complex square matrix such that
(i) \begin{equation} A^T = A \end{equation}
(ii) $\lambda, \mu \in \mathbb{C}$ are each a D1207: Complex number
(iii) \begin{equation} \lambda \neq \mu \end{equation}
(iv) $z, w \in \mathbb{C}^{N \times 1} \setminus \{ \boldsymbol{0} \}$ are each a D6214: Complex matrix standard eigenvector sequence
(v) \begin{equation} A z = \lambda z \end{equation}
(vi) \begin{equation} A w = \mu w \end{equation}
Then \begin{equation} z^T w = 0 \end{equation}
Proofs
Proof 0
Let $A \in \mathbb{C}^{N \times N}$ be a D6159: Complex square matrix such that
(i) \begin{equation} A^T = A \end{equation}
(ii) $\lambda, \mu \in \mathbb{C}$ are each a D1207: Complex number
(iii) \begin{equation} \lambda \neq \mu \end{equation}
(iv) $z, w \in \mathbb{C}^{N \times 1} \setminus \{ \boldsymbol{0} \}$ are each a D6214: Complex matrix standard eigenvector sequence
(v) \begin{equation} A z = \lambda z \end{equation}
(vi) \begin{equation} A w = \mu w \end{equation}
We have \begin{equation} \lambda z^T w = (\lambda z)^T w = (A z)^T w = z^T A^T w = z^T (A w) = z^T (\mu w) = \mu z^T w \end{equation}