The eigenvectors of distinct eigenvalues of Hermitian matrices are orthogonal.
Theorem
Let $A$ be a Hermitian matrix of size $n \times n$. Let the eigenvectors corresponding to two distinct eigenvalues $\lambda , \mu$ of $A$ be $\mathbf{x}$ and $\mathbf{y}$, that is,
$$ \begin{align*} A \mathbf{x} =& \lambda \mathbf{x} \quad \\ A \mathbf{y} =& \mu \mathbf{y} \end{align*} $$
Then, the two eigenvectors are orthogonal to each other.
$$ \mathbf{x} \perp \mathbf{y} $$
Explanation
Hermitian matrices have the property that not only all eigenvalues are real but also the eigenvectors corresponding to these eigenvalues are orthogonal to each other. These properties can clearly be useful in some proof somewhere. Considering the concept of eigenvalues, it seems obvious, but if you look carefully at the definition, it is not at all obvious.
Proof
Assuming that $\mathbf{x}$ is an eigenvector of $\lambda$, and $\mathbf{y}$ is an eigenvector of $\mu$. When multiplying both sides of $A \mathbf{x} = \lambda \mathbf{x}$ by $\mathbf{y}^{\ast}$ to the left, it follows that:
$$ \begin{equation} \mathbf{y}^{\ast} A \mathbf{x} = \lambda \mathbf{y}^{\ast} \mathbf{x} \label{lambda} \end{equation} $$
Similarly, when multiplying both sides of $A \mathbf{y} = \mu \mathbf{y}$ by $\mathbf{x}^{\ast}$ to the left, it follows that:
$$ \mathbf{x}^{\ast} A \mathbf{y} = \mu \mathbf{x}^{\ast} \mathbf{y} $$
Since $\mu$ is real and $A$ is a Hermitian matrix, taking the conjugate transpose $^{\ast}$ of both sides of the equation yields:
$$ \begin{align} && \left( \mathbf{x}^{\ast} A \mathbf{y} \right)^{\ast} =& \left( \mu \mathbf{x}^{\ast} \mathbf{y} \right)^{\ast} \notag{} \\ \implies && \mathbf{y}^{\ast} A^{\ast} \mathbf{x} =& \mu^{\ast} \mathbf{y}^{\ast} \mathbf{x} \notag{} \\ \implies && \mathbf{y}^{\ast} A \mathbf{x} =& \mu \mathbf{y}^{\ast} \mathbf{x} \label{mu} \end{align} $$
Then, by $\eqref{lambda}$ and $\eqref{mu}$, the following equation holds:
$$ \lambda \mathbf{y}^{\ast} \mathbf{x} = \mathbf{y}^{\ast} A \mathbf{x} = \mu \mathbf{y}^{\ast} \mathbf{x} $$
Organizing to the left-hand side, it follows that:
$$ (\lambda - \mu ) \mathbf{y}^{\ast} \mathbf{x} = 0 $$
Assuming that $\lambda$ and $\mu$ are distinct real numbers,
$$ \mathbf{y}^{\ast} \mathbf{x} = 0 \implies \mathbf{y} \cdot \mathbf{x} = 0 $$
Therefore,
$$ \mathbf{x} \perp \mathbf{y} $$
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