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本帖最后由 hbghlyj 于 2023-1-5 20:58 编辑 实对称矩阵的特征多项式的所有复根都是实数。
Lemma 8.19. If $\lambda$ is an eigenvalue of a self-adjoint linear operator then $\lambda \in \mathbb{R}$.
Proof. Assume $w \neq 0$ and $T(w)=\lambda w$ for some $\lambda \in \mathbb{C}$. Then
\begin{aligned}
\lambda\langle w, w\rangle & =\langle w, \lambda w\rangle=\langle w, T(w)\rangle=\left\langle T^*(w), w\right\rangle \\
& =\langle T(w), w\rangle=\langle\lambda w, w\rangle=\bar{\lambda}\langle w, w\rangle .
\end{aligned}Hence, as $\langle w, w\rangle \neq 0, \lambda=\bar{\lambda}$ and $\lambda \in \mathbb{R}$.
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