Chapter 26: real symmetric matrix diagonalizable

https://ccjou.wordpress.com/2011/02/09/實對稱矩陣可正交對角化的證明/

https://tex.stackexchange.com/questions/30619/what-is-the-best-symbol-for-vector-matrix-transpose

Theorem 26.1

實對稱矩陣的特徵值皆是實數，且對應特徵向量是實向量。

$$\begin{array}{c} \begin{cases} \begin{cases} A\in\mathcal{M}_{n\times n}\left(\mathbb{R}\right) & \text{real matrix}\\ A^{\intercal}=A & \text{symmetric matrix} \end{cases} & \text{real symmetric matrix}\\ A\boldsymbol{x}=\lambda\boldsymbol{x} & \begin{cases} \lambda\in\mathbb{C} & \text{complex eigenvalue}\\ \boldsymbol{0}\ne\boldsymbol{x}\in\mathbb{C}^{n} & \text{complex eigenvector} \end{cases} \end{cases}\\ \Downarrow\\ \begin{cases} \lambda\in\mathbb{R} & \text{real eigenvalue}\left(1\right)\\ \boldsymbol{x}\in\mathbb{R}^{n} & \text{real eigenvector}\left(2\right) \end{cases} \end{array}$$

Proof. $\left(1\right)$

$$\begin{aligned} A\boldsymbol{x}= & \lambda\boldsymbol{x}\\ \overline{A}\overline{\boldsymbol{x}}=\overline{A\boldsymbol{x}}= & \overline{\lambda\boldsymbol{x}}=\overline{\lambda}\overline{\boldsymbol{x}}\\ \overline{\boldsymbol{x}}^{\intercal}\overline{A}^{\intercal}=\left(\overline{A}\overline{\boldsymbol{x}}\right)^{\intercal}= & \left(\overline{\lambda}\overline{\boldsymbol{x}}\right)^{\intercal}=\overline{\lambda}\overline{\boldsymbol{x}}^{\intercal}\\ \overline{\boldsymbol{x}}^{\intercal}A\overset{\text{symmetric}}{=}\overline{\boldsymbol{x}}^{\intercal}A^{\intercal}\overset{\text{real}}{=}\\ \overline{\boldsymbol{x}}^{\intercal}A= & \overline{\lambda}\overline{\boldsymbol{x}}^{\intercal}\\ \lambda\overline{\boldsymbol{x}}^{\intercal}\boldsymbol{x}=\overline{\boldsymbol{x}}^{\intercal}\left(\lambda\boldsymbol{x}\right)\underset{A\boldsymbol{x}=\lambda\boldsymbol{x}}{\overset{\cdot\boldsymbol{x}}{=}}\overline{\boldsymbol{x}}^{\intercal}A\boldsymbol{x}= & \overline{\lambda}\overline{\boldsymbol{x}}^{\intercal}\boldsymbol{x}\\ \lambda\overline{\boldsymbol{x}}^{\intercal}\boldsymbol{x}= & \overline{\lambda}\overline{\boldsymbol{x}}^{\intercal}\boldsymbol{x}\\ \left(\lambda-\overline{\lambda}\right)\overline{\boldsymbol{x}}^{\intercal}\boldsymbol{x}= & 0\wedge\begin{cases} \overline{\boldsymbol{x}}^{\intercal}\boldsymbol{x}=\sum\limits _{i=1}^{n}\left|x_{i}\right|^{2}\\ \boldsymbol{x}\ne\boldsymbol{0} \end{cases}\Rightarrow\overline{\boldsymbol{x}}^{\intercal}\boldsymbol{x}\ne0\\ \lambda-\overline{\lambda}= & 0\\ \lambda= & \overline{\lambda}\Leftrightarrow\lambda\in\mathbb{R} \end{aligned}$$

Proof. $\left(1\right)$ fast concept

$$\begin{aligned} \color{orange}{\left(\overline{A}\overline{\boldsymbol{x}}\right)^{\intercal}\boldsymbol{x}}=\left(\overline{\boldsymbol{x}}^{\intercal}\overline{A}^{\intercal}\right)\boldsymbol{x}\overset{\text{symmetric}}{=} & \left(\overline{\boldsymbol{x}}^{\intercal}\overline{A}\right)\boldsymbol{x}=\color{orange}{\overline{\boldsymbol{x}}^{\intercal}\left(\overline{A}\boldsymbol{x}\right)}\\ \left(L\right)=\color{orange}{\left(\overline{A}\overline{\boldsymbol{x}}\right)^{\intercal}\boldsymbol{x}=} & \color{orange}{\overline{\boldsymbol{x}}^{\intercal}\left(\overline{A}\boldsymbol{x}\right)}=\left(R\right)\\ \left(L\right)=\color{orange}{\left(\overline{A}\overline{\boldsymbol{x}}\right)^{\intercal}\boldsymbol{x}}\overset{A\boldsymbol{x}=\lambda\boldsymbol{x}}{=} & \left(\overline{\lambda}\overline{\boldsymbol{x}}\right)^{\intercal}\boldsymbol{x}=\overline{\lambda}\overline{\boldsymbol{x}}^{\intercal}\boldsymbol{x}\\ \left(R\right)=\color{orange}{\overline{\boldsymbol{x}}^{\intercal}\left(\overline{A}\boldsymbol{x}\right)}\overset{\text{real}}{=}\overline{\boldsymbol{x}}^{\intercal}\left(A\boldsymbol{x}\right)\overset{A\boldsymbol{x}=\lambda\boldsymbol{x}}{=} & \overline{\boldsymbol{x}}^{\intercal}\left(\lambda\boldsymbol{x}\right)=\lambda\overline{\boldsymbol{x}}^{\intercal}\boldsymbol{x}\\ \overline{\lambda}\overline{\boldsymbol{x}}^{\intercal}\boldsymbol{x}=\color{orange}{\left(\overline{A}\overline{\boldsymbol{x}}\right)^{\intercal}\boldsymbol{x}=} & \color{orange}{\overline{\boldsymbol{x}}^{\intercal}\left(\overline{A}\boldsymbol{x}\right)}=\lambda\overline{\boldsymbol{x}}^{\intercal}\boldsymbol{x}\\ \overline{\lambda}\overline{\boldsymbol{x}}^{\intercal}\boldsymbol{x}= & \lambda\overline{\boldsymbol{x}}^{\intercal}\boldsymbol{x} \end{aligned}$$

Proof. $\left(2\right)$

???

推論特徵空間 $N(A-\lambda I)$ ($A-\lambda I$ 的零空間) 為 $\mathbb{R}^n$ 的子空間，故 $\boldsymbol{x}\in N(A-\lambda I)$ 是一個非零實向量。

Theorem 26.2

實對稱矩陣對應相異特徵值的特徵向量互為正交。

$$\begin{array}{c} \begin{cases} \begin{cases} A\in\mathcal{M}_{n\times n}\left(\mathbb{R}\right) & \text{real matrix}\\ A^{\intercal}=A & \text{symmetric matrix} \end{cases} & \text{real symmetric matrix}\\ A\boldsymbol{x}=\lambda\boldsymbol{x} & \ref{real-sym-real-eigen}\begin{cases} \lambda\in\mathbb{R} & \text{real eigenvalue}\\ \boldsymbol{x}\in\mathbb{R}^{n} & \text{real eigenvector} \end{cases}\\ \begin{cases} A\boldsymbol{x}_{{\scriptscriptstyle 1}}=\lambda_{{\scriptscriptstyle 1}}\boldsymbol{x}_{{\scriptscriptstyle 1}} & \left(e_{{\scriptscriptstyle 1}}\right)\\ A\boldsymbol{x}_{{\scriptscriptstyle 2}}=\lambda_{{\scriptscriptstyle 2}}\boldsymbol{x}_{{\scriptscriptstyle 2}} & \left(e_{{\scriptscriptstyle 2}}\right) \end{cases} & \lambda_{{\scriptscriptstyle 1}}\ne\lambda_{{\scriptscriptstyle 2}} \end{cases}\\ \Downarrow\\ \boldsymbol{x}_{{\scriptscriptstyle 1}}^{\intercal}\boldsymbol{x}_{{\scriptscriptstyle 2}}=0\Leftrightarrow\boldsymbol{x}_{{\scriptscriptstyle 1}}\perp\boldsymbol{x}_{{\scriptscriptstyle 2}} \end{array}$$

Proof. $\left(1\right)$

$$\begin{aligned} A\boldsymbol{x}_{{\scriptscriptstyle 2}}= & \lambda_{{\scriptscriptstyle 2}}\boldsymbol{x}_{{\scriptscriptstyle 2}}\\ \boldsymbol{x}_{{\scriptscriptstyle 1}}^{\intercal}A\boldsymbol{x}_{{\scriptscriptstyle 2}}\overset{\boldsymbol{x}_{{\scriptscriptstyle 1}}^{\intercal}\cdot}{=} & \boldsymbol{x}_{{\scriptscriptstyle 1}}^{\intercal}\lambda_{{\scriptscriptstyle 2}}\boldsymbol{x}_{{\scriptscriptstyle 2}}=\lambda_{{\scriptscriptstyle 2}}\boldsymbol{x}_{{\scriptscriptstyle 1}}^{\intercal}\boldsymbol{x}_{{\scriptscriptstyle 2}}=\left(1\right)\\ A\boldsymbol{x}_{{\scriptscriptstyle 1}}= & \lambda_{{\scriptscriptstyle 1}}\boldsymbol{x}_{{\scriptscriptstyle 1}}\\ \boldsymbol{x}_{{\scriptscriptstyle 1}}^{\intercal}A^{\intercal}=\left(A\boldsymbol{x}_{{\scriptscriptstyle 1}}\right)^{\intercal}= & \left(\lambda_{{\scriptscriptstyle 1}}\boldsymbol{x}_{{\scriptscriptstyle 1}}\right)^{\intercal}=\lambda_{{\scriptscriptstyle 1}}\boldsymbol{x}_{{\scriptscriptstyle 1}}^{\intercal}\\ \boldsymbol{x}_{{\scriptscriptstyle 1}}^{\intercal}A^{\intercal}= & \lambda_{{\scriptscriptstyle 1}}\boldsymbol{x}_{{\scriptscriptstyle 1}}^{\intercal}\\ \boldsymbol{x}_{{\scriptscriptstyle 1}}^{\intercal}A\boldsymbol{x}_{{\scriptscriptstyle 2}}\overset{\text{symmetric}}{=}\boldsymbol{x}_{{\scriptscriptstyle 1}}^{\intercal}A^{\intercal}\boldsymbol{x}_{{\scriptscriptstyle 2}}\overset{\cdot\boldsymbol{x}_{{\scriptscriptstyle 2}}}{=} & \lambda_{{\scriptscriptstyle 1}}\boldsymbol{x}_{{\scriptscriptstyle 1}}^{\intercal}\boldsymbol{x}_{{\scriptscriptstyle 2}}=\left(2\right)\\ \lambda_{{\scriptscriptstyle 2}}\boldsymbol{x}_{{\scriptscriptstyle 1}}^{\intercal}\boldsymbol{x}_{{\scriptscriptstyle 2}}\overset{\left(1\right)}{=}\boldsymbol{x}_{{\scriptscriptstyle 1}}^{\intercal}A\boldsymbol{x}_{{\scriptscriptstyle 2}}\overset{\left(2\right)}{=} & \lambda_{{\scriptscriptstyle 1}}\boldsymbol{x}_{{\scriptscriptstyle 1}}^{\intercal}\boldsymbol{x}_{{\scriptscriptstyle 2}}\\ \lambda_{{\scriptscriptstyle 2}}\boldsymbol{x}_{{\scriptscriptstyle 1}}^{\intercal}\boldsymbol{x}_{{\scriptscriptstyle 2}}= & \lambda_{{\scriptscriptstyle 1}}\boldsymbol{x}_{{\scriptscriptstyle 1}}^{\intercal}\boldsymbol{x}_{{\scriptscriptstyle 2}}\\ \left(\lambda_{{\scriptscriptstyle 2}}-\lambda_{{\scriptscriptstyle 1}}\right)\boldsymbol{x}_{{\scriptscriptstyle 1}}^{\intercal}\boldsymbol{x}_{{\scriptscriptstyle 2}}= & 0\wedge\lambda_{{\scriptscriptstyle 1}}\ne\lambda_{{\scriptscriptstyle 2}}\\ \boldsymbol{x}_{{\scriptscriptstyle 1}}^{\intercal}\boldsymbol{x}_{{\scriptscriptstyle 2}}= & 0 \end{aligned}$$

Proof. $\left(1\right)$ fast concept

$$\begin{aligned} \color{orange}{\left(A\boldsymbol{x}_{{\scriptscriptstyle 1}}\right)^{\intercal}\boldsymbol{x}_{{\scriptscriptstyle 2}}}=\left(\boldsymbol{x}_{{\scriptscriptstyle 1}}^{\intercal}A^{\intercal}\right)\boldsymbol{x}_{{\scriptscriptstyle 2}}\overset{\text{symmetric}}{=} & \left(\boldsymbol{x}_{{\scriptscriptstyle 1}}^{\intercal}A\right)\boldsymbol{x}_{{\scriptscriptstyle 2}}=\color{orange}{\boldsymbol{x}_{{\scriptscriptstyle 1}}^{\intercal}\left(A\boldsymbol{x}_{{\scriptscriptstyle 2}}\right)}\\ \left(L\right)=\color{orange}{\left(A\boldsymbol{x}_{{\scriptscriptstyle 1}}\right)^{\intercal}\boldsymbol{x}_{{\scriptscriptstyle 2}}=} & \color{orange}{\boldsymbol{x}_{{\scriptscriptstyle 1}}^{\intercal}\left(A\boldsymbol{x}_{{\scriptscriptstyle 2}}\right)} =\left(R\right)\\ \left(L\right)=\color{orange}{\left(A\boldsymbol{x}_{{\scriptscriptstyle 1}}\right)^{\intercal}\boldsymbol{x}_{{\scriptscriptstyle 2}}} \overset{\left(e_{{\scriptscriptstyle 1}}\right)}{=} & \left(\lambda_{{\scriptscriptstyle 1}}\boldsymbol{x}_{{\scriptscriptstyle 1}}\right)^{\intercal}\boldsymbol{x}_{{\scriptscriptstyle 2}}=\lambda_{{\scriptscriptstyle 1}}\boldsymbol{x}_{{\scriptscriptstyle 1}}^{\intercal}\boldsymbol{x}_{{\scriptscriptstyle 2}}\\ \left(R\right)=\color{orange}{\boldsymbol{x}_{{\scriptscriptstyle 1}}^{\intercal}\left(A\boldsymbol{x}_{{\scriptscriptstyle 2}}\right)}\overset{\left(e_{{\scriptscriptstyle 2}}\right)}{=} & \boldsymbol{x}_{{\scriptscriptstyle 1}}^{\intercal}\left(\lambda_{{\scriptscriptstyle 2}}\boldsymbol{x}_{{\scriptscriptstyle 2}}\right)=\lambda_{{\scriptscriptstyle 2}}\boldsymbol{x}_{{\scriptscriptstyle 1}}^{\intercal}\boldsymbol{x}_{{\scriptscriptstyle 2}}\\ \lambda_{{\scriptscriptstyle 1}}\boldsymbol{x}_{{\scriptscriptstyle 1}}^{\intercal}\boldsymbol{x}_{{\scriptscriptstyle 2}}=\color{orange}{\left(A\boldsymbol{x}_{{\scriptscriptstyle 1}}\right)^{\intercal}\boldsymbol{x}_{{\scriptscriptstyle 2}}=} & \color{orange}{\boldsymbol{x}_{{\scriptscriptstyle 1}}^{\intercal}\left(A\boldsymbol{x}_{{\scriptscriptstyle 2}}\right)}=\lambda_{{\scriptscriptstyle 2}}\boldsymbol{x}_{{\scriptscriptstyle 1}}^{\intercal}\boldsymbol{x}_{{\scriptscriptstyle 2}}\\ \lambda_{{\scriptscriptstyle 1}}\boldsymbol{x}_{{\scriptscriptstyle 1}}^{\intercal}\boldsymbol{x}_{{\scriptscriptstyle 2}}= & \lambda_{{\scriptscriptstyle 2}}\boldsymbol{x}_{{\scriptscriptstyle 1}}^{\intercal}\boldsymbol{x}_{{\scriptscriptstyle 2}} \end{aligned}$$

Theorem 26.3

$$\begin{array}{c} \begin{cases} \begin{cases} A\in\mathcal{M}_{n\times n}\left(\mathbb{R}\right) & \text{real matrix}\\ A^{\intercal}=A & \text{symmetric matrix} \end{cases} & \text{real symmetric matrix}\\ A\boldsymbol{x}_{{\scriptscriptstyle 1}}=\lambda\boldsymbol{x}_{{\scriptscriptstyle 1}} & \left(e\right)\\ \boldsymbol{x}_{{\scriptscriptstyle 2}}^{\intercal}\boldsymbol{x}_{{\scriptscriptstyle 1}}=0\Leftrightarrow\boldsymbol{x}_{{\scriptscriptstyle 2}}\perp\boldsymbol{x}_{{\scriptscriptstyle 1}} & \left(o\right) \end{cases}\\ \Downarrow\\ A\boldsymbol{x}_{{\scriptscriptstyle 2}}\perp\boldsymbol{x}_{{\scriptscriptstyle 1}}\Leftrightarrow\left(A\boldsymbol{x}_{{\scriptscriptstyle 2}}\right)^{\intercal}\boldsymbol{x}_{{\scriptscriptstyle 1}}=0 \end{array}$$

Proof. $$\begin{aligned} \left(A\boldsymbol{x}_{{\scriptscriptstyle 2}}\right)^{\intercal}\boldsymbol{x}_{{\scriptscriptstyle 1}}= & \left(\boldsymbol{x}_{{\scriptscriptstyle 2}}^{\intercal}A^{\intercal}\right)\boldsymbol{x}_{{\scriptscriptstyle 1}}\overset{\text{symmetric}}{=}\left(\boldsymbol{x}_{{\scriptscriptstyle 2}}^{\intercal}A\right)\boldsymbol{x}_{{\scriptscriptstyle 1}}\\ = & \boldsymbol{x}_{{\scriptscriptstyle 2}}^{\intercal}\left(A\boldsymbol{x}_{{\scriptscriptstyle 1}}\right)\overset{\left(e\right)}{=}\boldsymbol{x}_{{\scriptscriptstyle 2}}^{\intercal}\left(\lambda\boldsymbol{x}_{{\scriptscriptstyle 1}}\right)\\ = & \lambda\boldsymbol{x}_{{\scriptscriptstyle 2}}^{\intercal}\boldsymbol{x}_{{\scriptscriptstyle 1}}\overset{\left(o\right)}{=}\lambda\cdot0=0\\ \left(A\boldsymbol{x}_{{\scriptscriptstyle 2}}\right)^{\intercal}\boldsymbol{x}_{{\scriptscriptstyle 1}}= & 0\Leftrightarrow A\boldsymbol{x}_{{\scriptscriptstyle 2}}\perp\boldsymbol{x}_{{\scriptscriptstyle 1}} \end{aligned}$$