前言：我只寫了 \(A\) 矩陣的正交對角化，另外兩個矩陣雖然有做出來，但使用 \(\LaTeX\) 打這麼多矩陣有點過於麻煩，因此我先寫到這邊，剩下兩個就交給同學們了

\(\begin{align} p_A(t) &=\det(A-tI)=\det \left( \begin{bmatrix} 3-t& 2& 2\\ 2& 2-t& 0\\ 2& 0& 4-t\end{bmatrix} \right)\\ &=(3-t)(2-t)(4-t)-4(2-t)-4(4-t) \\ &=-t^3+9t^2-18t \\ &=-t(t-3)(t-6)\end{align}\)

\(\implies\) \(A\) 的 eigenvalues 是 \(0,3,6\)

接著尋找它們對應的 eigenvectors \(\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3\) 且滿足 \(\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3\) 為 \(\mathbb{R}^3\)  的一組 orthonormal basis 

1. \(\lambda_1=0 \implies (A-0I)\mathbf{v}_1=\mathbf{0}\)

   用 row operations 把 \(A-0I\) 變為 echelon form:

   \(\begin{bmatrix}3& 2& 2\\ 2& 2& 0\\ 2& 0& 4\end{bmatrix} \to \begin{bmatrix}3& 2& 2\\ 1& 1& 0\\ 1& 0& 2\end{bmatrix} \to \begin{bmatrix}3& 2& 2\\ 1& 1& 0\\ 0& -1& 2\end{bmatrix} \to \begin{bmatrix}0& -1& 2\\ 1& 1& 0\\ 0& -1& 2\end{bmatrix} \to \begin{bmatrix}0& 0& 0\\ 1& 1& 0\\ 0& -1& 2\end{bmatrix}  \to \begin{bmatrix}1& 1& 0\\ 0& -1& 2\\ 0& 0& 0\end{bmatrix}\)

   \(\implies \mathbf{v}_1=t\begin{bmatrix} -2\\ 2\\ 1\end{bmatrix} \implies \mathbf{u}_1=\frac{1}{\sqrt{(-2)^2+2^2+1^2}}\begin{bmatrix} -2\\ 2\\ 1\end{bmatrix}=\begin{bmatrix} -\frac{2}{3}\\ \frac{2}{3}\\ \frac{1}{3}\end{bmatrix}\)

2. \(\lambda_2=3 \implies (A-3I)\mathbf{v}_2=\mathbf{0}\)

   用 row operations 把 \(A-3I\) 變為 echelon form:

   \(\begin{align} A-3I &=\begin{bmatrix}3-3& 2& 2\\ 2& 2-3& 0\\ 2& 0& 4-3\end{bmatrix} =\begin{bmatrix}0& 2& 2\\ 2& -1& 0\\ 2& 0& 1\end{bmatrix} \to \begin{bmatrix}0& 2& 2\\ 2& -1& 0\\ 0& 1& 1\end{bmatrix} \to \begin{bmatrix}0& 0& 0\\ 2& -1& 0\\ 0& 1& 1\end{bmatrix}\\ &\to \begin{bmatrix}2& -1& 0\\  0& 1& 1\\ 0& 0& 0\end{bmatrix} \end{align}\)

   \(\implies \mathbf{v}_2=t\begin{bmatrix} -1\\ -2\\ 2\end{bmatrix} \implies \mathbf{u}_2= \frac{1}{\sqrt{(-1)^2+(-2)^2+2^2}}\begin{bmatrix} -1\\ -2\\ 2\end{bmatrix}=\begin{bmatrix} -\frac{1}{3}\\ -\frac{2}{3}\\ \frac{2}{3}\end{bmatrix}\)

3. \(\lambda_3=6 \implies (A-6I)\mathbf{v}_3=\mathbf{0}\)

   用 row operations 把 \(A-3I\) 變為 echelon form:

   \(\begin{align} A-6I &=\begin{bmatrix}3-6& 2& 2\\ 2& 2-6& 0\\ 2& 0& 4-6\end{bmatrix} =\begin{bmatrix}-3& 2& 2\\ 2& -4& 0\\ 2& 0& -2\end{bmatrix} \to \begin{bmatrix}-3& 2& 2\\ 2& -4& 0\\ 0& 4& -2\end{bmatrix} \to \begin{bmatrix}-3& 2& 2\\ 6& -12& 0\\ 0& 4& -2\end{bmatrix}\\ &\to \begin{bmatrix}-3& 2& 2\\  0& -8& 4\\ 0& 4& -2\end{bmatrix} \to \begin{bmatrix}-3& 2& 2\\  0& 0& 0\\ 0& 4& -2\end{bmatrix} \to \begin{bmatrix}-3& 6& 0\\  0& 0& 0\\ 0& 2& -1\end{bmatrix} \to \begin{bmatrix}-1& 2& 0\\  0& 2& -1\\ 0& 0& 0\end{bmatrix} \end{align}\)

   \(\implies \mathbf{v}_3=t\begin{bmatrix} 2\\ 1\\ 2\end{bmatrix} \implies \mathbf{u}_2= \frac{1}{\sqrt{2^2+1^2+2^2}}\begin{bmatrix} 2\\ 1\\ 2\end{bmatrix}=\begin{bmatrix} \frac{2}{3}\\ \frac{1}{3}\\ \frac{2}{3}\end{bmatrix}\)

令 \(P=\begin{bmatrix} | & | & | \\ \mathbf{u}_1 & \mathbf{u}_2 & \mathbf{u}_3 \\ | & | & | \end{bmatrix} = \begin{bmatrix} -\frac{2}{3} & -\frac{1}{3} & \frac{2}{3} \\ \frac{2}{3} & -\frac{2}{3} & \frac{1}{3} \\ \frac{1}{3} & \frac{2}{3} & \frac{2}{3} \end{bmatrix} \implies P^{-1}=P^t=\begin{bmatrix} -\frac{2}{3} & \frac{2}{3} & \frac{1}{3} \\ -\frac{1}{3} & -\frac{2}{3} & \frac{2}{3} \\ \frac{2}{3} & \frac{1}{3} & \frac{2}{3}\end{bmatrix}\)

(因為 \(P\) 是一個 orthogonal matrix，所以 \( P^{-1}=P^t\))

令 \(D =\begin{bmatrix} \lambda_1 & 0 & 0 \\ 0 & \lambda_2 & 0 \\ 0 & 0 & \lambda_3 \end{bmatrix} =\begin{bmatrix} 0 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 6 \end{bmatrix}\)

則 \(P\) 即為將其對角化的 orthogonal matrix 而 \(D\) 即為其對應的對角矩陣，可以用他們將 \(A\) 寫成正交對角化的形式：

\(A=PDP^{-1}=\begin{bmatrix} -\frac{2}{3} & -\frac{1}{3} & \frac{2}{3} \\ \frac{2}{3} & -\frac{2}{3} & \frac{1}{3} \\ \frac{1}{3} & \frac{2}{3} & \frac{2}{3} \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 6 \end{bmatrix} \begin{bmatrix} -\frac{2}{3} & \frac{2}{3} & \frac{1}{3} \\ -\frac{1}{3} & -\frac{2}{3} & \frac{2}{3} \\ \frac{2}{3} & \frac{1}{3} & \frac{2}{3}\end{bmatrix}\)

$B = \begin{bmatrix} 1 & -2 & 2 \\ -2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}$, 令 $p_B(t) = \det(B - tI_3) = \det \begin{bmatrix} 1-t & -2 & 2 \\ -2 & 1-t & 2 \\ 2 & 2 & 1-t \end{bmatrix}$

$= 2 \det \begin{bmatrix} -2 & 2 \\ 1-t & 2 \end{bmatrix} - 2 \det \begin{bmatrix} 1-t & 2 \\ -2 & 2 \end{bmatrix} + (1-t) \det \begin{bmatrix} 1-t & -2 \\ -2 & 1-t \end{bmatrix} $

$= 2(-4 - 2 + 2t) - 2(2 - 2t + 4) + (1-t)\left((1-t)^2 - 4\right) = -(t-3)^2(t+3)$,

得 eigenvalue $\lambda_{1} = 3$(重根), $\lambda_{2} = -3$

(i) 考慮 $(B - 3I_3)x = 0 \Rightarrow \begin{bmatrix} -2 & -2 & 2 \\ -2 & -2 & 2 \\ 2 & 2 & -2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$, 而 $B - 3I_3$ 之 1, 2 column 為 linear dependent, 由 1st column 可得 $x_3 = x_1 + x_2$

令 $v_1 = \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = w_1, v_2 = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}$, 而 $\langle v_1, v_2 \rangle = 1 \neq 0$, $v_1, v_2$ 不垂直,

使用Gram-Schmidt, 令 $w_2 = v_2 - \text{Proj}_{w_1}(v_2) = v_2 - \frac{\langle v_2, w_1 \rangle}{\|w_1\|^2} w_1$

其中 $\|w_1\|^2 = 2, \langle v_2, w_1 \rangle = 1$, 得 $w_2 = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} - \frac{1}{2} \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} \frac{1}{2} \\ \frac{1}{2} \\ 1 \end{bmatrix}$, 

得 $\|w_1\| = \sqrt{2}$, 取 unit vector $u_1 = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix}$,

而 $\|w_2\| = \frac{\sqrt{6}}{2}$, 取 unit vector $u_2 = \frac{1}{\sqrt{6}} \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}$
 

(ii) 考慮 $(B + 3I_3)x = 0 \Rightarrow \begin{bmatrix} 4 & -2 & 2 \\ -2 & 4 & 2 \\ 2 & 2 & 4 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}, x = t\begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix}, t \in \mathbb{R}$, 令 unit vector $u_3 = \frac{1}{\sqrt{3}} \begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix}$

綜合上述，得 $Q = \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}} \\ 0 & \frac{2}{\sqrt{6}} & -\frac{1}{\sqrt{3}} \end{bmatrix} $,

而$Q^T = \begin{bmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 \\ \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{6}} & \frac{2}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{3}} \end{bmatrix}, Q^{-1} = \begin{bmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 \\ \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{6}} & \frac{2}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{3}} \end{bmatrix}, Q^T = Q^{-1}$

因此有
$B = QDQ^{-1} = \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}} \\ 0 & \frac{2}{\sqrt{6}} & -\frac{1}{\sqrt{3}} \end{bmatrix} \begin{bmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & -3 \end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 \\ \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{6}} & \frac{2}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{3}} \end{bmatrix} = \begin{bmatrix} 1 & -2 & 2 \\ -2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}$, 可對角化