施密特正交化(Gram-Schmidt Orthogonalization)

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1 Gram-Schmidt的计算公式推导

问 ：以三维情况为例，已知三个线性无关的向量$\mathbf{a}$、$\mathbf{b}$、$\mathbf{c}$，如何能找到三个正交向量${\mathbf{\alpha }}_1$、${\mathbf{\alpha }}_2$、${\mathbf{\alpha }}_3$，在归一化后能形成标准正交基：${\mathbf{e}}_1$、${\mathbf{e}}_2$、${\mathbf{e}}_3$ ?

公式:

• 对三个线性无关的向量$\mathbf{a}$、$\mathbf{b}$、$\mathbf{c}$进行Gram-Schmidt正交化，所得的正交向量${\mathbf{\alpha }}_1$、${\mathbf{\alpha }}_2$、${\mathbf{\alpha }}_3$分别为:
  $$\begin{array}{rl}{\mathbf{\alpha }}_1& =\mathbf{a}\\ {\mathbf{\alpha }}_2& =\mathbf{b}-\frac{\mathbf{b}{\mathbf{\alpha }}_1}{|{\mathbf{\alpha }}_1{|}^2}{\mathbf{\alpha }}_1\\ {\mathbf{\alpha }}_3& =\mathbf{c}-\frac{\mathbf{c}{\mathbf{\alpha }}_1}{|{\mathbf{\alpha }}_1{|}^2}{\mathbf{\alpha }}_1-\frac{\mathbf{c}{\mathbf{\alpha }}_2}{|{\mathbf{\alpha }}_2{|}^2}{\mathbf{\alpha }}_2\end{array}$$
• 对$n$个线性无关的向量$\mathbf{a}$、$\mathbf{b}$、$\cdots$、$\mathbf{x}$进行Gram-Schmidt正交化，所得的正交向量${\mathbf{\alpha }}_1$、${\mathbf{\alpha }}_2$、$\cdots$、${\mathbf{\alpha }}_{\mathbf{n}}$分别为:
  α 1 = a α 2 = b − b   α 1 ∣ α 1 ∣ 2   α 1 α 3 = c − c   α 1 ∣ α 1 ∣ 2   α 1 − c   α 2 ∣ α 2 ∣ 2   α 2 ⋮ α n = x − x   α 1 ∣ α 1 ∣ 2   α 1 − x   α 2 ∣ α 2 ∣ 2   α 2   −   ⋯ −   x   α n − 1 ∣ α n − 1 ∣ 2   α n − 1 \begin{aligned} \bm{\alpha_1} &= \mathbf{a} \\ \bm{\alpha_2} &= \mathbf{b}-\frac{\mathbf{b} \ \bm{\alpha_1}}{|\bm{\alpha_1}|^2} \ \bm{\alpha_1} \\ \bm{\alpha_3} &= \mathbf{c}-\frac{\mathbf{c} \ \bm{\alpha_1}}{|\bm{\alpha_1}|^2} \ \bm{\alpha_1}-\frac{\mathbf{c} \ \bm{\alpha_2}}{|\bm{\alpha_2}|^2} \ \bm{\alpha_2} \\ \vdots \\ \bm{\alpha_n} &= \mathbf{x}-\frac{\mathbf{x} \ \bm{\alpha_1}}{|\bm{\alpha_1}|^2} \ \bm{\alpha_1}-\frac{\mathbf{x} \ \bm{\alpha_2}}{|\bm{\alpha_2}|^2} \ \bm{\alpha_2} \ - \ \cdots - \ \frac{\mathbf{x} \ \bm{\alpha_{n-1}}}{|\bm{\alpha_{n-1}}|^2} \ \bm{\alpha_{n-1}} \end{aligned} α1​α2​α3​⋮αn​​=a=b−∣α1​∣2b α1​​ α1​=c−∣α