施密特正交化（Gram-Schmidt method to orthonormalize vectors）

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本文详细介绍施密特正交化算法，通过逐步实例演示如何将给定的非正交向量组转化为欧几里得空间的正交基，并最终实现向量组的规范化，形成正交归一化基。涉及步骤详细且实用，适合初学者理解向量空间的构造过程。

施密特正交化算法

Give an arbitrary basis $\begin{Bmatrix} u_1, u_2,\cdots ,u_n \end{Bmatrix}$ for an n-dimensional inner product space V, we can constructs an orthogonal basis $\begin{Bmatrix} v_1, v_2,\cdots ,v_n \end{Bmatrix}$ for V.

the Gram-Schmidt algorithm is:

Step 1: Let $v_1 = u_1$ ；

Step 2: Let $v_2=u_2-\frac{\left \langle u_2,v_1 \right \rangle}{\left \| v_1 \right \|^2}v_1$ ；

Step 3: Let $v_3=u_3-\frac{\left \langle u_3,v_1 \right \rangle}{\left \| v_1 \right \|^2}v_1--\frac{\left \langle u_3,v_2 \right \rangle}{\left \| v_2 \right \|^2}v_2$

计算举例

Let $V=R^3$ with the Euclidean inner product. We will apply the Gram-Schmidt algorithm to orthogonalize the basis $\begin{Bmatrix} (1,-1,1), (1,0,1),(1,1,2) \end{Bmatrix}$ .

Step 1

Let $v_1 = u_1$ , so $v_1=(1, -1, 1)$ .

Step 2

Let $v_2=u_2-\frac{\left \langle u_2,v_1 \right \rangle}{\left \| v_1 \right \|^2}v_1$ .

$v_2=(1,0,1)-\frac{\left \langle (1,0,1)\cdot (1,-1,1) \right \rangle}{\left \| (1,-1,1) \right \|^2}(1,-1,1)=(1,0,1)-\frac{2}{3}(1,-1,1)=(\frac{1}{3},\frac{2}{3},\frac{1}{3})$ .

Step 3

Let $v_3=u_3-\frac{\left \langle u_3,v_1 \right \rangle}{\left \| v_1 \right \|^2}v_1--\frac{\left \langle u_3,v_2 \right \rangle}{\left \| v_2 \right \|^2}v_2$ .

$v_3=(1,1,2)-\frac{\left \langle (1,1,2)\cdot (1,-1,1) \right \rangle}{\left \| (1,-1,1) \right \|^2}(1,-1,1)--\frac{\left \langle (1,1,2)\cdot (\frac{1}{3},\frac{2}{3},\frac{1}{3}) \right \rangle}{\left \| (\frac{1}{3},\frac{2}{3},\frac{1}{3}) \right \|^2}(\frac{1}{3},\frac{2}{3},\frac{1}{3})=(1,1,2)-\frac{2}{3}(1,-1,1)-\frac{5}{2}(\frac{1}{3},\frac{2}{3},\frac{1}{3})=(-\frac{1}{2},0,\frac{1}{2})$

You can verify that $\begin{Bmatrix} (1,-1,1),(\frac{1}{3},\frac{2}{3},\frac{1}{3}),(-\frac{1}{2},0,\frac{1}{2}) \end{Bmatrix}$ forms an orthogonal basis for $R^3$ . Normalizing the vectors in the orthogonal basis, we obtain the orthonormal basis

$\begin{Bmatrix} (\frac{\sqrt{3}}{3},-\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3}),(\frac{\sqrt{6}}{6},\frac{\sqrt{6}}{3},\frac{\sqrt{6}}{6}),(-\frac{\sqrt{2}}{2},0,\frac{\sqrt{2}}{2}) \end{Bmatrix}$ .