向量的施密特正交化

先看一个例子：
设三个向量分别为：

$$\begin{align}\alpha_1 &=(1,-1,0)^T\\ \alpha_2 & = (1,0,-1)^T\\ \alpha_3 & = (1,1,1)^T \end{align}$$
那么对 $\alpha_1,\alpha_2$正交化：
$$\begin{align} \beta_1 & = \alpha_1=(1,-1,0)^T\\ \beta_2 & = \alpha_2-\frac{(\alpha_2, \beta_1)}{(\beta_1, \beta_1)}\beta_1\\ & = (1,0,-1)^T-\frac{1\times1+0\times(-1)+(-1)\times0}{1\times1+(-1)\times(-1)+0\times0}(1,-1,0)^T\\ & = (1,0,-1)^T - \frac{1}{2}(1,-1,0)^T\\ & = \frac{1}{2}(1,1,-2)^T \end{align}$$
再对 $\beta_1,\beta_2,\beta_3$单位化：
$$\begin{align} \gamma_1 & = \frac{1}{\sqrt{2}}(1,-1,0)^T\\ \gamma_2 & = \frac{1}{\sqrt{6}}(1,1,-2)^T\\ \gamma_3 & = \frac{1}{\sqrt{3}}(1,1,1)^T \end{align}$$

以下施密特正交化的标准定义引自Wiki

We define the projection operator by

$$proj_u(v) = \frac{ \langle v,u\rangle }{\langle u, u \rangle}u$$
where $\langle v, u \rangle$denotes the inner product of the vectors $\mathbf v$and $\mathbf u$. This operator projects the vector $\mathbf v$orthogonally onto the line spanned by vector $\mathbf u$. If $\mathbf u=0$, we define $proj_0(v):=0$. i.e., the procjection map $proj_0$is the zero map, sending every vector to the zero vector.
The Gram-Schmidt process then works as follows:
$$\begin{align} u_1 & =v_1 & e_1=\frac{u_1}{\Vert u_1 \Vert}\\ u_2 & =v_2-proj_{u_1}(v_2) & e_2=\frac{u_2}{\Vert u_2 \Vert}\\ u_3 & =v_3-proj_{u_1}(v_3)-proj_{u_2}(v_3), & e_3=\frac{u_3}{\Vert u_3 \Vert}\\ u_4 & =v_4-proj_{u_1}(v_4)-proj_{u_2}(v_4)-proj_{u_3}(v_4), & e_3=\frac{u_4}{\Vert u_4 \Vert}\\ &\vdots \\ u_k & =v_k-\sum_{j=1}^{k-1}proj_{u_j}(v_k), & e_k=\frac{u_k}{\Vert u_k \Vert} \end{align}$$
The sequence $u_1, ..., u_k$is the required system of orthogonal vectors, and the normalized vectors $e_1, ..., e_k$from an orthonormal set. The calculation of the sequence $u_1, ..., u_k$is known as Gram-Schmidt rothogonalization, while the calculation of the sequence $e_1, ..., e_k$is known as Gram-Schmidt orthonormalization as the vectors are normalized.

部分符号和国内教材上有差异。