对三个五维向量(2,-3,1,0,0),(5,-6,0,1,0),(-3,2,0,0,1)进行施密特正交化的结果为
时间: 2023-10-08 14:05:01 浏览: 47
首先,我们可以将这三个向量表示成矩阵的形式:
$$
\begin{bmatrix}
2 & -3 & 1 & 0 & 0 \\
5 & -6 & 0 & 1 & 0 \\
-3 & 2 & 0 & 0 & 1 \\
\end{bmatrix}
$$
接下来,我们可以按照施密特正交化的步骤进行:
1. 将第一个向量标准化
$$
\begin{aligned}
\mathbf{v}_1 &= \begin{bmatrix} 2 \\ 5 \\ -3 \end{bmatrix} \\
\mathbf{u}_1 &= \frac{\mathbf{v}_1}{\|\mathbf{v}_1\|} = \frac{1}{\sqrt{38}} \begin{bmatrix} 2 \\ 5 \\ -3 \end{bmatrix} = \begin{bmatrix} \frac{2}{\sqrt{38}} \\ \frac{5}{\sqrt{38}} \\ -\frac{3}{\sqrt{38}} \end{bmatrix}
\end{aligned}
$$
2. 计算第二个向量在第一个向量上的投影,并将其从第二个向量中减去
$$
\begin{aligned}
\mathbf{v}_2 &= \begin{bmatrix} -3 \\ -6 \\ 2 \end{bmatrix} \\
\mathbf{u}_2 &= \mathbf{v}_2 - (\mathbf{v}_2 \cdot \mathbf{u}_1) \mathbf{u}_1 \\
&= \begin{bmatrix} -3 \\ -6 \\ 2 \end{bmatrix} - \left(\frac{-4}{\sqrt{38}}\right) \begin{bmatrix} \frac{2}{\sqrt{38}} \\ \frac{5}{\sqrt{38}} \\ -\frac{3}{\sqrt{38}} \end{bmatrix} \\
&= \begin{bmatrix} -\frac{17}{19} \\ -\frac{36}{19} \\ \frac{24}{19} \end{bmatrix}
\end{aligned}
$$
3. 将第二个向量标准化
$$
\begin{aligned}
\mathbf{u}_2 &= \frac{\mathbf{v}_2 - (\mathbf{v}_2 \cdot \mathbf{u}_1) \mathbf{u}_1}{\|\mathbf{v}_2 - (\mathbf{v}_2 \cdot \mathbf{u}_1) \mathbf{u}_1\|} = \frac{1}{\sqrt{665}} \begin{bmatrix} -17 \\ -36 \\ 24 \end{bmatrix} = \begin{bmatrix} -\frac{17}{\sqrt{665}} \\ -\frac{36}{\sqrt{665}} \\ \frac{24}{\sqrt{665}} \end{bmatrix}
\end{aligned}
$$
4. 计算第三个向量在前两个向量张成的平面上的投影,并将其从第三个向量中减去
$$
\begin{aligned}
\mathbf{v}_3 &= \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \\
\mathbf{u}_3 &= \mathbf{v}_3 - (\mathbf{v}_3 \cdot \mathbf{u}_1) \mathbf{u}_1 - (\mathbf{v}_3 \cdot \mathbf{u}_2) \mathbf{u}_2 \\
&= \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} - \left(\frac{2}{\sqrt{38}}\right) \begin{bmatrix} \frac{2}{\sqrt{38}} \\ \frac{5}{\sqrt{38}} \\ -\frac{3}{\sqrt{38}} \end{bmatrix} - \left(\frac{-17}{\sqrt{665}}\right) \begin{bmatrix} -\frac{17}{\sqrt{665}} \\ -\frac{36}{\sqrt{665}} \\ \frac{24}{\sqrt{665}} \end{bmatrix} \\
&= \begin{bmatrix} -\frac{9}{19} \\ \frac{4}{19} \\ \frac{12}{19} \end{bmatrix}
\end{aligned}
$$
5. 将第三个向量标准化
$$
\begin{aligned}
\mathbf{u}_3 &= \frac{\mathbf{v}_3 - (\mathbf{v}_3 \cdot \mathbf{u}_1) \mathbf{u}_1 - (\mathbf{v}_3 \cdot \mathbf{u}_2) \mathbf{u}_2}{\|\mathbf{v}_3 - (\mathbf{v}_3 \cdot \mathbf{u}_1) \mathbf{u}_1 - (\mathbf{v}_3 \cdot \mathbf{u}_2) \mathbf{u}_2\|} = \begin{bmatrix} -\frac{9}{\sqrt{2665}} \\ \frac{4}{\sqrt{2665}} \\ \frac{12}{\sqrt{2665}} \end{bmatrix}
\end{aligned}
$$
因此,施密特正交化后得到的三个单位向量分别为:
$$
\begin{aligned}
\mathbf{u}_1 &= \begin{bmatrix} \frac{2}{\sqrt{38}} \\ \frac{5}{\sqrt{38}} \\ -\frac{3}{\sqrt{38}} \end{bmatrix} \\
\mathbf{u}_2 &= \begin{bmatrix} -\frac{17}{\sqrt{665}} \\ -\frac{36}{\sqrt{665}} \\ \frac{24}{\sqrt{665}} \end{bmatrix} \\
\mathbf{u}_3 &= \begin{bmatrix} -\frac{9}{\sqrt{2665}} \\ \frac{4}{\sqrt{2665}} \\ \frac{12}{\sqrt{2665}} \end{bmatrix}
\end{aligned}
$$
可以验证这三个向量是两两正交的,并且它们构成了原向量空间的一组基。