在内积空间 $R^{4}$ 中已给

α_{1} = (1, 1, 1, 1)

α_{2} = (3, 3, 1, 1)

α_{3} = (1, 9, 1, 9)

α_{4} = (4, 0, 0, 0)

把它们标准正交化.

解

首先将其正交化.

取 $β_{1} = α_{1} = (1, 1, 1, 1)$ .

β_{2} = α_{2} - \frac{( α _{2} , β _{1} )}{( β _{1} , β _{1} )} β_{1} = α_{2} - \frac{8}{4} β_{1} = (1, 1, - 1, - 1) .

β_{3} = α_{3} - \frac{( α _{3} , β _{1} )}{( β _{1} , β _{1} )} β_{1} - \frac{( α _{3} , β _{2} )}{( β _{2} , β _{2} )} β_{2} = α_{3} - \frac{20}{4} β_{1} - \frac{0}{4} β_{2} = (- 4, 4, - 4, 4) .

β_{4} = α_{4} - \frac{( α _{4} , β _{1} )}{( β _{1} , β _{1} )} β_{1} - \frac{( α _{4} , β _{2} )}{( β _{2} , β _{2} )} β_{2} - \frac{( α _{4} , β _{3} )}{( β _{3} , β _{3} )} β_{3} = α_{4} - \frac{4}{4} β_{1} - \frac{4}{4} β_{2} + \frac{16}{64} β_{3} = (1, - 1, - 1, 1) .

由此可得正交向量组

β_{1} β_{2} β_{3} β_{4} = (1, 1, 1, 1) = (1, 1, - 1, - 1) = (- 4, 4, - 4, 4) = (1, - 1, - 1, 1)

其次,将 $β_{1}, β_{2}, β_{3}, β_{4}$ 单位化,得

η_{1} η_{2} η_{3} η_{4} = \frac{1}{2} (1, 1, 1, 1) = \frac{1}{2} (1, 1, - 1, - 1) = \frac{1}{2} (- 1, 1, - 1, 1) = \frac{1}{2} (1, - 1, - 1, 1)

以 $η_{1}, η_{2}, η_{3}, η_{4}$ 为列向量,得到一个 4 阶矩阵,即

A = \frac{1}{2} \frac{1}{2} \frac{1}{2} \frac{1}{2} \frac{1}{2} \frac{1}{2} - \frac{1}{2} - \frac{1}{2} - \frac{1}{2} \frac{1}{2} - \frac{1}{2} \frac{1}{2} \frac{1}{2} - \frac{1}{2} - \frac{1}{2} \frac{1}{2}

它的列两两正交,容易验证 $A^{T} A = A A^{T} = E$ ,我们称它是一个正交矩阵.

Youliang Zhong