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线性代数:正交矩阵的总结

正交矩阵的定义: $AA^T =0$ 那么 $A$ 是正交矩阵.

性质:

  1. 逆等于转置
  2. 行列式等于 $\pm1$
  3. 其转置, 逆, 伴随矩阵都是正交矩阵
  4. 有限个正交矩阵的乘积一定是正交矩阵
  5. (充要条件)所有行(列)向量都是正交单位向量组
  6. 它的定义表明, 每行每列的平方和都是 1, 它的任意两行之间, 任意两列之间的内积都是0, 它的 X 列乘以它的转置的非 X 行的积是 0

求正交矩阵 $P$ 使得 $P$ 的第一行为 $\alpha_{1}=\frac{1}{2}(1,1,1,1)$

根据正交矩阵的性质, 它的任意两行之间的内积是 0, 不妨设任意一行为 $\alpha = (x_1, x_2, x_3, x_4)$

那么 $\alpha \cdot \alpha1 = 0$ 推出 $x{1}+x{2}+x{3}+x_{4}=0$

把 $x{1}+x{2}+x{3}+x{4}=0$ 看作一个齐次线性方程组.

$\left{x_1 = -x_2-x_3-x_4\right.$

分别令

$\left[\begin{array}{c}{x{2}} \ {x{3}} \ {x_{4}}\end{array}\right]=\left[\begin{array}{l}{1} \ {0} \ {0}\end{array}\right],\left[\begin{array}{c}{0} \ {1} \ {0}\end{array}\right] ,\left[\begin{array}{c}{0} \ {0} \ {1}\end{array}\right]$

解得

$\left[\begin{array}{c}{x{1}} \ {x{2}} \ {x{3}} \ {x{4}}\end{array}\right] =\left[\begin{array}{c}{-1} \ {1} \ {0} \ {0}\end{array}\right]\left[\begin{array}{c}{-1} \ {0} \ {1} \ {0}\end{array}\right]\left[\begin{array}{c}{-1} \ {0} \ {0} \ {1}\end{array}\right]$

基础解系

$\left[\begin{array}{lll}{\varepsilon{1}} & {\varepsilon{2}} & {c_{3}}\end{array}\right]=\left[\begin{array}{ccc}{-1} & {-1} & {-1} \ {1} & {0} & {0} \ {0} & {1} & {0} \ {0} & {0} & {1}\end{array}\right]$

下面进行正交化

$\beta{1}=\varepsilon{1}=\left[\begin{array}{c}{-1} \ {1} \ {0} \ {0}\end{array}\right]$

$\beta{2}=\varepsilon{2}-\frac{\varepsilon{2} \beta{1}}{\beta{1} \beta{1}} \beta_{1}$ $=\left[\begin{array}{c}{-1} \ {0} \ {1} \ {0}\end{array}\right]-\frac{1}{2}\left[\begin{array}{c}{-1} \ {1} \ {0} \ {0}\end{array}\right]$ $=\left[\begin{array}{c}{-\frac{1}{2}} \ {-\frac{1}{2}} \ {1} \ {0}\end{array}\right]=\frac{1}{2}\left[\begin{array}{c}{-1} \ {-1} \ {2} \ {0}\end{array}\right]$

$\beta{3}=\varepsilon{3}-\frac{\varepsilon{3} \beta{1}}{\beta{1} \beta{1}} \beta1$ $-\frac{\varepsilon{1} \beta{2}}{\beta{2} \beta{2}} \beta{2}$ $=\varepsilon_{1}-\frac{1}{2}\left[\begin{array}{c}{-1} \ {1} \ {0} \ {0}\end{array}\right]$ $-\frac{1}{3} \cdot \frac{1}{2}\left[\begin{array}{c}{-1} \ {-1} \ {2} \ {0}\end{array}\right]$ $=\left[\begin{array}{cccc}{-1} & {+\frac{1}{2}} & {+} & {\frac{1}{6}} \ {0} & {-\frac{1}{2}} & {+\frac{1}{6}} & {} \ {0} & {-0} & {-\frac{1}{3}} & {} \ {1} & {-0} & {-0} & {}\end{array}\right]$ $=\left[\begin{array}{r}{-\frac{1}{3}} \ {-\frac{1}{3}} \ {-\frac{1}{3}} \ {1}\end{array}\right]=\frac{1}{3}\left[\begin{array}{c}{-1} \ {-1} \ {-1} \ {3}\end{array}\right]$

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