若n阶方阵A满足
A
T
A
=
E
A^TA = E
A T A = E A是正交阵的充要条件：A的列(行)向量都是单位向量，且两两正交。
若A为正交矩阵，x为向量，则Ax称为正交变换
正交变换不改变向量的长度
y
=
A
x
,
y
T
y
=
(
A
x
)
T
A
x
=
x
T
A
T
A
x
=
x
T
E
x
=
x
T
x
y=Ax, y^Ty = (Ax)^TAx = x^TA^TAx = x^TEx = x^Tx
y = A x , y T y = ( A x ) T A x = x T A T A x = x T E x = x T x
正交矩阵的性质
若A为正交矩阵，则逆矩阵
A
−
1
A^{-1}
A − 1
若P、Q为正交矩阵，那么
P
∗
Q
P*Q
P ∗ Q
对于m*n的列满秩矩阵A, 必有,
A
m
∗
n
=
Q
m
∗
m
⋅
R
m
∗
n
A_{m*n} = Q_{m*m} · R_{m*n}
A m ∗ n = Q m ∗ m ⋅ R m ∗ n
其中Q为正交矩阵，R为非奇异上三角矩阵，当要求R的对角线元素为正的时候，该分解唯一。
该分解叫做QR分解，常用语求解A的特征值、A的逆，最小二乘等问题
QR分解是将矩阵分解为一个正交矩阵与上三角矩阵的乘积
备注：图片托管于github，请确保网络的可访问性
这其中，Q为正交矩阵，
Q
T
Q
=
l
Q^TQ = l
Q T Q = l
实际中，QR分解经常被用来解线性最小二乘问题。
把一组线性无关向量组化为规范正交向量组，继而得到正交阵
η
1
=
β
1
∣
∣
β
1
∣
∣
,
η
2
=
β
2
∣
∣
β
2
,
⋯
,
η
r
=
β
r
∣
∣
β
r
∣
∣
\eta_1 = \frac{\beta_1}{||\beta_1||}, \eta_2 = \frac{\beta_2}{||\beta_2}, \cdots, \eta_r = \frac{\beta_r}{||\beta_r||}
η 1 = ∣ ∣ β 1 ∣ ∣ β 1 , η 2 = ∣ ∣ β 2 β 2 , ⋯ , η r = ∣ ∣ β r ∣ ∣ β r
α
1
,
α
2
,
.
.
.
,
α
r
\alpha_1, \alpha_2, ..., \alpha_r
α 1 , α 2 , . . . , α r
设
α
1
,
α
2
,
.
.
.
,
α
r
\alpha_1, \alpha_2, ..., \alpha_r
α 1 , α 2 , . . . , α r
β
1
=
α
1
,
β
2
=
α
2
−
[
β
1
,
α
2
]
[
β
1
,
β
1
]
β
1
,
β
3
=
α
3
−
[
β
1
,
α
3
]
β
1
,
β
2
β
1
−
[
β
2
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α
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[
β
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β
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⋯
\beta_1 = \alpha_1, \beta_2 = \alpha_2 - \frac{[\beta_1, \alpha_2]}{[\beta_1, \beta_1]} \beta_1, \beta_3 = \alpha_3 - \frac{[\beta_1, \alpha_3]}{\beta_1, \beta_2} \beta_1 - \frac{[\beta_2, \alpha_3]}{[\beta_2, \beta_2]} \beta_2 \cdots \cdots
β 1 = α 1 , β 2 = α 2 − [ β 1 , β 1 ] [ β 1 , α 2 ] β 1 , β 3 = α 3 − β 1 , β 2 [ β 1 , α 3 ] β 1 − [ β 2 , β 2 ] [ β 2 , α 3 ] β 2 ⋯ ⋯
β
r
=
α
r
−
[
β
1
,
α
r
]
[
β
1
,
β
1
]
β
1
−
[
β
2
,
α
r
]
[
β
2
,
β
2
]
β
2
−
⋯
−
[
β
r
−
1
α
r
]
[
β
r
−
1
,
β
r
−
1
]
β
r
−
1
\beta_r = \alpha_r - \frac{[\beta_1, \alpha_r]}{[\beta_1, \beta_1]}\beta_1 - \frac{[\beta_2, \alpha_r]}{[\beta_2, \beta_2]} \beta_2 - \cdots - \frac{[\beta_{r-1} \alpha_r]}{[\beta_{r-1}, \beta_{r-1}]} \beta_{r-1}
β r = α r − [ β 1 , β 1 ] [ β 1 , α r ] β 1 − [ β 2 , β 2 ] [ β 2 , α r ] β 2 − ⋯ − [ β r − 1 , β r − 1 ] [ β r − 1 α r ] β r − 1 则
β
1
,
β
2
,
⋯
,
β
r
\beta_1, \beta_2, \cdots, \beta_r
β 1 , β 2 , ⋯ , β r
α
1
,
α
2
,
⋯
,
α
r
\alpha_1, \alpha_2, \cdots, \alpha_r
α 1 , α 2 , ⋯ , α r
例1
求矩阵
A
=
(
1
1
−
1
1
0
0
0
1
0
0
0
1
)
A=\left (\begin{array}{cccc}1 & 1 & -1 \\1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1\end{array} \right )
A = ⎝ ⎜ ⎜ ⎛ 1 1 0 0 1 0 1 0 − 1 0 0 1 ⎠ ⎟ ⎟ ⎞
分析
容易判断出
A
∈
C
3
4
×
3
A \in C_3^{4×3}
A ∈ C 3 4 × 3
将
A
=
[
α
1
,
α
2
,
α
3
]
A = [\alpha_1, \alpha_2, \alpha_3]
A = [ α 1 , α 2 , α 3 ]
β
1
=
α
1
=
[
1
1
0
0
]
T
\beta_1 = \alpha_1 = [1 \ \ 1 \ \ 0 \ \ 0]^T
β 1 = α 1 = [ 1 1 0 0 ] T
β
2
=
α
2
−
(
α
2
,
β
1
)
β
1
,
β
1
β
1
=
α
2
−
1
2
β
1
=
[
1
2
−
1
2
1
0
]
T
\beta_2 = \alpha_2 - \frac{(\alpha_2, \beta_1)}{\beta_1, \beta_1} \beta_1 = \alpha_2 - \frac{1}{2} \beta_1 = [\frac{1}{2} \ \ \frac{-1}{2} \ \ 1 \ \ 0]^T
β 2 = α 2 − β 1 , β 1 ( α 2 , β 1 ) β 1 = α 2 − 2 1 β 1 = [ 2 1 2 − 1 1 0 ] T
β
3
=
α
3
−
(
α
3
,
β
1
)
β
1
,
β
1
β
1
−
(
α
3
,
β
2
)
β
2
,
β
2
β
2
=
α
3
+
1
2
β
1
+
1
3
β
2
=
[
−
1
3
1
3
1
3
1
]
T
\beta_3 = \alpha_3 - \frac{(\alpha_3, \beta_1)}{\beta_1, \beta_1} \beta_1 - \frac{(\alpha_3, \beta_2)}{\beta_2, \beta_2} \beta_2 = \alpha_3 + \frac{1}{2} \beta_1 + \frac{1}{3} \beta_2 = [\frac{-1}{3} \ \ \frac{1}{3} \ \ \frac{1}{3} \ \ 1]^T
β 3 = α 3 − β 1 , β 1 ( α 3 , β 1 ) β 1 − β 2 , β 2 ( α 3 , β 2 ) β 2 = α 3 + 2 1 β 1 + 3 1 β 2 = [ 3 − 1 3 1 3 1 1 ] T 再将其单位化，得到一组标准正交向量组
η
1
=
1
∣
∣
β
1
∣
∣
β
1
=
[
2
2
2
2
0
0
]
T
\eta_1 = \frac{1}{||\beta_1||} \beta_1 = [\frac{\sqrt{2}}{2} \ \ \frac{\sqrt{2}}{2} \ \ 0 \ \ 0]^T
η 1 = ∣ ∣ β 1 ∣ ∣ 1 β 1 = [ 2 2
2 2
0 0 ] T
η
2
=
1
∣
∣
β
2
∣
∣
β
2
=
[
6
6
−
6
3
6
3
0
]
T
\eta_2 = \frac{1}{||\beta_2||} \beta_2 = [\frac{\sqrt{6}}{6} \ \ -\frac{\sqrt{6}}{3} \ \ \frac{\sqrt{6}}{3} \ \ 0]^T
η 2 = ∣ ∣ β 2 ∣ ∣ 1 β 2 = [ 6 6
− 3 6
3 6
0 ] T
η
3
=
1
∣
∣
β
3
∣
∣
β
3
=
[
−
3
6
3
6
3
6
3
2
]
T
\eta_3 = \frac{1}{||\beta_3||} \beta_3 = [-\frac{\sqrt{3}}{6} \ \ \frac{\sqrt{3}}{6} \ \ \frac{\sqrt{3}}{6} \ \ \frac{\sqrt{3}}{2}]^T
η 3 = ∣ ∣ β 3 ∣ ∣ 1 β 3 = [ − 6 3
6 3
6 3
2 3
] T
⇒
Q
(
η
1
,
η
2
,
η
3
)
=
[
2
2
6
6
−
3
6
2
2
−
6
6
3
6
0
6
3
3
6
0
0
3
2
]
\Rightarrow Q(\eta_1, \eta_2, \eta_3) = \left [\begin{array}{cccc}\frac{\sqrt{2}}{2} & \frac{\sqrt{6}}{6} & -\frac{\sqrt{3}}{6} \\\frac{\sqrt{2}}{2} & -\frac{\sqrt{6}}{6} & \frac{\sqrt{3}}{6} \\0 & \frac{\sqrt{6}}{3} & \frac{\sqrt{3}}{6} \\0 & 0 & \frac{\sqrt{3}}{2}\end{array} \right ]
⇒ Q ( η 1 , η 2 , η 3 ) = ⎣ ⎢ ⎢ ⎢ ⎡ 2 2
2 2
0 0 6 6
− 6 6
3 6
0 − 6 3
6 3
6 3
2 3
⎦ ⎥ ⎥ ⎥ ⎤
β
1
=
α
1
=
[
1
1
0
0
]
T
\beta_1 = \alpha_1 = [1 \ \ 1 \ \ 0 \ \ 0]^T
β 1 = α 1 = [ 1 1 0 0 ] T
β
2
=
α
2
−
(
α
2
,
β
1
)
(
β
1
,
β
1
)
β
1
=
α
2
−
1
2
β
1
=
[
1
2
−
1
2
1
0
]
T
\beta_2 = \alpha_2 - \frac{(\alpha_2, \beta_1)}{(\beta_1, \beta_1)} \beta_1 = \alpha_2 - \frac{1}{2} \beta_1 = [\frac{1}{2} \ \ \frac{-1}{2} \ \ 1 \ \ 0]^T
β 2 = α 2 − ( β 1 , β 1 ) ( α 2 , β 1 ) β 1 = α 2 − 2 1 β 1 = [ 2 1 2 − 1 1 0 ] T
β
3
=
α
3
−
(
α
3
,
β
1
)
β
1
,
β
1
β
1
−
(
α
3
,
β
2
)
β
2
,
β
2
β
2
=
α
3
+
1
2
β
1
+
1
3
β
2
=
[
−
1
3
1
3
1
3
1
]
T
\beta_3 = \alpha_3 - \frac{(\alpha_3, \beta_1)}{\beta_1, \beta_1}\beta_1 - \frac{(\alpha_3, \beta_2)}{\beta_2, \beta_2} \beta_2 = \alpha_3 + \frac{1}{2}\beta_1 + \frac{1}{3}\beta_2 = [\frac{-1}{3} \ \ \frac{1}{3} \ \ \frac{1}{3} \ \ 1]^T
β 3 = α 3 − β 1 , β 1 ( α 3 , β 1 ) β 1 − β 2 , β 2 ( α 3 , β 2 ) β 2 = α 3 + 2 1 β 1 + 3 1 β 2 = [ 3 − 1 3 1 3 1 1 ] T
⇒
\Rightarrow
⇒
α
1
=
β
1
\alpha_1 = \beta_1
α 1 = β 1
α
2
=
1
2
β
1
+
β
2
\alpha_2 = \frac{1}{2}\beta_1 + \beta_2
α 2 = 2 1 β 1 + β 2
α
3
=
−
1
2
β
1
−
1
3
β
2
+
β
3
\alpha_3 = -\frac{1}{2}\beta_1 - \frac{1}{3}\beta_2 + \beta_3
α 3 = − 2 1 β 1 − 3 1 β 2 + β 3
再将其单位化，得到一组标准正交向量组
由
β
1
=
∣
∣
β
1
∣
∣
η
1
β
2
=
∣
∣
β
2
∣
∣
η
2
β
3
=
∣
∣
β
3
∣
∣
η
3
\left.\begin{array}{cccc}\beta_1 = ||\beta_1|| \eta_1 \\ \beta_2 = ||\beta_2|| \eta_2 \\ \beta_3 = ||\beta_3|| \eta_3\end{array} \right.
β 1 = ∣ ∣ β 1 ∣ ∣ η 1 β 2 = ∣ ∣ β 2 ∣ ∣ η 2 β 3 = ∣ ∣ β 3 ∣ ∣ η 3
α
1
=
β
1
α
2
=
1
2
β
1
+
β
2
α
3
=
−
1
2
β
1
−
1
3
β
2
+
β
3
\left. \begin{array}{cccc} \alpha_1 = \beta_1 \\ \alpha_2 = \frac{1}{2}\beta_1 + \beta_2 \\ \alpha_3 = -\frac{1}{2}\beta_1 - \frac{1}{3}\beta_2 + \beta_3 \end{array} \right.
α 1 = β 1 α 2 = 2 1 β 1 + β 2 α 3 = − 2 1 β 1 − 3 1 β 2 + β 3
⇒
α
1
=
2
η
1
α
2
=
6
2
η
2
+
2
2
η
1
α
3
=
2
3
3
η
3
−
6
6
η
2
−
2
2
η
1
⇒
R
=
[
2
2
2
−
2
2
0
6
2
6
6
0
0
2
3
3
]
\Rightarrow \left.\begin{array}{cccc}\alpha_1 = \sqrt{2} \eta_1 \\\alpha_2 = \frac{\sqrt{6}}{2} \eta_2 + \frac{\sqrt{2}}{2} \eta_1 \\\alpha_3 = \frac{2\sqrt{3}}{3} \eta_3 - \frac{\sqrt{6}}{6} \eta_2 - \frac{\sqrt{2}}{2} \eta_1 \\ \end{array} \right. \Rightarrow R = \left [ \begin{array}{cccc} \sqrt{2} & \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \\ 0 & \frac{\sqrt{6}}{2} & \frac{\sqrt{6}}{6} \\ 0 & 0 & \frac{2\sqrt{3}}{3} \end{array} \right ]
⇒ α 1 = 2
η 1 α 2 = 2 6
η 2 + 2 2
η 1 α 3 = 3 2 3
η 3 − 6 6
η 2 − 2 2
η 1 ⇒ R = ⎣ ⎢ ⎡ 2
0 0 2 2
2 6
0 − 2 2
6 6
3 2 3
⎦ ⎥ ⎤ 故得到A矩阵的QR分解如下:
A
=
(
α
1
α
2
α
3
)
=
Q
R
=
[
2
2
6
6
−
3
6
2
2
−
6
6
3
6
0
6
3
3
6
0
0
3
2
]
[
2
2
2
−
2
2
0
6
2
6
6
0
6
3
3
6
0
0
2
3
3
]
A = (\alpha_1 \ \ \alpha_2 \ \ \alpha_3) = QR =\left [\begin{array}{cccc}\frac{\sqrt{2}}{2} & \frac{\sqrt{6}}{6} & -\frac{\sqrt{3}}{6} \\\frac{\sqrt{2}}{2} & -\frac{\sqrt{6}}{6} &\frac{\sqrt{3}}{6} \\ 0 & \frac{\sqrt{6}}{3} & \frac{\sqrt{3}}{6} \\ 0 & 0 & \frac{\sqrt{3}}{2} \end{array} \right ] \left [ \begin{array}{cccc} \sqrt{2} & \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \\ 0 & \frac{\sqrt{6}}{2} & \frac{\sqrt{6}}{6} \\ 0 & \frac{\sqrt{6}}{3} & \frac{\sqrt{3}}{6} \\ 0 & 0 & \frac{2\sqrt{3}}{3} \end{array} \right ]
A = ( α 1 α 2 α 3 ) = Q R = ⎣ ⎢ ⎢ ⎢ ⎡ 2 2
2 2
0 0 6 6
− 6 6
3 6
0 − 6 3
6 3
6 3
2 3
⎦ ⎥ ⎥ ⎥ ⎤ ⎣ ⎢ ⎢ ⎢ ⎡ 2
0 0 0 2 2
2 6
3 6
0 − 2 2
6 6
6 3
3 2 3
⎦ ⎥ ⎥ ⎥ ⎤ 简写为：
A
4
×
3
=
Q
R
=
Q
4
×
3
R
3
×
3
A_{4×3} = QR = Q_{4×3} R_{3×3}
A 4 × 3 = Q R = Q 4 × 3 R 3 × 3
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