文章目录
矩阵的因子分解1-奇异值分解
题型：对 A ∈ C m × n A \in \mathbb{C}^{m \times n} A∈Cm×n 进行奇异值分解 A = U Σ V H A = U \Sigma V^H A=UΣVH
题目中为简化计算，都是取 C m × n \mathbb{C}^{m\times n} Cm×n的特殊情形： R m × n \mathbb{R}^{m\times n} Rm×n，如下也是按照 R m × n \mathbb{R}^{m\times n} Rm×n 来展开的
求法归纳
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求 A H A A^HA AHA 的特征值和特征向量 α 1 , α 2 , … {\alpha_1,\alpha_2,\dots} α1,α2,…
单位化特征向量得到 V V V -
用非零特征值求 ： A A A 的奇异值 和 将奇异值按从大到小的顺序排列并形成对角矩阵 Σ \Sigma Σ
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求 A A H AA^H AAH 的特征值和特征向量 β 1 , β 2 , … {\beta_1,\beta_2,\dots} β1,β2,…
单位化特征向量得到 U U U -
A = U ( Σ 0 0 0 ) V H A =U \begin{pmatrix} \Sigma&0\\ 0&0 \end{pmatrix} V^H A=U(Σ000)VH
注：
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A H A A^HA AHA 和 A A H AA^H AAH 均为对称矩阵，特征值均非负，且二者的非零特征值相同；不同特征值对应的特征向量正交
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计算量大但推荐，不用通过 Gram-Schmidt 正交化方法补充单位向量
例1. 对矩阵 A = ( 0 1 − 1 0 0 2 1 0 ) A = \begin{pmatrix} 0 & 1 \\ -1 & 0 \\ 0 & 2 \\ 1 & 0 \end{pmatrix} A= 0−1011020 进行奇异值分解
1. 计算 A H A A^H A AHA 的特征值和特征向量
A H A = ( 0 − 1 0 1 1 0 2 0 ) ( 0 1 − 1 0 0 2 1 0 ) = ( 2 0 0 5 ) A^H A = \begin{pmatrix} 0 & -1 & 0 & 1 \\ 1 & 0 & 2 & 0 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ -1 & 0 \\ 0 & 2 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 2 & 0 \\ 0 & 5 \end{pmatrix} AHA=(01−100210) 0−1011020 =(2005)
特征值为：
λ 1 = 5 , λ 2 = 2 \lambda_1 = 5, \quad \lambda_2 = 2 λ1=5,λ2=2
对应的特征向量为：
α 1 = ( 0 1 ) , α 2 = ( 1 0 ) \alpha_1 = \begin{pmatrix} 0 \\ 1 \end{pmatrix}, \quad \alpha_2 = \begin{pmatrix} 1 \\ 0 \end{pmatrix} α1=(01),α2=(10)
将特征向量单位化：
v 1 = α 1 ∥ α 1 ∥ = ( 0 1 ) , v 2 = α 2 ∥ α 2 ∥ = ( 1 0 ) v_1 = \frac{\alpha_1}{\|\alpha_1\|} = \begin{pmatrix} 0 \\ 1 \end{pmatrix}, \quad v_2 = \frac{\alpha_2}{\|\alpha_2\|} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} v1=∥α1∥α1=(01),v2=∥α2∥α2=(10)
V = ( 0 1 1 0 ) V = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} V=(0110)
2. 将奇异值按从大到小排列，并构造对角矩阵 Σ \Sigma Σ
奇异值是特征值的平方根
σ 1 = 5 , σ 2 = 2 \sigma_1 = \sqrt{5}, \quad \sigma_2 = \sqrt{2} σ1=5,σ2=2
则
Σ = ( 5 0 0 2 ) \Sigma = \begin{pmatrix} \sqrt{5} & 0 \\ 0 & \sqrt{2} \end{pmatrix} Σ=(5002)
3. 计算 A A H A A^H AAH 的特征值和特征向量
A A H = ( 0 1 − 1 0 0 2 1 0 ) ( 0 − 1 0 1 1 0 2 0 ) = ( 1 0 2 0 0 1 0 − 1 2 0 4 0 0 − 1 0 1 ) A A^H = \begin{pmatrix} 0 & 1 \\ -1 & 0 \\ 0 & 2 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0 & -1 & 0 & 1 \\ 1 & 0 & 2 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 2 & 0 \\ 0 & 1 & 0 & -1 \\ 2 & 0 & 4 & 0 \\ 0 & -1 & 0 & 1 \end{pmatrix} AAH= 0−1011020 (01−100210)= 1020010−120400−101
特征值为：
λ 1 = 5 , λ 2 = 2 , λ 3 = 0 , λ 4 = 0 \lambda_1 = 5, \quad \lambda_2 = 2, \quad \lambda_3 = 0, \quad \lambda_4 = 0 λ1=5,λ2=2,λ3=0,λ4=0
对应的特征向量为：
β 1 = ( 1 0 2 0 ) , β 2 = ( 0 − 1 0 1 ) , β 3 = ( 0 1 0 1 ) , β 4 = ( − 2 0 1 0 ) \beta_1 = \begin{pmatrix} 1 \\ 0 \\ 2 \\ 0 \end{pmatrix}, \quad \beta_2 = \begin{pmatrix} 0 \\ -1 \\ 0 \\ 1 \end{pmatrix}, \quad \beta_3 = \begin{pmatrix} 0 \\ 1 \\ 0 \\ 1 \end{pmatrix}, \quad \beta_4 = \begin{pmatrix} -2 \\ 0 \\ 1 \\ 0 \end{pmatrix} β1= 1020 ,β2= 0−101 ,β3= 0101 ,β4= −2010
将特征向量单位化：
u 1 = β 2 ∥ β 2 ∥ = ( 1 5 0 2 5 0 ) , u 2 = β 1 ∥ β 1 ∥ = ( 0 − 1 2 0 1 2 ) , u 3 = β 3 ∥ β 3 ∥ = ( 0 1 2 0 1 2 ) , u 4 = β 4 ∥ β 4 ∥ = ( − 2 5 0 1 5 0 ) u_1 = \frac{\beta_2}{\|\beta_2\|} = \begin{pmatrix} \frac{1}{\sqrt{5}} \\ 0 \\ \frac{2}{\sqrt{5}} \\ 0 \end{pmatrix}, \quad u_2 = \frac{\beta_1}{\|\beta_1\|} = \begin{pmatrix} 0 \\ -\frac{1}{\sqrt{2}} \\ 0 \\ \frac{1}{\sqrt{2}} \end{pmatrix}, \quad \\ u_3 = \frac{\beta_3}{\|\beta_3\|} = \begin{pmatrix} 0 \\ \frac{1}{\sqrt{2}} \\ 0 \\ \frac{1}{\sqrt{2}} \end{pmatrix}, \quad u_4 = \frac{\beta_4}{\|\beta_4\|} = \begin{pmatrix} -\frac{2}{\sqrt{5}} \\ 0 \\ \frac{1}{\sqrt{5}} \\ 0 \end{pmatrix} u1=∥β2∥β2= 510520 ,u2=∥β1∥β1= 0−21021 ,u3=∥β3∥β3= 021021 ,u4=∥β4∥β4= −520510
U = ( 1 5 0 0 − 2 5 0 − 1 2 1 2 0 2 5 0 0 1 5 0 1 2 1 2 0 ) U = \begin{pmatrix} \frac{1}{\sqrt{5}} & 0 & 0 & -\frac{2}{\sqrt{5}} \\ 0 & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ \frac{2}{\sqrt{5}} &0 & 0 & \frac{1}{\sqrt{5}} \\ 0 &\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \end{pmatrix} U= 5105200−21021021021−520510
4. 构造分解结果
根据奇异值分解公式：
A = U ( Σ 0 0 0 ) V H A = U \begin{pmatrix} \Sigma & 0 \\ 0 & 0 \end{pmatrix} V^H A=U(Σ000)VH
其中：
Σ = ( 5 0 0 2 ) , ( Σ 0 0 0 ) = ( 5 0 0 2 0 0 0 0 ) \Sigma = \begin{pmatrix} \sqrt{5} & 0 \\ 0 & \sqrt{2} \end{pmatrix}, \quad \begin{pmatrix} \Sigma & 0 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} \sqrt{5} & 0 \\ 0 & \sqrt{2} \\ 0 & 0 \\ 0 & 0 \end{pmatrix} Σ=(5002),(Σ000)= 50000200
因此，分解结果为：
A = U ( 5 0 0 2 0 0 0 0 ) V H A = U \begin{pmatrix} \sqrt{5} & 0\\ 0 & \sqrt{2} \\ 0 & 0 \\ 0 & 0 \end{pmatrix} V^H A=U 50000200 VH
