共轭梯度法的推导

在数值线性代数中，共轭梯度法是一种求解对称正定线性方程组

${\boldsymbol {Ax}}={\boldsymbol {b}}$

的迭代方法。共轭梯度法可以从不同的角度推导而得，包括作为求解最优化问题的共轭方向法的特例，以及作为求解特征值问题的Arnoldi/Lanczos迭代的变种。

本条目记述这些推导方法中的重要步骤。

从共轭方向法推导

共轭梯度法可以看作是应用于二次函数最小化的共轭方向法的特例

$f({\boldsymbol {x}})={\boldsymbol {x}}^{\mathrm {T} }{\boldsymbol {A}}{\boldsymbol {x}}-2{\boldsymbol {b}}^{\mathrm {T} }{\boldsymbol {x}}{\text{.}}$

共轭方向法

在共轭方向上求最小化：

$f({\boldsymbol {x}})={\boldsymbol {x}}^{\mathrm {T} }{\boldsymbol {A}}{\boldsymbol {x}}-2{\boldsymbol {b}}^{\mathrm {T} }{\boldsymbol {x}}{\text{.}}$

从最初的猜测${\boldsymbol {x}}_{0}$开始，以及相应的残差 ${\boldsymbol {r}}_{0}={\boldsymbol {b}}-{\boldsymbol {Ax}}_{0}$, 并通过公式计算迭代和残差

${\begin{aligned}\alpha _{i}&={\frac {{\boldsymbol {p}}_{i}^{\mathrm {T} }{\boldsymbol {r}}_{i}}{{\boldsymbol {p}}_{i}^{\mathrm {T} }{\boldsymbol {Ap}}_{i}}}{\text{,}}\\{\boldsymbol {x}}_{i+1}&={\boldsymbol {x}}_{i}+\alpha _{i}{\boldsymbol {p}}_{i}{\text{,}}\\{\boldsymbol {r}}_{i+1}&={\boldsymbol {r}}_{i}-\alpha _{i}{\boldsymbol {Ap}}_{i}\end{aligned}}$

其中，${\boldsymbol {p}}_{0},{\boldsymbol {p}}_{1},{\boldsymbol {p}}_{2},\ldots$ 为一系列相互共轭的方向，例如：

${\boldsymbol {p}}_{i}^{\mathrm {T} }{\boldsymbol {Ap}}_{j}=0$，对于任何$i\neq j$

由于共轭方向法没有给出用于选择方向的公式，因此该方法具有误差 ${\boldsymbol {p}}_{0},{\boldsymbol {p}}_{1},{\boldsymbol {p}}_{2},\ldots$

而共轭方向法也有多种方法分支，包括共轭梯度法和高斯消元法。

从Arnoldi/Lanczos迭代推导

共轭梯度法可以看作Arnoldi/Lanczos迭代应用于求解线性方程组时的一个变种。

一般Arnoldi方法

Arnoldi迭代从一个向量${\boldsymbol {r}}_{0}$开始，通过定义${\boldsymbol {v}}_{i}={\boldsymbol {w}}_{i}/\lVert {\boldsymbol {w}}_{i}\rVert _{2}$，其中

${\boldsymbol {w}}_{i}={\begin{cases}{\boldsymbol {r}}_{0}&{\text{if }}i=1{\text{,}}\\{\boldsymbol {Av}}_{i-1}-\sum _{j=1}^{i-1}({\boldsymbol {v}}_{j}^{\mathrm {T} }{\boldsymbol {Av}}_{i-1}){\boldsymbol {v}}_{j}&{\text{if }}i>1{\text{,}}\end{cases}}$

逐步建立Krylov子空间

${\mathcal {K}}({\boldsymbol {A}},{\boldsymbol {r}}_{0})=\{{\boldsymbol {r}}_{0},{\boldsymbol {Ar}}_{0},{\boldsymbol {A}}^{2}{\boldsymbol {r}}_{0},\ldots \}$

的一组标准正交基$\{{\boldsymbol {v}}_{1},{\boldsymbol {v}}_{2},{\boldsymbol {v}}_{3},\ldots \}$。

换言之，对于$i>1$，${\boldsymbol {v}}_{i}$由将${\boldsymbol {Av}}_{i-1}$相对于$\{{\boldsymbol {v}}_{1},{\boldsymbol {v}}_{2},\ldots ,{\boldsymbol {v}}_{i-1}\}$进行Gram-Schmidt正交化然后归一化得到。

写成矩阵形式，迭代过程可以表示为

${\boldsymbol {AV}}_{i}={\boldsymbol {V}}_{i+1}{\boldsymbol {\tilde {H}}}_{i}{\text{,}}$

其中

${\begin{aligned}{\boldsymbol {V}}_{i}&={\begin{bmatrix}{\boldsymbol {v}}_{1}&{\boldsymbol {v}}_{2}&\cdots &{\boldsymbol {v}}_{i}\end{bmatrix}}{\text{,}}\\{\boldsymbol {\tilde {H}}}_{i}&={\begin{bmatrix}h_{11}&h_{12}&h_{13}&\cdots &h_{1,i}\\h_{21}&h_{22}&h_{23}&\cdots &h_{2,i}\\&h_{32}&h_{33}&\cdots &h_{3,i}\\&&\ddots &\ddots &\vdots \\&&&h_{i,i-1}&h_{i,i}\\&&&&h_{i+1,i}\end{bmatrix}}={\begin{bmatrix}{\boldsymbol {H}}_{i}\\h_{i+1,i}{\boldsymbol {e}}_{i}^{\mathrm {T} }\end{bmatrix}}{\text{,}}\\h_{ji}&={\begin{cases}{\boldsymbol {v}}_{j}^{\mathrm {T} }{\boldsymbol {Av}}_{i}&{\text{if }}j\leq i{\text{,}}\\\lVert {\boldsymbol {w}}_{i+1}\rVert _{2}&{\text{if }}j=i+1{\text{,}}\\0&{\text{if }}j>i+1{\text{.}}\end{cases}}\end{aligned}}$

当应用于求解线性方程组时，Arnoldi迭代对应于初始解${\boldsymbol {x}}_{0}$的残量${\boldsymbol {r}}_{0}={\boldsymbol {b}}-{\boldsymbol {Ax}}_{0}$开始迭代，在每一步迭代之后计算${\boldsymbol {y}}_{i}={\boldsymbol {H}}_{i}^{-1}(\lVert {\boldsymbol {r}}_{0}\rVert _{2}{\boldsymbol {e}}_{1})$和新的近似解${\boldsymbol {x}}_{i}={\boldsymbol {x}}_{0}+{\boldsymbol {V}}_{i}{\boldsymbol {y}}_{i}$.

直接Lanzcos方法

在余下的讨论中，我们假定${\boldsymbol {A}}$是对称正定矩阵。由于${\boldsymbol {A}}$的对称性, 上Hessenberg矩阵${\boldsymbol {H}}_{i}={\boldsymbol {V}}_{i}^{\mathrm {T} }{\boldsymbol {AV}}_{i}$变成对阵三对角矩阵。于是它可以被更明确地表示为

${\boldsymbol {H}}_{i}={\begin{bmatrix}a_{1}&b_{2}\\b_{2}&a_{2}&b_{3}\\&\ddots &\ddots &\ddots \\&&b_{i-1}&a_{i-1}&b_{i}\\&&&b_{i}&a_{i}\end{bmatrix}}{\text{.}}$

这使得迭代中的${\boldsymbol {v}}_{i}$有一个简短的三项递推式。Arnoldi迭代从而简化为Lanczos迭代。

由于${\boldsymbol {A}}$对称正定，${\boldsymbol {H}}_{i}$同样也对称正定。因此，${\boldsymbol {H}}_{i}$可以通过不选主元的LU分解分解为

${\boldsymbol {H}}_{i}={\boldsymbol {L}}_{i}{\boldsymbol {U}}_{i}={\begin{bmatrix}1\\c_{2}&1\\&\ddots &\ddots \\&&c_{i-1}&1\\&&&c_{i}&1\end{bmatrix}}{\begin{bmatrix}d_{1}&b_{2}\\&d_{2}&b_{3}\\&&\ddots &\ddots \\&&&d_{i-1}&b_{i}\\&&&&d_{i}\end{bmatrix}}{\text{,}}$

其中$c_{i}$和$d_{i}$有简单的递推式：

${\begin{aligned}c_{i}&=b_{i}/d_{i-1}{\text{,}}\\d_{i}&={\begin{cases}a_{1}&{\text{if }}i=1{\text{,}}\\a_{i}-c_{i}b_{i}&{\text{if }}i>1{\text{.}}\end{cases}}\end{aligned}}$

改写${\boldsymbol {x}}_{i}={\boldsymbol {x}}_{0}+{\boldsymbol {V}}_{i}{\boldsymbol {y}}_{i}$为

${\begin{aligned}{\boldsymbol {x}}_{i}&={\boldsymbol {x}}_{0}+{\boldsymbol {V}}_{i}{\boldsymbol {H}}_{i}^{-1}(\lVert {\boldsymbol {r}}_{0}\rVert _{2}{\boldsymbol {e}}_{1})\\&={\boldsymbol {x}}_{0}+{\boldsymbol {V}}_{i}{\boldsymbol {U}}_{i}^{-1}{\boldsymbol {L}}_{i}^{-1}(\lVert {\boldsymbol {r}}_{0}\rVert _{2}{\boldsymbol {e}}_{1})\\&={\boldsymbol {x}}_{0}+{\boldsymbol {P}}_{i}{\boldsymbol {z}}_{i}{\text{,}}\end{aligned}}$

其中

${\begin{aligned}{\boldsymbol {P}}_{i}&={\boldsymbol {V}}_{i}{\boldsymbol {U}}_{i}^{-1}{\text{,}}\\{\boldsymbol {z}}_{i}&={\boldsymbol {L}}_{i}^{-1}(\lVert {\boldsymbol {r}}_{0}\rVert _{2}{\boldsymbol {e}}_{1}){\text{.}}\end{aligned}}$

此时需要观察到

${\begin{aligned}{\boldsymbol {P}}_{i}&={\begin{bmatrix}{\boldsymbol {P}}_{i-1}&{\boldsymbol {p}}_{i}\end{bmatrix}}{\text{,}}\\{\boldsymbol {z}}_{i}&={\begin{bmatrix}{\boldsymbol {z}}_{i-1}\\\zeta _{i}\end{bmatrix}}{\text{.}}\end{aligned}}$

实际上，${\boldsymbol {p}}_{i}$和$\zeta _{i}$同样有简短的递推式：

${\begin{aligned}{\boldsymbol {p}}_{i}&={\frac {1}{d_{i}}}({\boldsymbol {v}}_{i}-b_{i}{\boldsymbol {p}}_{i-1}){\text{,}}\\\alpha _{i}&=-c_{i}\zeta _{i-1}{\text{.}}\end{aligned}}$

通过这个形式，我们得到${\boldsymbol {x}}_{i}$的一个简单的递推式：

${\begin{aligned}{\boldsymbol {x}}_{i}&={\boldsymbol {x}}_{0}+{\boldsymbol {P}}_{i}{\boldsymbol {z}}_{i}\\&={\boldsymbol {x}}_{0}+{\boldsymbol {P}}_{i-1}{\boldsymbol {z}}_{i-1}+\zeta _{i}{\boldsymbol {p}}_{i}\\&={\boldsymbol {x}}_{i-1}+\zeta _{i}{\boldsymbol {p}}_{i}{\text{.}}\end{aligned}}$

以上的递推关系立即导出比共轭梯度法稍微更复杂的直接Lanczos方法。

从正交性和共轭性导出共轭梯度法

如果我们允许缩放${\boldsymbol {p}}_{i}$并通过常数因子补偿缩放的系数，我们可能可以得到以下形式的更简单的递推式：

${\begin{aligned}{\boldsymbol {x}}_{i}&={\boldsymbol {x}}_{i-1}+\alpha _{i-1}{\boldsymbol {p}}_{i-1}{\text{,}}\\{\boldsymbol {r}}_{i}&={\boldsymbol {r}}_{i-1}-\alpha _{i-1}{\boldsymbol {Ap}}_{i-1}{\text{,}}\\{\boldsymbol {p}}_{i}&={\boldsymbol {r}}_{i}+\beta _{i-1}{\boldsymbol {p}}_{i-1}{\text{.}}\end{aligned}}$

作为简化的前提，我们现在推导${\boldsymbol {r}}_{i}$的正交性和${\boldsymbol {p}}_{i}$的共轭性，即对于$i\neq j$,

${\begin{aligned}{\boldsymbol {r}}_{i}^{\mathrm {T} }{\boldsymbol {r}}_{j}&=0{\text{,}}\\{\boldsymbol {p}}_{i}^{\mathrm {T} }{\boldsymbol {Ap}}_{j}&=0{\text{.}}\end{aligned}}$

各个残量之间满足正交性的原因是${\boldsymbol {r}}_{i}$实际上可由${\boldsymbol {v}}_{i+1}$乘上一个系数而得。这是因为对于$i=0$，${\boldsymbol {r}}_{0}=\lVert {\boldsymbol {r}}_{0}\rVert _{2}{\boldsymbol {v}}_{1}$，对于$i>0$，

${\begin{aligned}{\boldsymbol {r}}_{i}&={\boldsymbol {b}}-{\boldsymbol {Ax}}_{i}\\&={\boldsymbol {b}}-{\boldsymbol {A}}({\boldsymbol {x}}_{0}+{\boldsymbol {V}}_{i}{\boldsymbol {y}}_{i})\\&={\boldsymbol {r}}_{0}-{\boldsymbol {AV}}_{i}{\boldsymbol {y}}_{i}\\&={\boldsymbol {r}}_{0}-{\boldsymbol {V}}_{i+1}{\boldsymbol {\tilde {H}}}_{i}{\boldsymbol {y}}_{i}\\&={\boldsymbol {r}}_{0}-{\boldsymbol {V}}_{i}{\boldsymbol {H}}_{i}{\boldsymbol {y}}_{i}-h_{i+1,i}({\boldsymbol {e}}_{i}^{\mathrm {T} }{\boldsymbol {y}}_{i}){\boldsymbol {v}}_{i+1}\\&=\lVert {\boldsymbol {r}}_{0}\rVert _{2}{\boldsymbol {v}}_{1}-{\boldsymbol {V}}_{i}(\lVert {\boldsymbol {r}}_{0}\rVert _{2}{\boldsymbol {e}}_{1})-h_{i+1,i}({\boldsymbol {e}}_{i}^{\mathrm {T} }{\boldsymbol {y}}_{i}){\boldsymbol {v}}_{i+1}\\&=-h_{i+1,i}({\boldsymbol {e}}_{i}^{\mathrm {T} }{\boldsymbol {y}}_{i}){\boldsymbol {v}}_{i+1}{\text{.}}\end{aligned}}$

要得到${\boldsymbol {p}}_{i}$的共轭性，只需证明${\boldsymbol {P}}_{i}^{\mathrm {T} }{\boldsymbol {AP}}_{i}$是对角矩阵：

${\begin{aligned}{\boldsymbol {P}}_{i}^{\mathrm {T} }{\boldsymbol {AP}}_{i}&={\boldsymbol {U}}_{i}^{-\mathrm {T} }{\boldsymbol {V}}_{i}^{\mathrm {T} }{\boldsymbol {AV}}_{i}{\boldsymbol {U}}_{i}^{-1}\\&={\boldsymbol {U}}_{i}^{-\mathrm {T} }{\boldsymbol {H}}_{i}{\boldsymbol {U}}_{i}^{-1}\\&={\boldsymbol {U}}_{i}^{-\mathrm {T} }{\boldsymbol {L}}_{i}{\boldsymbol {U}}_{i}{\boldsymbol {U}}_{i}^{-1}\\&={\boldsymbol {U}}_{i}^{-\mathrm {T} }{\boldsymbol {L}}_{i}\end{aligned}}$

是对称的下三角矩阵，因而必然是对角矩阵。

现在我们可以单纯由${\boldsymbol {r}}_{i}$的正交性和${\boldsymbol {p}}_{i}$的共轭性推导相对于缩放后的${\boldsymbol {p}}_{i}$的常数因子$\alpha _{i}$和$\beta _{i}$。

由于${\boldsymbol {r}}_{i}$的正交性，必然有${\boldsymbol {r}}_{i+1}^{\mathrm {T} }{\boldsymbol {r}}_{i}=({\boldsymbol {r}}_{i}-\alpha _{i}{\boldsymbol {Ap}}_{i})^{\mathrm {T} }{\boldsymbol {r}}_{i}=0$。于是

${\begin{aligned}\alpha _{i}&={\frac {{\boldsymbol {r}}_{i}^{\mathrm {T} }{\boldsymbol {r}}_{i}}{{\boldsymbol {r}}_{i}^{\mathrm {T} }{\boldsymbol {Ap}}_{i}}}\\&={\frac {{\boldsymbol {r}}_{i}^{\mathrm {T} }{\boldsymbol {r}}_{i}}{({\boldsymbol {p}}_{i}-\beta _{i-1}{\boldsymbol {p}}_{i-1})^{\mathrm {T} }{\boldsymbol {Ap}}_{i}}}\\&={\frac {{\boldsymbol {r}}_{i}^{\mathrm {T} }{\boldsymbol {r}}_{i}}{{\boldsymbol {p}}_{i}^{\mathrm {T} }{\boldsymbol {Ap}}_{i}}}{\text{.}}\end{aligned}}$

类似地，由于${\boldsymbol {p}}_{i}$的共轭性，必然有${\boldsymbol {p}}_{i+1}^{\mathrm {T} }{\boldsymbol {Ap}}_{i}=({\boldsymbol {r}}_{i+1}+\beta _{i}{\boldsymbol {p}}_{i})^{\mathrm {T} }{\boldsymbol {Ap}}_{i}=0$。于是

${\begin{aligned}\beta _{i}&=-{\frac {{\boldsymbol {r}}_{i+1}^{\mathrm {T} }{\boldsymbol {Ap}}_{i}}{{\boldsymbol {p}}_{i}^{\mathrm {T} }{\boldsymbol {Ap}}_{i}}}\\&=-{\frac {{\boldsymbol {r}}_{i+1}^{\mathrm {T} }({\boldsymbol {r}}_{i}-{\boldsymbol {r}}_{i+1})}{\alpha _{i}{\boldsymbol {p}}_{i}^{\mathrm {T} }{\boldsymbol {Ap}}_{i}}}\\&={\frac {{\boldsymbol {r}}_{i+1}^{\mathrm {T} }{\boldsymbol {r}}_{i+1}}{{\boldsymbol {r}}_{i}^{\mathrm {T} }{\boldsymbol {r}}_{i}}}{\text{.}}\end{aligned}}$

推导至此完成。

参考文献

1. Hestenes, M. R.; Stiefel, E. Methods of conjugate gradients for solving linear systems (PDF). Journal of Research of the National Bureau of Standards. 1952年12月, 49 (6) [2010-06-20]. （原始内容 (PDF)存档于2010-05-05）. 
2. Saad, Y. Chapter 6: Krylov Subspace Methods, Part I. Iterative methods for sparse linear systems 2nd. SIAM. 2003. ISBN 978-0898715347.