共軛梯度法的推導

在數值線性代數中，共軛梯度法是一種求解對稱正定線性方程組

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

的迭代方法。共軛梯度法可以從不同的角度推導而得，包括作為求解最優化問題的共軛方向法的特例，以及作為求解特徵值問題的Arnoldi/Lanczos迭代的變種。

本條目記述這些推導方法中的重要步驟。

從Arnoldi/Lanczos迭代推導

共軛梯度法可以看作Arnoldi/Lanczos迭代應用於求解線性方程組時的一個變種。

一般Arnoldi方法

Arnoldi迭代從一個向量${\boldsymbol {r}}_{0}$開始，通過定義${\boldsymbol {v}}_{i}={\boldsymbol {w}}_{i}/\lVert {\boldsymbol {w}}_{i}\rVert _{2}$，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子空間

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

的一組標準正交基$\{{\boldsymbol {v}}_{1},{\boldsymbol {v}}_{2},{\boldsymbol {v}}_{3},\ldots \}$。

換言之，對於$i>1$，${\boldsymbol {v}}_{i}$由將${\boldsymbol {Av}}_{i-1}$相對於$\{{\boldsymbol {v}}_{1},{\boldsymbol {v}}_{2},\ldots ,{\boldsymbol {v}}_{i-1}\}$進行Gram-Schmidt正交化然後歸一化得到。

寫成矩陣形式，迭代過程可以表示為

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

其中

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

當應用於求解線性方程組時，Arnoldi迭代對應於初始解${\boldsymbol {x}}_{0}$的殘量${\boldsymbol {r}}_{0}={\boldsymbol {b}}-{\boldsymbol {Ax}}_{0}$開始迭代，在每一步迭代之後計算${\boldsymbol {y}}_{i}={\boldsymbol {H}}_{i}^{-1}(\lVert {\boldsymbol {r}}_{0}\rVert _{2}{\boldsymbol {e}}_{1})$和新的近似解${\boldsymbol {x}}_{i}={\boldsymbol {x}}_{0}+{\boldsymbol {V}}_{i}{\boldsymbol {y}}_{i}$.

直接Lanzcos方法

在餘下的討論中，我們假定${\boldsymbol {A}}$是對稱正定矩陣。由於${\boldsymbol {A}}$的對稱性, 上Hessenberg矩陣${\boldsymbol {H}}_{i}={\boldsymbol {V}}_{i}^{\mathrm {T} }{\boldsymbol {AV}}_{i}$變成對陣三對角矩陣。於是它可以被更明確地表示為

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

這使得迭代中的${\boldsymbol {v}}_{i}$有一個簡短的三項遞推式。Arnoldi迭代從而簡化為Lanczos迭代。

由於${\boldsymbol {A}}$對稱正定，${\boldsymbol {H}}_{i}$同樣也對稱正定。因此，${\boldsymbol {H}}_{i}$可以通過不選主元的LU分解分解為

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

其中$c_{i}$和$d_{i}$有簡單的遞推式：

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

改寫${\boldsymbol {x}}_{i}={\boldsymbol {x}}_{0}+{\boldsymbol {V}}_{i}{\boldsymbol {y}}_{i}$為

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

其中

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

此時需要觀察到

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

實際上，${\boldsymbol {p}}_{i}$和$\zeta _{i}$同樣有簡短的遞推式：

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

通過這個形式，我們得到${\boldsymbol {x}}_{i}$的一個簡單的遞推式：

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

以上的遞推關係立即導出比共軛梯度法稍微更複雜的直接Lanczos方法。

從正交性和共軛性導出共軛梯度法

如果我們允許縮放${\boldsymbol {p}}_{i}$並通過常數因子補償縮放的係數，我們可能可以得到以下形式的更簡單的遞推式：

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

作為簡化的前提，我們現在推導${\boldsymbol {r}}_{i}$的正交性和${\boldsymbol {p}}_{i}$的共軛性，即對於$i\neq j$,

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

各個殘量之間滿足正交性的原因是${\boldsymbol {r}}_{i}$實際上可由${\boldsymbol {v}}_{i+1}$乘上一個係數而得。這是因為對於$i=0$，${\boldsymbol {r}}_{0}=\lVert {\boldsymbol {r}}_{0}\rVert _{2}{\boldsymbol {v}}_{1}$，對於$i>0$，

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

要得到${\boldsymbol {p}}_{i}$的共軛性，只需證明${\boldsymbol {P}}_{i}^{\mathrm {T} }{\boldsymbol {AP}}_{i}$是對角矩陣：

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

是對稱的下三角矩陣，因而必然是對角矩陣。

現在我們可以單純由${\boldsymbol {r}}_{i}$的正交性和${\boldsymbol {p}}_{i}$的共軛性推導相對於縮放後的${\boldsymbol {p}}_{i}$的常數因子$\alpha _{i}$和$\beta _{i}$。

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

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

類似地，由於${\boldsymbol {p}}_{i}$的共軛性，必然有${\boldsymbol {p}}_{i+1}^{\mathrm {T} }{\boldsymbol {Ap}}_{i}=({\boldsymbol {r}}_{i+1}+\beta _{i}{\boldsymbol {p}}_{i})^{\mathrm {T} }{\boldsymbol {Ap}}_{i}=0$。於是

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

推導至此完成。

參考文獻

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]. （原始內容 (PDF)存檔於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.