Sherman-Morrison公式
Sherman-Morrison公式以 Jack Sherman 和 Winifred J. Morrison命名，在线性代数中，是求解逆矩阵的一种方法。本篇博客将介绍该公式及其应用，首先我们来看一下该公式的内容及其证明。
(Sherman-Morrison公式)假设(Ainmathbb{R}^{n imes n})为可逆矩阵，(u,vinmathbb{R}^{n})为列向量，则(A+uv^{T})可逆当且仅当(1+v^{T}A^{-1}u
eq 0), 且当(A+uv^{T})可逆时，该逆矩阵由以下公式给出：
证明：
((Leftarrow))当(1+v^{T}A^{-1}u
eq 0)时，令(X=A+uv^{T}, Y=A^{-1}-{A^{-1}uv^{T}A^{-1} over 1+v^{T}A^{-1}u}),则只需证明(XY=YX=I)即可，其中(I)为n阶单位矩阵。
同理，有(YX=I).因此，当(1+v^{T}A^{-1}u
eq 0)时，((A+uv^{T})^{-1}=A^{-1}-{A^{-1}uv^{T}A^{-1} over 1+v^{T}A^{-1}u}.)
((Rightarrow))当(u=0)时，显然有(1+v^{T}A^{-1}u=1
eq 0.)当(u
eq0)时,用反正法证明该命题成立。假设(A+uv^{T})可逆，但(1+v^{T}A^{-1}u = 0)，则有
因为(A+uv^{T})可逆，故(A^{-1})u=0,又因为(A^{-1})可逆，故(u=0),此与假设(u eq 0)矛盾。因此，当(A+uv^{T})可逆时，有(1+v^{T}A^{-1}u eq 0.)
Sherman-Morrison公式的应用
应用1：(A=I)时的Sherman-Morrison公式
在Sherman-Morrison公式中，令(A=I),则有：(I+uv^{T})可逆当且仅当(1+v^{T}u eq 0), 且当(I+uv^{T})可逆时，该逆矩阵由以下公式给出：
再令(v=u),则(1+u^{T}u > 0), 因此，(I+uu^{T})可逆，且
应用2：BFGS算法
Sherman-Morrison公式在BFGS算法中的应用，可用来求解BFGS算法中近似Hessian矩阵的逆。本篇博客并不打算给出Sherman-Morrison公式在BFGS算法中的应用，将会再写篇博客介绍BFGS算法，到时再给出该公式的应用，并会在之后补上该博客的链接（因为笔者还没写）。
应用3：循环三对角线性方程组的求解
本篇博客将详细讲述Sherman-Morrison公式在循环三对角线性方程组的求解中的应用。
首先给给出理论知识介绍部分。
对于(Ainmathbb{R}^{n imes n})为可逆矩阵，(u,vinmathbb{R}^{n})为列向量，(1+v^{T}A^{-1}u
eq 0)，需要求解方程((A+uv^{T})x=b.)对此，我们可以先求解以下两个方程：
{displaystyle
{egin{aligned}
(A+uv^{T})x&=(A+uv^{T})(y-frac{v^{T}y}{1+v^{T}z}z)
&=Ay+uv^{T}y-frac{v^{T}y}{1+v^{T}z}Az-frac{v^{T}y}{1+v^{T}z}uv^{T}z
&=b+uv^{T}y-frac{v^{T}yu+v^{T}yuv^{T}z}{1+v^{T}z}
&=b+uv^{T}y-frac{(1+v^{T}z)v^{T}yu}{1+v^{T}z}
&=b+uv^{T}y-uv^{T}y
&=bend{aligned}}}
A=egin{bmatrix}
b_1&c_1&0&cdots&0&a_1
a_2&b_2&c_2&0&vdots&0
0&ddots&ddots&ddots&0&vdots
vdots&vdots&a_{n-2}&b_{n-2}&c_{n-2}&0
0&cdots&cdots&a_{n-1}&b_{n-1}&c_{n-1}
c_n&0&cdots&0&a_n&b_nend{bmatrix}
A^{'}=egin{bmatrix}
b_1-gamma&c_1&0&cdots&0&0
a_2&b_2&c_2&0&vdots&0
0&ddots&ddots&ddots&0&vdots
vdots&vdots&a_{n-2}&b_{n-2}&c_{n-2}&0
0&cdots&cdots&a_{n-1}&b_{n-1}&c_{n-1}
0&0&cdots&0&a_n&b_n-frac{a_1c_n}{gamma}end{bmatrix}
然后，就能根据(y,z)求出(x).而以上两个方程为三对角线性方程组，可以用追赶法（或Thomas法）求解，具体算法可以参考博客：三对角线性方程组(tridiagonal systems of equations)的求解 。
综上，我们利用Sherman-Morrison公式的思想，可以将循环三对角线性方程组转化为三对角线性方程组求解。我们将会在下面给出该算法的Python语言实现。
Python实现
我们要解的循环三对角线性方程组如下：
用Python实现解该方程的Python完整代码如下：
# use Sherman-Morrison Formula and Thomas Method to solve cyclic tridiagonal linear equation
import numpy as np
# Thomas Method for soling tridiagonal linear equation Ax=d
# parameter: a,b,c,d are list-like of same length
# b: main diagonal of matrix A
# a: main diagonal below of matrix A
# c: main diagonal upper of matrix A
# d: Ax=d
# return: x(type=list), the solution of Ax=d
def TDMA(a,b,c,d):
try:
n = len(d) # order of tridiagonal square matrix
# use a,b,c to create matrix A, which is not necessary in the algorithm
A = np.array([[0]*n]*n, dtype='float64')
for i in range(n):
A[i,i] = b[i]
if i > 0:
A[i, i-1] = a[i]
if i < n-1:
A[i, i+1] = c[i]
# new list of modified coefficients
c_1 = [0]*n
d_1 = [0]*n
for i in range(n):
if not i:
c_1[i] = c[i]/b[i]
d_1[i] = d[i] / b[i]
else:
c_1[i] = c[i]/(b[i]-c_1[i-1]*a[i])
d_1[i] = (d[i]-d_1[i-1]*a[i])/(b[i]-c_1[i-1] * a[i])
# x: solution of Ax=d
x = [0]*n
for i in range(n-1, -1, -1):
if i == n-1:
x[i] = d_1[i]
else:
x[i] = d_1[i]-c_1[i]*x[i+1]
x = [round(_, 4) for _ in x]
return x
except Exception as e:
return e
# Sherman-Morrison Fomula for soling cyclic tridiagonal linear equation Ax=d
# parameter: a,b,c,d are list-like of same length
# b: main diagonal of matrix A
# a: main diagonal below of matrix A
# c: main diagonal upper of matrix A
# d: Ax=d
# return: x(type=list), the solution of Ax=d
def Cyclic_Tridiagnoal_Linear_Equation(a,b,c,d):
try:
# use a,b,c to create cyclic tridiagonal matrix A
n = len(d)
A = np.array([[0] * n] * n, dtype='float64')
for i in range(n):
A[i, i] = b[i]
if i > 0:
A[i, i - 1] = a[i]
if i < n - 1:
A[i, i + 1] = c[i]
A[0, n - 1] = a[0]
A[n - 1, 0] = c[n - 1]
gamma = 1 # gamma can be set freely
u = [gamma] + [0] * (n - 2) + [c[n - 1]]
v = [1] + [0] * (n - 2) + [a[0] / gamma]
# modify the coefficient to form A'
b[0] -= gamma
b[n - 1] -= a[0] * c[n - 1] / gamma
a[0] = 0
c[n - 1] = 0
# solve A'y=d, A'z=u by using Thomas Method
y = np.array(TDMA(a, b, c, d))
z = np.array(TDMA(a, b, c, u))
# use y and z to calculate x
# x = y-(v·y)/(1+v·z) *z
# x is the solution of Ax=d
x = y - (np.dot(np.array(v), y)) / (1 + np.dot(np.array(v), z)) * z
x = [round(_, 3) for _ in x]
return x
except Exception as e:
return e
def main():
'''
equation:
A = [[4,1,0,0,2],
[1,4,1,0,0],
[0,1,4,1,0],
[0,0,1,4,1],
[3,0,0,1,4]]
d = [7,6,6,6,8]
solution x should be [1,1,1,1,1]
'''
a = [2, 1, 1, 1, 1]
b = [4, 4, 4, 4, 4]
c = [1, 1, 1, 1, 3]
d = [7, 6, 6, 6, 8]
x = Cyclic_Tridiagnoal_Linear_Equation(a,b,c,d)
print('The solution is %s'%x)
main()
输出结果如下：
The solution is [1.0, 1.0, 1.0, 1.0, 1.0]
参考文献
- https://en.wikipedia.org/wiki/Sherman–Morrison_formula
- http://wwwmayr.in.tum.de/konferenzen/Jass09/courses/2/Soldatenko_paper.pdf
- https://scicomp.stackexchange.com/questions/10137/solving-system-of-linear-equations-with-cyclic-tridiagonal-matrix
- https://blog.csdn.net/jclian91/article/details/80251244
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