Sherman-Morrison公式及其应用

摘要：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。应用循环三对角线性方程组的求解本篇博客将详细讲述公式在循环三对角线性方程组的求解中的应用。

  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：
$$(A+uv^{T})^{-1}=A^{-1}-{A^{-1}uv^{T}A^{-1} over 1+v^{T}A^{-1}u}.$$
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。

$$ {displaystyle {egin{aligned} XY&=(A+uv^{T})left(A^{-1}-{A^{-1}uv^{T}A^{-1} over 1+v^{T}A^{-1}u} ight) &=AA^{-1}+uv^{T}A^{-1}-{AA^{-1}uv^{T}A^{-1}+uv^{T}A^{-1}uv^{T}A^{-1} over 1+v^{T}A^{-1}u} &=I+uv^{T}A^{-1}-{uv^{T}A^{-1}+uv^{T}A^{-1}uv^{T}A^{-1} over 1+v^{T}A^{-1}u} &=I+uv^{T}A^{-1}-{u(1+v^{T}A^{-1}u)v^{T}A^{-1} over 1+v^{T}A^{-1}u} &=I+uv^{T}A^{-1}-uv^{T}A^{-1} &=Iend{aligned}}} $$

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
$$(A+uv^{T})A^{-1}u=u+u(v^{T}A^{-1}u)=(1+v^{T}A^{-1}u)u=0.$$
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$1+v^{T}A^{-1}u eq 0.$

  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：
$$(I+uv^{T})^{-1}=I-{uv^{T} over 1+v^{T}u}.$$
  再令$v=u$,则$1+u^{T}u > 0$, 因此，$I+uu^{T}$可逆，且
$$(I+uu^{T})^{-1}=I-{uu^{T} over 1+u^{T}u}.$$

  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）。

  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：
$$Ay=b,qquad Az=u$$.
然后令$x=y-frac{v^{T}y}{1+v^{T}z}z$,该解即为原方程的解，验证如下：

$$ {displaystyle {egin{aligned} (A+uv^{T})x&=(A+uv^{T})(y-frac{v^{T}y}{1+v^{T}z}z) &=Ay+uv^{T}y-frac{v^{T}y}{1+v^{T}z}Az-frac{v^{T}y}{1+v^{T}z}uv^{T}z &=b+uv^{T}y-frac{v^{T}yu+v^{T}yuv^{T}z}{1+v^{T}z} &=b+uv^{T}y-frac{(1+v^{T}z)v^{T}yu}{1+v^{T}z} &=b+uv^{T}y-uv^{T}y &=bend{aligned}}} $$

  这样将原方程$ (A+uv^{T})x=b$分成两个方程，可以在一定程度上简化原方程。接下来，我们将介绍循环三对角线性方程组的求解。
  所谓循环三对角线性方程组，指的是系数矩阵为如下形式：

$$ A=egin{bmatrix} b_1&c_1&0&cdots&0&a_1 a_2&b_2&c_2&0&vdots&0 0&ddots&ddots&ddots&0&vdots vdots&vdots&a_{n-2}&b_{n-2}&c_{n-2}&0 0&cdots&cdots&a_{n-1}&b_{n-1}&c_{n-1} c_n&0&cdots&0&a_n&b_nend{bmatrix} $$

循环三对角线性方程组可写成$Ax=d$,其中$d=(d_{1},d_2,...,d_{n})^{T}.$
  对于此方程的求解，我们令$u=(gamma, 0,0,...,c_{n})^{T}, v=(1,0,0,...,frac{a_1}{gamma})^{T}$, 且$A=A^{"}+uv^{T}$,其中$A^{"}$如下：

$$ A^{"}=egin{bmatrix} b_1-gamma&c_1&0&cdots&0&0 a_2&b_2&c_2&0&vdots&0 0&ddots&ddots&ddots&0&vdots vdots&vdots&a_{n-2}&b_{n-2}&c_{n-2}&0 0&cdots&cdots&a_{n-1}&b_{n-1}&c_{n-1} 0&0&cdots&0&a_n&b_n-frac{a_1c_n}{gamma}end{bmatrix} $$

$A^{"}$为三对角矩阵。根据以上的理论知识，我们只需要求解以下两个方程
$$A^{"}y=d,qquad A^{"}z=u，$$
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。

  我们要解的循环三对角线性方程组如下：

$$ {egin{bmatrix} 4&1&{0}&{0}&{2} {1}&{4}&{1}&{0}&{0} {0}&{1}&{4}&{1}&{0} {0}&{0}&{1}&{4}&{1} {3}&{0}&{0}&{1}&{4} end{bmatrix}} {egin{bmatrix}{x_{1}}{x_{2}}{x_{3}}{x_{4}} {x_{5}}end{bmatrix}}={egin{bmatrix}{76 668}end{bmatrix}} $$

  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]

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