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矩阵指数

此示例说明 19 种矩阵指数计算方法中的 3 种。

有关矩阵指数计算的背景信息,请参阅:

Moler, Cleve, and Charles Van Loan.“Nineteen Dubious Ways to Compute the Exponential of a Matrix, Twenty-Five Years Later.”SIAM Review 45, no. 1 (January 2003):3–49. https://doi.org/10.1137/S00361445024180.

首先创建矩阵 A

A = [0 1 2; 0.5 0 1; 2 1 0]
A = 3×3

         0    1.0000    2.0000
    0.5000         0    1.0000
    2.0000    1.0000         0

Asave = A;

方法 1:加权平方

expmdemo1 是以下著作中算法 11.3.1 的实现:

Golub, Gene H. and Charles Van Loan.Matrix Computations, 3rd edition.Baltimore, MD:Johns Hopkins University Press, 1996.

% Scale A by power of 2 so that its norm is < 1/2 .
[f,e] = log2(norm(A,'inf'));
s = max(0,e+1);
A = A/2^s;

% Pade approximation for exp(A)
X = A;
c = 1/2;
E = eye(size(A)) + c*A;
D = eye(size(A)) - c*A;
q = 6;
p = 1;
for k = 2:q
   c = c * (q-k+1) / (k*(2*q-k+1));
   X = A*X;
   cX = c*X;
   E = E + cX;
   if p
     D = D + cX;
   else
     D = D - cX;
   end
   p = ~p;
end
E = D\E;

% Undo scaling by repeated squaring
for k = 1:s
    E = E*E;
end

E1 = E
E1 = 3×3

    5.3091    4.0012    5.5778
    2.8088    2.8845    3.1930
    5.1737    4.0012    5.7132

方法 2:泰勒级数

expmdemo2 使用矩阵指数的经典定义,表示为幂级数

eA=k=01k!Ak.

A0 是与 A 具有相同维度的单位矩阵。作为一种实用的数值方法,如果 norm(A) 太大,此方法将很慢且不准确。

A = Asave;

% Taylor series for exp(A)
E = zeros(size(A));
F = eye(size(A));
k = 1;

while norm(E+F-E,1) > 0
   E = E + F;
   F = A*F/k;
   k = k+1;
end

E2 = E
E2 = 3×3

    5.3091    4.0012    5.5778
    2.8088    2.8845    3.1930
    5.1737    4.0012    5.7132

方法 3:特征值和特征向量

expmdemo3 假定矩阵包含一组完整的特征向量 V,使得 A=VDV-1。矩阵指数可以通过对特征值的对角矩阵求幂来计算:

eA=VeDV-1.

作为一种实际的数值方法,准确性由特征向量矩阵的条件确定。

A = Asave;

[V,D] = eig(A);
E = V * diag(exp(diag(D))) / V;

E3 = E
E3 = 3×3

    5.3091    4.0012    5.5778
    2.8088    2.8845    3.1930
    5.1737    4.0012    5.7132

比较结果

对于此示例中的矩阵,所有三种方法都同样有效。

E = expm(Asave);
err1 = E - E1
err1 = 3×3
10-14 ×

    0.3553    0.1776    0.0888
    0.0888    0.1332   -0.0444
         0         0   -0.2665

err2 = E - E2
err2 = 3×3
10-14 ×

         0         0   -0.1776
   -0.0444         0   -0.0888
    0.1776         0    0.0888

err3 = E - E3
err3 = 3×3
10-13 ×

   -0.0711   -0.0444   -0.0799
   -0.0622   -0.0488   -0.0933
   -0.0711   -0.0533   -0.1066

泰勒级数失败

对于某些矩阵,泰勒级数中的项在变为零之前变得非常大。因此,expmdemo2 失败。

A = [-147 72; -192 93];
E1 = expmdemo1(A)
E1 = 2×2

   -0.0996    0.0747
   -0.1991    0.1494

E2 = expmdemo2(A)
E2 = 2×2
106 ×

   -1.1985   -0.5908
   -2.7438   -2.0442

E3 = expmdemo3(A)
E3 = 2×2

   -0.0996    0.0747
   -0.1991    0.1494

特征值和特征向量失败

以下是不包含一组完整的特征向量的矩阵。因此,expmdemo3 失败。

A = [-1 1; 0 -1];
E1 = expmdemo1(A)
E1 = 2×2

    0.3679    0.3679
         0    0.3679

E2 = expmdemo2(A)
E2 = 2×2

    0.3679    0.3679
         0    0.3679

E3 = expmdemo3(A)
E3 = 2×2

    0.3679         0
         0    0.3679

另请参阅