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
是与 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 ,使得 。矩阵指数可以通过对特征值的对角矩阵求幂来计算:
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