Matrix Exponentials
This example shows three of the 19 ways to compute the exponential of a matrix.
For background on the computation of matrix exponentials, see:
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.
Start by creating a matrix 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;
Method 1: Scaling and Squaring
expmdemo1
is an implementation of algorithm 11.3.1 in the book:
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
Method 2: Taylor Series
expmdemo2
uses the classic definition for the matrix exponential given by the power series
is the identity matrix with the same dimensions as . As a practical numerical method, this approach is slow and inaccurate if norm(A)
is too large.
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
Method 3: Eigenvalues and Eigenvectors
expmdemo3
assumes that the matrix has a full set of eigenvectors such that . The matrix exponential can be calculated by exponentiating the diagonal matrix of eigenvalues:
As a practical numerical method, the accuracy is determined by the condition of the eigenvector matrix.
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
Compare Results
For the matrix in this example, all three methods work equally well.
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
Taylor Series Failure
For some matrices the terms in the Taylor series become very large before they go to zero. Consequently, expmdemo2
fails.
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
Eigenvalues and Eigenvectors Failure
Here is a matrix that does not have a full set of eigenvectors. Consequently, expmdemo3
fails.
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