Matrix Exponentials

This example shows 3 of the 19 ways to compute the exponential of a matrix.

For background on the computation of matrix exponentials, see:

Moler, C. and C. Van Loan. "Nineteen dubious ways to compute the exponential of a matrix, twenty-five years later." SIAM Review. Vol. 45, Number 1, 2003, pp. 3-49.

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

$e^{A} = \sum_{k = 0}^{\infty} \frac{1}{k!} A^{k} .$

$A^{0}$ is the identity matrix with the same dimensions as $A$ . 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 $V$ such that $A = VD V^{- 1}$ . The matrix exponential can be calculated by exponentiating the diagonal matrix of eigenvalues:

$e^{A} = V e^{D} V^{- 1} .$

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

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