# 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;

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

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

${\mathit{A}}^{0}$ is the identity matrix with the same dimensions as $\mathit{A}$. 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 $\mathit{V}$ such that $\mathit{A}=\mathrm{VD}{\mathit{V}}^{-1}$. 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
10-14 ×

-0.7105   -0.5329   -0.7105
-0.6661   -0.5773   -0.8882
-0.7105   -0.7105   -0.9770

### 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