Calculate expm(m*A*t/n) from expm(A*t/n)

Is it possible to efficiently calculate M = expm(m*A*T/n) given N = expm(A*T/n)? Here m, n, and T are scalars, m <= n.

The idea is to calculcate N only once, and thereafter use it to quickly approximate M for any m <= n.

A formula to find expm(A*T/m) given expm(A*T) is also interesting, as is a formula to find expm(A*T) as some combination of fractions of expm(A*T ).

I see there is now a function expmv(A,b,tvals), but its internal factors do not seem to be available.

Answer 1

[Torsten]

Your question is how to get expm(c*A) given A for a scalar c.

Diagonalize A such that A*V = V*D.

It follows that (c*A)*V = V*(c*D) and - if V is invertible - c*A = V*(c*D)*inv(V).

Thus expm(c*A) = V*diag(exp(diag(c*D)))*inv(V) .

Summarizing: Saving V and inv(V) (if V is invertible) gives you expm(c*A) from a diagonalization of the matrix A as A = V*D*inv(V).

A = [1 2; 3 4];
[V,D] = eig(A)

V = 2×2

-0.8246 -0.4160 0.5658 -0.9094

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D = 2×2

-0.3723 0 0 5.3723

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expm(A)

ans = 2×2

51.9690 74.7366 112.1048 164.0738

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V*diag(exp(diag(D)))*inv(V)

ans = 2×2

51.9690 74.7366 112.1048 164.0738

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c = 2;
expm(c*A)

ans = 2×2

1.0e+04 * 1.1079 1.6146 2.4219 3.5299

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V*diag(exp(diag(c*D)))*inv(V)

ans = 2×2

1.0e+04 * 1.1079 1.6146 2.4219 3.5299

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[marcel hendrix]

Yes, this approach is exactly what I need (my V's are [made] invertible). In addition to quick computation it provides all the eigenvalues, opening the possibility to do modelreduction in a very straightforward way (I need the code for circuit simulation). Also, the eigenvectors should provide valuable insights for discrete control.

Thanks a lot!