08/27/2007, 12:43 PM

bo198214 Wrote:Gottfried Wrote:But it seems as if A must have a diagonal of 1.bo198214 Wrote:Did you anyway already realize that instead ofNo, but it looks very good. I'll give it a deeper look, thanks for the hint!

you can directly use the binomial series for computation?

So probably you have to convert A into a Jordan normal form first as you must do with the logarithm (or how do you compute the matrix logarithm?)

Second try.

You say

Let

if A is diagonalizable.

Also, since

it is also

Then the above means, for example t = 3

Let's look at the parenthese only as the diagonal matrix E. The entries e_k can directly be computed:

by binomial-theorem

and

So the rhs is

and the lhs is

and the both sides are equal.

Here was no assumption needed concerning the actual eigenvalues of A, except the assumption, that it is diagonalizable at all. It is perhaps important to note, that symbolically we may write formulae with equal eigenvalues even if a numerical solver would not produce a solution because of the mentioned indeterminacy of the eigenvectors. But this does not affect the analytical description.

Well, here I assumed A is diagonalizable (with no assumptions about the actual value of its eigenvalues). If it is not, then at least it is possible to derive the Jordan-canonical form, and in A = Q*J*Q^-1 we get J non-diagonal, but in its Jordan-canonical form. All what was said about the diagonal form is still valid in principle, only that powers and functions on J have not the simple form as in the true diagonal case. But still it is a simpler computation than with the original matrix, especially if infinite series are assumed.

Gottfried

Gottfried Helms, Kassel