Arbitrary function of a matrix - power-series representation

[bjnartowt]

1. The problem statement, all variables and given/known data

I am trying to evaluate exp(i*f(A)), where A is a matrix whose eigenvalues are known and real.

2. Relevant equations
You can expand functions of a matrix in a power-series. I think that's the way to get started on this problem. I foresee the exponential of a power-series expansion of f(A) suddenly becoming a product of exponentials of a matrix, which, themselves, are power-series. It sounds ugly, but perhaps some pleasant collapse will happen.

3. The attempt at a solution
Okay, let's try finding power-series expansion of f(A):
f(A) = \sum\limits_{n = 0}^{n \to \infty } {\frac{{{f^{(n)}}(0)}}{{n!}}{A^n}} = {\rm{oh dear}}...{\rm{unaware of derivatives of f at 0}}{\rm{.}}

Hmmm...maybe my idea above is not so great.

I think I am thwarted by the arbitrariness of the function "f(A)". Prior experience suggests arbitrariness is nothing to be afraid of...just an opportunity for higher generality...which is neato.

When you, dear reader, peruse this cry for help, do any suggestions come to your mind for a starting point to evaluating exp(i*f(A))? I just can't seem to bring "A" outside of "f" in order to take advantage of the fact that we know its eigenvalues.
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution

[lanedance]

do you know anything about f?

otherwise I would just expand the exponential as
e^{i f(A)} = \sum_{k=0} \frac{(if(A))^k}{k!}

[lanedance]

not sure this helps but playing with stuff

start by making the assumption A is diagonalisable - is it...? may lead to somthing interesting...

then there exists some S such that
S^{-1}AS = \lambda
where lambda is diagonal, with main entries the eigenvalues of A

then
f(A)
= f(S \lambda S^{-1})
= \sum_{n = 0} {\frac{f^{(n)}(0)}{n!}{(S \lambda S^{-1})^n}}
= \sum_{n = 0} {\frac{f^{(n)}(0)}{n!}{(S \lambda^n S^{-1})}}
= S \left( \sum_{n = 0} {\frac{f^{(n)}(0)}{n!}{( \lambda^n )}} \right)S^{-1}
= Sf(\lambda)S^{-1}

[lanedance]

then applying same idea in the exponential
e^{i f(A)} = e^{i Sf(\lambda)S^{-1} }
= \sum_{k=0} \frac{(i S f(\lambda)S^{-1} )^k}{k!}
= S \left( \sum_{k=0} \frac{(if(\lambda))^k}{k!} \right) S^{-1}
= S e^{i f(\lambda)}S^{-1}

and i'm not sure whether thats helpful...

[lanedance]

then you could also note for a diagonal matrix lambda
f(\lambda) = \begin{pmatrix}
f(\lambda_1) & 0 & ... \\
0 & f(\lambda_2) & ... \\
... & ... & ... \end{pmatrix}

which should be apparent from post #3