Trace of Matrix Exponential closed form expression I would like to compute the $\mathbf{tr}(e^A)$ where $A$ is some square matrix with entries that have only values of either $0$ or $1$, and $\mathbf{tr}$ is the trace operator.  Are there closed form expressions for this or some way to compute it easily without having to compute the actual matrix exponential for every element in A?
I have gotten this far:
$$ \mathbf{tr}(e^A) = \sum_{k=0}^\infty \frac{1}{k!} \mathbf{tr}(A^k) $$ due to the property that
$$ \mathbf{tr}(A+B) = \mathbf{tr}(A) + \mathbf{tr}(B) $$
but not sure how to go further and perhaps get a simpler expression?  I would like some expression form of this because I am going to use this as a constraint in an optimization problem... otherwise I wouldn't worry about the computational cost of computing all the elements of $e^A$.

There is a general form for this expression, but it relies on the fact that every square matrix has a Jordan canonical form.  That is, for every $A$ there is invertible matrix $Q$ and some block diagonal $J$ with 1's on the super-diagonal such that $A = Q^{-1}JQ$.  Notice now that $A^k = (Q^{-1}JQ)^k = Q^{-1}J^kQ$ and so your sum for the matrix exponential reduces to
$$
\exp(A) \;\; =\;\; Q^{-1} \left ( \sum_{k=0}^\infty \frac{1}{k!}J^k \right ) Q \;\; =\;\; Q^{-1} \exp(J) Q.
$$
Now, given the fact that $Tr(ABC) = Tr(BCA) = Tr(CAB)$ we have that $Tr(\exp(A)) = Tr(\exp(J))$. Because you're only computing the trace, you only need to worry about the diagonal values in the Jordan block matrix $J$.  Therefore 
$$
Tr(\exp(A)) \;\; =\;\; \sum_{\lambda \in \Lambda(A)} m_\lambda e^\lambda
$$
where $\Lambda(A)$ is the set of distinct generalized eigenvalues of $A$, and $m_\lambda$ is the multiplicity corresponding to $\lambda \in \Lambda(A)$.
