rank one update Given a matrix $X$, we can compute its matrix exponential $e^X$. Now one entry of $X$ (say $x_{i,j}$) is changed to $b$, the updated matrix is denoted by $X'$. My problem is how to compute $e^{X'}$ from $e^X$ in a fast way? 
PS: I know that if our goal is to calculate the matrix inversion (not matrix exponential), we can use Sherman–Morrison formula to compute ${X'}^{-1}$ from $X^{-1}$ easily, but currently I have not found a way to deal with the matrix exponential. Hope you can give me a hand. Thanks!
 A: To expand on my comment, observe that
$$ vw^* A^n vw^* = \left( w^* A^n v \right) vw^* $$
so in any product of $A$'s and $vw^*$'s with more than one copy of $vw^*$, we can convert the middle part to a scalar and extract it.
Applying this and grouping like terms gives the formula
$$\begin{align}
(A + vw^*)^n = A^n + \sum_{i=0}^{n-1} A^i v w^* A^{n-1-i}
+ \sum_{i=0}^{n-2} \sum_{j=0}^{n-2-i} A^i v w^* A^j \left( w^* (A + vw^*)^{n-2-i-j} v \right)
\end{align}$$
Summing this to get $\exp(A + vw^*)$, the second term yields
$$ \sum_{n=1}^{+\infty} \frac{1}{n!} \sum_{i=0}^{n-1} A^i v w^* A^{n-1-i} 
= \sum_{i=0}^{+\infty} A^i v w^* \sum_{n=i+1}^{+\infty} \frac{1}{n!}A^{n-1-i} $$
I'm not particularly inclined to deal with truncated exponentials of $A$. :( The third term also involves truncated exponentials of $A + vw^*$.
The path forward with this idea is not clear. I only see two ideas, and both promise to be irritating:


*

*Try to come up with a simplified formula for all truncated exponentials, hoping the complicated terms cancel or otherwise collect together or have a nice recursion

*Use combinatorics to further simplify $w^* (A + vw^*)^{n-2-i-j} v$ and hope something nice falls out.

A: What you are looking for is the Baker–Campbell–Hausdorff formula.
You are trying to compute
$$e^{X+B}$$ where $B=X'-X$ has only one non-vanishing entry. If you are lucky then $X$ and $B$ commute and you can apply the standard formula
$$e^{X+B}=e^X\cdot e^B,$$ where $e^B$ is easy to compute.
In gnereal you will get higher terms which involve (iterated) commutators. It is then up to the form of $X$ and $B$ whether you can find suitable bounds and appoximate the series with a finite sum.
