# Matrix exponential question

Let $X$ be an $n\times n$ matrix, $u,v$ are two vectors. Can we express $e^{X+uv^T}$ in terms of $e^X$ and $e^{uv^T}$? Is there a concise formula? I know there is a Lie product formula http://en.wikipedia.org/wiki/Lie_product_formula, but it depends on a limit.

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What do we know about the vectors? Nothing? I assume you know that if $[X,Y]=0$ then $e^{X+Y}=e^Xe^Y$. – Alex Youcis Nov 1 '11 at 1:25
There's also the Baker-Campbell-Hausdorff formula. You need to know some relation between $u$, $v$, and $X$ to hope to simplify further. – Michael Joyce Nov 1 '11 at 1:28
What about assuming $u=e_i$, $v=e_j$, where $e_i$ means the $i$th column of the identity matrix? Is their a concise form in this case? – Sunni Nov 1 '11 at 2:19
If $i \neq j$ then, writing $Y = u v^T$, we have $[Y,[Y,Z]]=0$ for any $Z$. This should make Baker-Campbell-Hausdorff simplify a lot. I suspect there are also simplifications available when $i = j$, but I don't see what they are right now. – David Speyer Nov 1 '11 at 15:35

I don't know if this directly helps, but computationally you could do the following:

1. Obtain a rational approximation to exp(X)
2. Use the following software: http://www.cs.cornell.edu/cv/ResearchPDF/GenSherMorr.htm which computes rank-1 updates to rational functions of matrices.

Naturally, for specialized choices of $X$, $u$, $v$, one can obtain better algorithms.

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"Obtain a rational approximation" - the Padé approximants are particularly convenient. Here is the paper discussing the generalization of the Sherman-Morrison-Woodbury formula to rational matrix functions. – J. M. Nov 5 '11 at 12:00