# How to expand this matrix exponential: $(I-AB^{-1})^{-n}$?

I want to kow how to expand this matrix exponential: $(I-AB^{-1})^{-n}$.

Also, can anyone name a few applications where you may want to consider this expansion?

Sorry if the question is too naive, please consider me as from some other planet coming to earth to study maths. I would be quick to learn if you guys can name me a few resources where I should look, or just keywords to search.

Thank you!

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What in this green earth do you mean by $A/B$? $B^{-1}A$, or $AB^{-1}$? – J. M. Nov 4 '11 at 1:48
AB^-1. Sorry I am a beginner on linear algebra, and LaTex. – SoManyProb_for_a_broken_heart. Nov 4 '11 at 1:49
The binomial expansion applies to matrices as well. – J. M. Nov 4 '11 at 1:51
Thank you J.M for your help on editing it. – SoManyProb_for_a_broken_heart. Nov 4 '11 at 2:11
See this link about autovectors maybe it's what you are looking for. For example, if you solve $X\lambda=\lambda v$, where $X=I-AB^{-1}$, then it's easy find the power $n$ or $-n$ of $X$. Because how explain this link if X is diagonalizable, then we can write $X$ as $YZY^{-1}$ too and it's easy see ${(YZY^{-1})}^n=YZ^nY^{-1}$. – GarouDan Nov 4 '11 at 2:33

An "easy" way, without Eigenvalues and eigenvectors:

${(I-C)}^{-n}={((I-C)^n)}^{(-1)}=(\displaystyle\sum_{i=0}^{n}\binom{n}{i}(-C)^n)^{(-1)}$

Taking $C=AB^{-1}$

$C^n=AB^{-1}AB^{-1}AB^{-1}\ldots AB^{-1}$ with $n$ produts of $AB^{-1}$

Then a way to expand this could be:

$(I-AB^{-1})^{-n}=(\sum_{i=0}^{n}\binom{n}{i}(-\prod_{j=1}^{n}{AB^{-1}})^{-1}$.

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