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The matrix exponential is defined as

$$ e^A = \sum_{n=0}^\infty \frac{1}{n!} A^n $$

However I would like to solve something similar:

$$ B = \sum_{n=1}^\infty \frac{1}{n!} A^{n-1} $$ (NOTE: starting index and the power)

I can transform that into

$$ B = A^{-1}\sum_{n=0}^\infty \frac{1}{n!} A^{n} -I = A^{-1} \left(e^A - I \right) $$

  • That works. But when $A^{-1}$ does not exist ($A$ is not invertible) it does not work.

  • Is it possible to make it work for general $A$? Has anyone thoughts on that? Thanks.

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    A generic matrix is invertible and the right and side exists. So if you should be unlucky and have a $B$ with $\det B =0$, then you simply add $\epsilon I$ to $B$ calculate the right hand side and then let $\epsilon \to 0$. – 2017-01-15

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Basically, your question asks how to apply the function $(e^x-1)/x$ to a matrix. One way to do it is using the integral representation $$\frac{e^x-1}{x}=\int_0^1 e^{tx}\ dt.$$ Thus we can define $B$ to be $$ B:=\int_0^1 e^{tA}\ dt. $$ Since you already know what the matrix exponential is, this expression makes sense: the integral of a matrix is computed by taking the integral of each of its entries.

Note also that this coincides with the power series definition you have written, since $$\int_0^1 \frac{(tA)^{n}}{n!}\ dt=\frac{A^{n}}{(n+1)!}.$$

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    That seems good, but I don't quite understand how I can get a feasible solution for $B$ from that... – 2017-01-15
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    The feasible solution for $B$ is $\int_0^1 e^{tA}\ dt$. For any matrix, this is a well-defined quantity that is straightforward to compute. – 2017-01-15