Solve matrix exponential power series

Question

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.

Answer 1

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 $$\int_0^1 \frac{(tA)^{n}}{n!}\ dt=\frac{A^{n}}{(n+1)!}.$$

Answer 2

Your matrix B is in literature called phi-function $\varphi_1$. The power series expansion of the general form is \begin{align} \varphi_k (A) := \sum_{n=0}^\infty \frac{A^n}{(n+k)!} \end{align} If an approximate solution is sufficient a Pade approximation $\varphi_1(A) \approx (1-A/2)^{-1}$ would be a possible solution.