# Exponential of singular matrix times inverse of singular matrix

I need to find a way to solve the following problem:

Suppose I have a singular matrix A. I need to find a solution to

$$\frac{e^{hA}-I}{A}$$

However, note that the inverse of A is infinity. Can anyone help with this problem? Maybe l'hospital rule could help? I am not really sure. Thanks in advance.

• Welcome to MSE. If you say that you want to find a solution, you should write down an equation. Moreover, let us know what you have tried already and where your interest in this question comes from. – user526015 Dec 4 '18 at 14:12

What you wrote doesn't make sense, however the function $$f(z) = \dfrac{e^{hz}-1}{z}$$ has a removable singularity at $$0$$, and after removing it you have an entire function, so you can define $$f(A)$$ by the functional calculus. For example, you could compute it using the power series $$f(A) = \sum_{n=1}^\infty \frac{h^n}{n!} A^{n-1}$$ Or if $$A$$ is diagonalizable: $$A = S^{-1} D S$$ where $$D$$ is diagonal, then $$f(A) = S^{-1} f(D) S$$ where $$f(D)$$ is diagonal with $$(f(D))_{jj} = f(D_{jj})$$.
• There are many ways to compute it besides the series. I noted one in the case where $A$ is diagonalizable. More generally you could use Jordan canonical form. – Robert Israel Dec 4 '18 at 16:53