The matrix function $\frac{e^x - 1}{x}$ for non-invertible matrices

$\def\mymatrix#1{ \left[\begin{array}{c}#1\end{array}\right] }$ Nick Higham has a recommendation for calculating such matrix functions.

Write a simple function, applicable to non-singular matrices.
For example, here is some Julia code

function phi(X)
   return (exp(X)-I) / X
end

For every singular matrix there is a nearby non-singular matrix. So evaluate the simple function by perturbing the singular matrix with a random matrix $(X+R)\,$ where $\,\|R\|\approx\sqrt{\varepsilon}\;$ where $\varepsilon$ is the machine epsilon. The function itself will be accurate to $\sqrt{\varepsilon}\approx10^{-8}$.

So for singular matrices, call the function like so

ε,X = 1e-8, rand(4,2)*rand(2,4)  # create a singular matrix
P = phi(X + ε*randn(4,4))        # evaluate the function
norm(I+X*P - exp(X))             # validate the function

If this simple method isn't good enough, there are rational (Pade) approximations for the exponential and phi-functions. Here is a fairly comprehensive paper.

Another suggestion is to evaluate the function using quad precision and a perturbation $\|R\|\approx10^{-16}.\;$ Then the function will be accurate to double precision (for singular matrix arguments).

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Update

Another approach is to calculate the exponential of a block-triangular matrix $$\eqalign{ \exp\left(\mymatrix{X&I\\0&0}\right) &= \mymatrix{e^X&\frac{e^X-e^0}{X}\\0&e^0} = \mymatrix{e^X&\frac{e^X-I}{X}\\0&I} }$$ then extract the upper-right block. I'm not sure of the numerical accuracy of this method, but it should be as accurate as the underlying $\,\exp\,$ function. Note that this result holds for any analytic function, not just the exponential, i.e. $$\eqalign{ f\left(\mymatrix{X&I\\0&0}\right) &= \mymatrix{f(X)&\frac{f(X)-f(0)}{X}\\0&f(0)} }$$