<s>[INST] Inverse of identity minus matrix exponential I am trying to analytically find the inverse of a matrix given by:
\begin{align}
   W = \left( I - \alpha e^A \right)^{-1}, 
\end{align}
where $I$ is the identity matrix of appropriate size, $e^A$ denotes matrix exponential of $A$ and
\begin{align}
    A = \begin{bmatrix}
           \frac{ (1-1)^2 }{ \sigma^2 } & \frac{ (2-1)^2 }{ \sigma^2 } & 
           \frac{ (3-1)^2 }{ \sigma^2 } & \dots  & \frac{ (n-1)^2 }{ \sigma^2 } \\
           \frac{ (1-2)^2 }{ \sigma^2 } & \frac{ (2-2)^2 }{ \sigma^2 } & 
           \frac{ (3-2)^2 }{ \sigma^2 } & \dots  & \frac{ (n-2)^2 }{ \sigma^2 } \\
            \vdots & \vdots & \vdots & \ddots & \vdots \\
           \frac{ (1-N)^2 }{ \sigma^2 } & \frac{ (2-N)^2 }{ \sigma^2 } & 
           \frac{ (3-N)^2 }{ \sigma^2 } & \dots  & \frac{ (N-N)^2 }{ \sigma^2 } \\
        \end{bmatrix}
\end{align}
Any help would be much appreciated! 
Thank you very much, 
Katie
 [/INST] Define the variables
$$\eqalign{
\lambda &= \log(\alpha) &\implies \alpha = e^\lambda \cr
X &= -\tfrac{1}{2}(A+\lambda I) \cr
}$$
Then
$$\eqalign{
W^{-1} &= \Big(I-e^{-2X}\Big) &= 2e^{-X}\sinh(X) \cr
W &= \Big(I-e^{-2X}\Big)^{-1} &= \tfrac{1}{2}e^{X}{\,\rm csch}(X) \cr
}$$
</s>