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 } &amp; \frac{ (2-1)^2 }{ \sigma^2 } &amp; 
           \frac{ (3-1)^2 }{ \sigma^2 } &amp; \dots  &amp; \frac{ (n-1)^2 }{ \sigma^2 } \\
           \frac{ (1-2)^2 }{ \sigma^2 } &amp; \frac{ (2-2)^2 }{ \sigma^2 } &amp; 
           \frac{ (3-2)^2 }{ \sigma^2 } &amp; \dots  &amp; \frac{ (n-2)^2 }{ \sigma^2 } \\
            \vdots &amp; \vdots &amp; \vdots &amp; \ddots &amp; \vdots \\
           \frac{ (1-N)^2 }{ \sigma^2 } &amp; \frac{ (2-N)^2 }{ \sigma^2 } &amp; 
           \frac{ (3-N)^2 }{ \sigma^2 } &amp; \dots  &amp; \frac{ (N-N)^2 }{ \sigma^2 } \\
        \end{bmatrix}
\end{align}

Any help would be much appreciated! 

Thank you very much, 

Katie
Define the variables
$$\eqalign{
\lambda &amp;= \log(\alpha) &amp;\implies \alpha = e^\lambda \cr
X &amp;= -\tfrac{1}{2}(A+\lambda I) \cr
}$$
Then
$$\eqalign{
W^{-1} &amp;= \Big(I-e^{-2X}\Big) &amp;= 2e^{-X}\sinh(X) \cr
W &amp;= \Big(I-e^{-2X}\Big)^{-1} &amp;= \tfrac{1}{2}e^{X}{\,\rm csch}(X) \cr
}$$
You can take out $\frac{1}{\sigma^2}$. Notice also that your matrix is symmetric, and of the form: Find rank of the matrix $a_{ij}=(i-j)^2$, $i,j=1,\dots, n$
You can make it upper triangular by a few transformations (after which it's easy to transform to identity).