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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

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Define the variables $$\eqalign{ \lambda &= \log(\alpha) &\implies \alpha = e^\lambda \cr X &= -\tfrac{1}{2}(A+\lambda I) \cr }$$ Then $$\eqalign{ W^{-1} &= \Big(I-e^{-2X}\Big) &= 2e^{-X}\sinh(X) \cr W &= \Big(I-e^{-2X}\Big)^{-1} &= \tfrac{1}{2}e^{X}{\,\rm csch}(X) \cr }$$

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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).

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