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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 elementwise 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} so that I'm trying to find \begin{align} W = \begin{bmatrix} 1 - \alpha \exp{ \left( \frac{ 0^2 }{ \sigma^2 } \right) } & - \alpha \exp{ \left( \frac{ 1^2 }{ \sigma^2 } \right) } & - \alpha \exp{ \left( \frac{ 2^2 }{ \sigma^2 } \right) } & \dots & - \alpha \exp{ \left( \frac{ (n-1)^2 }{ \sigma^2 } \right) } \\ - \alpha \exp{ \left( \frac{ 1^2 }{ \sigma^2 } \right) } & 1 - \alpha \exp{ \left( \frac{ 0^2 }{ \sigma^2 } \right) } & - \alpha \exp{ \left( \frac{ 1^2 }{ \sigma^2 } \right) } & \dots & - \alpha \exp{ \left( \frac{ (n-2)^2 }{ \sigma^2 } \right) } \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ - \alpha \exp{ \left( \frac{ (1-N)^2 }{ \sigma^2 } \right) } & - \alpha \exp{ \left( \frac{ (2-N)^2 }{ \sigma^2 } \right) } & - \alpha \exp{ \left( \frac{ (3-N)^2 }{ \sigma^2 } \right) } & \dots & 1 - \alpha \exp{ \left( \frac{ (N-N)^2 }{ \sigma^2 } \right) } \\ \end{bmatrix}^{-1}. \end{align}

Any help would be much appreciated!

Thank you very much,

Katie

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  • $\begingroup$ Are you trying to find $W^{-1}$? That would be the last matrix because $(A^{-1})^{-1}$... $\endgroup$ – Viktor Glombik Dec 20 '18 at 21:08
  • $\begingroup$ Your matrix $W$ has only zeros on the diagonal and is symmetric, have you tried diagonalisation? $\endgroup$ – Viktor Glombik Dec 20 '18 at 21:09
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    $\begingroup$ How do you define $e^A$ ? $\endgroup$ – Damien Dec 20 '18 at 21:21
  • $\begingroup$ Thank you everyone for your interest in my question. I'm trying to find W, as opposed to $W^{-1}$, as that would simply be $I - \alpha e^A$, which would be wonderfully easy! W unfortunately does not have 0's on the diagonal, but rather $1-\alpha$. I have looked at diagonalisation but haven't made much progress. Regarding the definition of $e^A$, I wondered if there were multiple definitions of a matrix exponential, which is why I included the expanded matrix at the end. I've simply taken the exponential of each matrix element. This may be wrong? Thanks again, Katie. $\endgroup$ – Katie Dec 20 '18 at 22:18
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    $\begingroup$ The classical definition of $e^A$ corresponds to en.wikipedia.org/wiki/Matrix_exponential $\endgroup$ – Damien Dec 21 '18 at 14:38

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