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I am trying to find the derivative of a matrix with respect to the inverse of the same matrix. The matrix in question is a non singular symmetric matrix. Any thoughts?

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The starting point is the von Neumann series $$(I-A)^{-1}=I+A+A^2+\dots$$ which implies that the matrix function $A\mapsto (I-A)$ has identity as its derivative at $A=0$.

The general case of $B\mapsto B^{-1}$ differentiated at $B_0$ can be reduced to the above by considering $B=(I-A)B_0$, hence $B^{-1}=B_0^{-1}(I-A)^{-1}$. The linear term here is $B_0^{-1}A$, and since $A=-(B-B_0)B_0^{-1}$, the derivative can be written as $\Delta B\mapsto -B_0^{-1}(\Delta B) B_0^{-1}$.

Which, not incidentally, simplifies to $(1/x)'=-1/x^2$ when everything commutes.

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