Computing the derivative of an inverse matrix

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SUMMARY

The discussion focuses on computing the derivative of the inverse of a matrix expression, specifically \(\frac{d}{dt}_{t=0}(A+tB)^{-1}\), where \(A\) is an invertible matrix and \(B\) is another matrix in Mat(n, R). The solution involves applying the formula \(\frac{d}{dt}(A+tB)^{-1}=-(A+tB)^{-1}\frac{d}{dt}((A+tB))(A+tB)^{-1}\) and evaluating at \(t=0\). Participants suggest using linearity of the derivative and consider alternative approaches such as power expansion of the matrix expression.

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Demon117
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Homework Statement


If A, B are elements of Mat(n, R) and A is invertible, compute

[tex]\frac{d}{dt}_{t=0}(A+tB)^{-1}[/tex]


The Attempt at a Solution



The derivative will be of the form

[tex]\frac{d}{dt}(A+tB)^{-1}=-(A+tB)^{-1}\frac{d}{dt}((A+tB))(A+tB)^{-1}[/tex]

but I need to evaluate this at t=0, so how do I simplify the expression on the right hand side? Since d/dt is a linear operator do I just attack each term individually, that is, take a derivative of A with respect to t plus a derivative of tB with respect to t?

This really is hard notation for me to follow.
 
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Just take the derivative in the middle and set t=0 on the outside factors. Another way to do it would be by writing A+tB as A(I+t*A^(-1)*B), taking the inverse and power expanding (I+t*A^(-1)*B)^(-1), but I seem to get the same thing.
 
Last edited:

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