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Let $A \in \mathbb{R}^{n \times n}$ a non-singular matrix, $I$ be the identity matrix and $c$ and $f \in \mathbb{R}^{n \times 1}$ two vectors. Define the function $H(s) = c^T (sI-A)^{-1} f$. I am wondering how I can calculate the derivative $\dfrac{d H(s)}{d s}$. In fact I have no idea how to do this because we have the inverse of $(sI-A)$. Can anyone help me?

many thanks, Koen

uniquesolution
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Koen
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  • $H(s)=-c^TA^{-1}(I-sA^{-1})^{-1}f$ and for many values of $s$ $(I-sA^{-1})^{-1}=I+sA^{-1}+s^2A^{-2}+s^3A^{-4}+...$. Take derivatives first in those places to see what it looks like. – OR. Jun 16 '17 at 09:56
  • I think I know how to do it. From https://math.stackexchange.com/questions/1471825/derivative-of-the-inverse-of-a-matrix we got that $( (sI- A)^{-1} )'= - (sI- A)^{-1}(sI- A)'(sI- A)^{-1} = - (sI- A)^{-1}I(sI- A)^{-1} $ and because $c$ and $f$ are independent of $s$ we get $dH(s)/ds = -c^t (sI- A)^{-1}(sI- A)^{-1} f$. Is that correct? – Koen Jun 16 '17 at 10:12
  • Yes, that is correct. You can simplify the expression by collecting powers $$\frac{dH}{ds}=-c^T(sI-A)^{-2}f$$ – lynn Jun 16 '17 at 19:04

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