15-01-DerivativeOfInverseMatrix.tex

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\pmtitle{derivative of inverse matrix}
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\pmtype{Theorem}
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\pmclassification{msc}{15-01}

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\begin{document}
\begin{thm}
Suppose $A$ is a square matrix depending on a real parameter $t$
taking values in an open set $I\subseteq \R$. Further, suppose all
component functions in $A$ are differentiable, and $A(t)$ is invertible
for all $t$. Then, in $I$, we have
$$
   \frac{dA^{-1}}{dt}=-A^{-1} \frac{dA}{dt} A^{-1}, 
$$
where $\frac{d}{dt}$ is the derivative.
\end{thm}

\begin{proof}
Suppose $a_{ij}(t)$ are the component functions for $A$, 
and $a^{jk}(t)$ are component functions for $A^{-1}(t)$. Then 
for each $t$ we have 
$$
   \sum_{j=1}^n a_{ij}(t) a^{jk}(t)=\delta_i^k
$$
where $n$ is the order of $A$, and 
$\delta_i^k$ is the Kronecker delta symbol. Hence
$$
 \sum_{j=1}^n \frac{da_{ij}}{dt} a^{jk} + a_{ij}\frac{da^{jk}}{dt} = 0,
$$
that is, 
$$
  \frac{dA}{dt} A^{-1} = - A \frac{dA^{-1}}{dt}
$$
from which the claim follows. 
\end{proof}
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