= Solution
For every invertible matrix $A$,
$$
(r_C\circ f\circ l_C)(A)
=r_C((CA)^{-1})
=A^{-1}C^{-1}C
=A^{-1}=f(A).
$$
The maps $l_C$ and $r_C$ are linear. Differentiate this identity at $A=I$ and use the <chain rule> and part (a):
$$
r_C\bigl(Df(C)(Ch)\bigr)=-h.
$$
Thus $Df(C)(Ch)C=-h$. Given an arbitrary increment $k$, set $h=C^{-1}k$ to obtain the <derivative of matrix inversion>
$$
\boxed{Df(C)(k)=-C^{-1}kC^{-1}}.
$$
This also proves differentiability at every $C\in V$.
Back to article page
