Past exam of the options II course of the University of Oxford / 2006 / F (Multivariable Calculus)

The Fréchet derivative at $a$ is the linear map $D{f}_a:{\mathbb{R}}^n\to {\mathbb{R}}^k$ satisfyingIt is unique. When both derivatives exist, the chain rule is $D(g\circ f{)}_a=D{g}_{f(a)}\circ D{f}_a$.

For the square map,The remainder has norm at most $\Vert \Delta {\Vert }^2$ in a submultiplicative matrix norm, hence is $o(\Vert \Delta \Vert )$. ThereforeMatrix multiplication is not commutative, so the two terms cannot generally be combined as $2A\Delta$.

For small $\Vert \Delta \Vert$, the matrix $A+\Delta =A(I+A^{-1}\Delta )$ remains invertible and its inverse is locally bounded, for example by the convergent Neumann series when $\Vert A^{-1}\Delta \Vert <1$. Applying the resolvent identity twice givesThe last term is $O(\Vert \Delta {\Vert }^2)$ by that local bound. Hence the inverse map is differentiable, withThis verifies the differentiability and its remainder, rather than only formally differentiating the identity $A{A}^{-1}=I$.

First simplify the composite: $(S\circ \mathcal{I}\circ S)(A)=((A^2{)}^{-1}{)}^2=A^{-4}$. Applying the product rule to four factors of $A^{-1}$, and the preceding inverse derivative to one factor at a time, givesThe four terms have left and right exponents $(1,4),(2,3),(3,2),(4,1)$. The same expression follows by composing $D{S}_{A^{-2}}$, $D{\mathcal{I}}_{A^2}$ and $D{S}_A$ in that order. Only when $\Delta$ commutes with $A$ can it be simplified to $-4A^{-5}\Delta$.