Past exam of the short options course of the University of Oxford / 2014 / 4

The Fréchet derivative at a point $a$ is the unique linear map $D{f}_a:{\mathbb{R}}^n\to {\mathbb{R}}^m$ such thatThe error divided by $\Vert h\Vert$ tends to zero as $h\to 0$. In standard coordinates this linear map is represented by the Jacobian matrix, but the derivative itself is the map acting on increments, not merely a list of directional rates.

Factor $A+H=A(I+A^{-1}H)$. The determinant expansion near the identity isthe linear terms in the permutation formula come from choosing one diagonal increment and all remaining identity entries. ThereforeThe required Fréchet derivative is the linear functionalThe quadratic remainder proves full differentiability, rather than only the existence of each directional derivative. This is the derivative of the determinant.

The regular level set theorem says that a $C^1$ map $f:{\mathbb{R}}^n\to {\mathbb{R}}^m$ with surjective derivative at every point of $f^{-1}(0)$ has a zero set that is an embedded $C^1$ submanifold of dimension $n-m$. At each point $a$ of that zero set,For a smooth map the submanifold is smooth. The full-rank hypothesis is essential for this kernel description when using a particular defining map.

Apply the regular level set theorem to $F(A)=\det A-1$ on the $n^2$-dimensional vector space of real matrices. At a point with determinant one, part (b) gives $D{F}_A(H)=\mathrm{tr}(A^{-1}H)$. Choosing $H=A/n$ gives $D{F}_A(H)=1$, so this derivative is surjective.

Hence the special linear group $\mathrm{SL}(n,\mathbb{R})$ is a smooth submanifold of dimension $n^2-1$. At the identity,This is the special linear Lie algebra. When $n=1$, the group is a single point and the tangent space is zero, consistent with the same formula.

Use the six-dimensional vector space of real symmetric $3\times 3$ matrices as the target, rather than all nine matrix entries independently. DefineIts derivative isThe given $J$ is symmetric with $J^2=I$. Every $A\in G$ is invertible, since taking determinants of $A^{T}JA=J$ gives $(\det A{)}^2=1$. For any symmetric target $S$, chooseThen $A^{T}JH=\frac{1}{2}JJS=S/2$, and its transpose is also $S/2$. Thus $D{F}_A(H)=S$ and the derivative is surjective at every point of $G$.

The regular level set theorem now givesIt also proves smooth embeddedness in the entire ambient vector space of matrices. The target symmetry is what gives the correct codimension.

The tangent space at the identity is the kernel of a linear map from part (i):This says precisely that $JH$ is skew-symmetric. Since $J^{-1}=J$, multiplication by $J$ identifies this tangent space with the three-dimensional space of skew-symmetric $3\times 3$ matrices. In entries it isSubstitution verifies the equation directly and displays its three independent parameters.