Proposition 35

Let E be a finite-dimensional vector space. Let $\phi$ be the morphism $g \mapsto \det g$ of the Lie group $\mathbf{GL}(E)$ into the Lie group $K^*$. The mapping $\mathbf{L}(\phi)$ of $\mathscr{L}(E)$ into K is the mapping $x \mapsto \operatorname{Tr} x$. The kernel $\mathbf{SL}(E)$ of $\phi$ is a Lie subgroup of $\mathbf{GL}(E)$ with Lie algebra $\mathfrak{sl}(E)$.

We choose a norm and a basis of E. The expansion of the determinant proves that

when $u$ tends to 0 in $\mathscr{L}(E)$. Hence, using Proposition 34, no. 9, for $x \in \mathscr{L}(E) = \mathbf{L}(\mathbf{GL}(E))$: It follows that $\phi$ is a submersion. Therefore, $\operatorname{Ker} \phi = \mathbf{SL}(E)$ is a Lie subgroup of $\mathbf{GL}(E)$ whose Lie algebra is $\operatorname{Ker} \mathbf{L}(\phi) = \mathfrak{sl}(E)$.

Let $E_1, \ldots, E_n$ be complete normable spaces and E their direct sum. Every $x \in \mathscr{L}(E)$ can be represented by a matrix $(x_{ij})_{1 \leq i, j \leq n}$, where $x_{ij} \in \mathscr{L}(E_i, E_j)$.