Lemma 1

*There exists an isomorphism $\theta$ from the vector space $F \otimes E'$ onto the space $\mathscr{L}_f(E; F)$ of all finite rank continuous linear mappings from E into F, characterized by $\theta(b \otimes a^*) = ba^*$ for $a \in E, b \in F$.

By A, II, § 4, No. 2, there exists an injective linear mapping $\theta$ from $F \otimes E'$ into $\mathscr{L}(E; F)$ and only one such, which transforms $b \otimes a'$ into the linear mapping $x \mapsto ba'(x)$ for $a' \in E', b \in F$. Evidently $\theta(b \otimes a^*) = ba^*$, and the image of $\theta$ is contained in $\mathscr{L}_f(E; F)$. However, let $u \in \mathscr{L}_f(E; F)$ and let $(e_1, ..., e_n)$ be an orthonormal basis of the image of $u$ in F. Let $f_i = u^*(e_i)$ for $1 \leq i \leq n$. For every $x \in E$, we have

hence $u = \sum_{i=1}^n e_i f_i^* = \theta(\sum_{i=1}^n e_i \otimes f_i^*)$. Therefore the image of $\theta$ is equal to $\mathscr{L}_f(E; F)$.

We shall henceforth assume that $E = F$, and we set $\mathscr{L}_f(E) = \mathscr{L}_f(E; E)$. By lemma 1, there exists a unique linear form $\tau$ on $\mathscr{L}_f(E)$, such that $\tau(\theta(a \otimes a')) = a'(a)$ for $a \in E, a' \in E'$; in other words, we have

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When E is finite dimensional, we have $\mathscr{L}_f(E) = \mathscr{L}(E)$ and $\tau(u)$ is the trace of the endomorphism $u$ of E (A, II, § 4, No. 3).