Let $H$ be a separable Hilbert space. Let $(e_i)_i$ be an orthonormal basis. For any bounded linear map $T$ we write, whenever possible
$\operatorname{tr} T := \sum_{i}^{\infty} \langle T e_i, e_i \rangle$
Now let $L(H)$ be the set of bounded operators, $N(H)$ the set of nuclear operators. We want to show
$i: L(H) \rightarrow N(H)' , \quad T \mapsto ( S \mapsto \operatorname{tr}(TS))$
is an isometric isomorphism.
By the ideal property of the nuclear operators, it is easy to see that $i$ is well-defined, linear and bounded with $|i| \leq 1$.
Recall that the operators $\langle \cdot, e_i \rangle e_j$ form a basis of $N(H)$. You easily see the operator is injective (if the induced functional is zero, it's preimage must have been zero,too) and surjective (simply build up an operator $T \in L(H)$ which induces a given functional in $N(H)$).
Question: How can I finish the proof with showing that $i$ is an isometry?