Another answer is that $M_n$, the space of $n\times n$ complex matrices, carries an operator norm where the norm of a matrix is its norm as a linear operator from $\mathbb{C}^n$ to itself (giving $\mathbb{C}^n$ euclidean norm). For some of us, this is the most natural and useful norm on $M_n$.
With operator norm, $M_n$ is a finite dimensional Banach space, so it has a dual space, which is just $M_n$ equipped with trace norm. In infinite dimensions the trace class operators on $H$, with trace norm, form the (unique) predual of $B(H)$.