Trace norm of a tensor product

If either $A$ or $B$ is zero, or a domain or codomain factor is zero-dimensional, then $A\otimes B=0$ and at least one factor trace norm is zero, so the formula holds. Otherwise choose SVDs $Az={\sum }_{i=1}^r{s}_i⟨z,v_i⟩u_i$ and $Bw={\sum }_{j=1}^q{t}_j⟨w,q_j⟩p_j$ as in [F2], with all $s_i,t_j>0$.

By [F3], on elementary tensors $(A\otimes B)(z\otimes w)={\sum }_{i=1}^r{\sum }_{j=1}^q{s}_i{t}_j⟨z\otimes w,v_i\otimes q_j⟩u_i\otimes p_j\mathrm{.}$ The families $(v_i\otimes q_j{)}_{i,j}$ and $(u_i\otimes p_j{)}_{i,j}$ are orthonormal because their pairings are products of the factor pairings. Since elementary tensors span, this is a singular-value decomposition of $A\otimes B$ with positive singular values exactly $s_i{t}_j$. Thus [F1] gives $\mathrm{\parallel }A\otimes B{\mathrm{\parallel }}_1={\sum }_{i,j}{s}_i{t}_j=\left({\sum }_i{s}_i\right)\left({\sum }_j{t}_j\right)=\mathrm{\parallel }A{\mathrm{\parallel }}_1\mathrm{\parallel }B{\mathrm{\parallel }}_1\mathrm{.}$