Trace norm of a finite classical block operator

Each $X_i$ is normal because $X_i^{†}=X_i$, so $X_i{X}_i^{†}=X_i^{†}{X}_i$; apply [F1] to choose an orthonormal eigenbasis $(e_{i,j}{)}_j$ with $X_i{e}_{i,j}={\lambda }_{i,j}{e}_{i,j}$ and ${\lambda }_{i,j}\in \mathbb{C}$. By [F4], $⟨X_i{e}_{i,j},e_{i,j}⟩=⟨e_{i,j},X_i{e}_{i,j}⟩=\overline{⟨X_i{e}_{i,j},e_{i,j}⟩}$, so this expectation is real. Since the inner product is linear in its first argument and $⟨e_{i,j},e_{i,j}⟩=1$, the expectation equals ${\lambda }_{i,j}$; hence ${\lambda }_{i,j}\in \mathbb{R}$. By [F3], the vectors $\mathrm{\mid }i⟩\otimes e_{i,j}$ over all $i,j$ form an orthonormal basis of ${\mathbb{C}}^I\otimes H_E$. The operator $X$ is diagonal in this basis with eigenvalues exactly the finite list $({\lambda }_{i,j}{)}_{i,j}$, so it too is self-adjoint.

Since $X$ and each $X_i$ are self-adjoint, their singular values are the absolute values of their eigenvalues: in each eigenbasis, $X^{†}X$ has eigenvalues ${\lambda }_{i,j}^2$. Applying [F2] to the displayed basis and to each block gives $\mathrm{\parallel }X{\mathrm{\parallel }}_1={\sum }_{i,j}\mathrm{\mid }{\lambda }_{i,j}\mathrm{\mid }={\sum }_i\mathrm{\parallel }{X}_i{\mathrm{\parallel }}_1$.