Every finite-dimensional channel has a finite isometric dilation

By [F1] and [F2], a finite Kraus family exists; take any such family $(K_a{)}_{a<N}$ for $\mathrm{\Phi }$. Since $\mathrm{\Phi }$ is trace-preserving, [F3] gives ${\sum }_{a<N}{K}_a^{†}{K}_a=I_H$. Take $E={\mathbb{C}}^N$ with its standard orthonormal basis $(u_a{)}_{a<N}$ and define $Vx={\sum }_{a<N}{K}_{a}x\otimes u_a$. Each summand is linear in $x$ by [F7], so $V$ is linear. Orthogonality of $(u_a)$, the product pairing [F6], the adjoint identity [F8], and the Kraus normalization give $\mathrm{\parallel }Vx{\mathrm{\parallel }}^2={\sum }_a\mathrm{\parallel }{K}_{a}x{\mathrm{\parallel }}^2=⟨x,({\sum }_a{K}_a^{†}{K}_a)x{⟩}_H=\mathrm{\parallel }x{\mathrm{\parallel }}^2$ for every $x$; thus $V$ is a linear isometry. By [F11], $V^{†}V=I_H$.

For each $b<N$, $y\in K$, and $x\in H$, the product pairing and [F8] give $⟨Vx,y\otimes u_b⟩={\sum }_a⟨K_{a}x,y⟩⟨u_a,u_b⟩=⟨K_{b}x,y⟩=⟨x,K_b^{†}y⟩$. By uniqueness of the adjoint, $V^{†}(y\otimes u_b)=K_b^{†}y$. Therefore, on each product vector $y\otimes u_b$, $VX{V}^{†}(y\otimes u_b)={\sum }_{a<N}{K}_{a}X{K}_b^{†}y\otimes u_a$. By [F10], taking the partial trace sums exactly the blocks with $a=b$, so ${\mathrm{Tr}}_E(VX{V}^{†})={\sum }_{a<N}{K}_{a}X{K}_a^{†}=\mathrm{\Phi }(X)$ by [F4] for every $X$. This is the representation in [L1], with $\dim E=N$. If $N=0$, [F3] forces $I_H=0$, hence $H=0$; the empty sums and unique map give the same identities.