Let $H$ be a finite-dimensional Hilbert space (so there is a canonical isomorphism $H\cong H^*$). For a Hilbert space $H$ define $B(H)$ to be the space of linear operators on $H$; we have $B(H)\cong H\otimes H^*$.
Suppose $\Lambda:B(H)\to B(H)$ is a CPTP (completely positive trace preserving) map. In particular it is linear, so we can view it as an element of $B(H)\otimes B(H)^*\cong (H\otimes H^*)\otimes (H\otimes H^*)^*$. There is an isomorphism $F$ from this space to $[(H\otimes H^*)\otimes (H\otimes H^*)]^*$, the space of bilinear (or quadratic) forms on $H\otimes H^*$. The map is defined by $F(|E_{kl}\rangle\langle E_{ij}|)=\langle E_{ki}|\otimes \langle E_{lj}|$. (I think--if another map makes more sense then replace this.) Note here we are viewing matrices as vectors in the space $B(H)$.
Question: What is the condition on $F(\Lambda)$ that is equivalent to $\Lambda$ being CPTP (and why might it intuitively hold), and is there a way to prove this directly? Is this a useful perspective?
By the Kraus representation, the CPTP maps $\Lambda$ are exactly those in the form $\Lambda(\rho)=\sum_i A_i\rho A_i^{\dagger}$. The above then maps $\Lambda$ to the quadratic form $\sum_i\langle A_i|\otimes \langle A_i|$, which is positive semidefinite quadratic form. In particular, for example, this tells us that we need at most $n^2$ matrices $A_i$. So $F(\Lambda)$ being positive semidefinite is a necessary condition for $\Lambda$ to be CPTP; what is a necessary and sufficient condition? Furthermore, this was proved using Kraus representation, but could it be proved directly (and hence the Krause representation would be a corollary)?