Proof that the adjoint operator of $\Phi$ is $\Phi(\sigma)=\sum_{i}A_{i}^{*}\sigma B_{i}$ in terms of Kraus operators. I am studying basics of quantum information theory and I am trying to prove the fact that the adjoint operator of $\Phi\in\mathcal{L}(\mathcal{L(\mathcal{H})},\mathcal{L(\mathcal{K})})$ is $\Phi^{*}(\sigma)=\sum_{i}A_{i}^{*}\sigma B_{i}$ where $\mathcal{H}$ and $\mathcal{K}$ are finite-dimensional Hilbert spaces. I know that $\Phi(\rho)=\sum_{i}A_{i}\rho B_{i}^{*}-(1)$ for finite set of Kraus operators $\{A_{i},B_{i}\in\mathcal{L}(\mathcal{H},\mathcal{K})\}$ for any $\rho\in\mathcal{H}$.
Now, I am trying to use the definition of adjoint operators that $\rho\cdot\Phi^{*}(\sigma)=\Phi(\rho)\cdot\sigma$ to prove it. I think I should plug in the formula $(1)$ into $\Phi(\rho)$ but after that I am stucked about how to calculate the inner product "$\cdot$". Which fact should I use to proceed the calculation?
 A: In the context of QIT, the inner-product over $\mathcal L(\mathcal H)$ is defined by
$$
\langle \rho_1,\rho_2 \rangle_{\mathcal L(\mathcal H)} = \operatorname{Tr}(\rho_1 \rho_2^*).
$$
With that in mind, we want to show that for any $\rho \in \mathcal L(\mathcal H)$ and $\sigma \in \mathcal {L(K)},$ your formulas for $\Phi$ and $\Phi^*$ are such that
$$
\langle \Phi(\rho),\sigma \rangle_{\mathcal {L(K)}} = \langle \rho, \Phi^*(\sigma)\rangle_{\mathcal {L(H)}}.
$$
To that end, we note that
\begin{align}
\langle \Phi(\rho),\sigma \rangle_{\mathcal {L(K)}} &= 
\operatorname{Tr}\left[ \left(\sum_i A_i \rho B_i^*\right) \sigma^* \right]
\\ & = \sum_i\operatorname{Tr}\left[ A_i \rho B_i^* \sigma^* \right]
\\ & = \sum_i\operatorname{Tr}\left[ \rho B_i^* \sigma^* A_i \right]
\\ & = \sum_i\operatorname{Tr}\left[ \rho \left(A_i^*\sigma B_i\right)^* \right]
\\ & = \operatorname{Tr}\left[ \rho \left(\sum_i A_i^*\sigma B_i\right)^* \right]
= \langle \rho, \Phi^*(\sigma) \rangle_{\mathcal{L(H)}}.
\end{align}
