Partial trace of action on density matrix This topic is from chapter 11 of Kitaev, Shen and Vyalyi's Classical and Quantum Computation. If $V$ is an isometric (i.e. preserves the inner product) embedding $V: \mathcal{N} \rightarrow \mathcal{N} \otimes \mathcal{F}$, and we define the superoperator $V \cdot V^\dagger: \rho \mapsto V\rho V^\dagger$, what is the trace over $\mathcal{F}$ of $V\rho V^\dagger$? I did a computation which seems to show that $Tr_{\mathcal{F}}(V\rho V^\dagger) = Tr_\mathcal{F}(\rho)$, but I don't think this result is right. Where did I go wrong in the following argument?
Let $\{\phi_i\}$ be an orthonormal basis for $\mathcal{F}$. Then
\begin{equation*}
\begin{split}
Tr_{\mathcal{F}}(V\rho V^\dagger) & = Tr_{\mathcal{F}}(V^\dagger V\rho) \\
& = \sum_i\langle\phi_i,V^\dagger V\rho\phi_i\rangle \\
& = \sum_i\langle V\phi_i,V\rho\phi_i\rangle \\
& = \sum_i\langle\phi_i,\rho\phi_i\rangle \\
& = Tr_\mathcal{F}(\rho)
\end{split}
\end{equation*} 
where the first step is from the invariance of trace under cyclic permutations and the removal of the $V$s from the inner product is because $V$ is an isometry.
 A: There are a number of things wrong with your computation.

*

*The expression "$\operatorname{Tr}_{\mathcal{F}}(\rho)$" doesn't make any sense, as $\rho$ is an operator on $\mathcal{N}$ and not on $\mathcal{N}\otimes\mathcal{F}$. The expressions "$\rho\phi_{i}$" likewise are meaningless, as $\phi_i\in\mathcal{F}$.


*The partial trace is not invariant under cyclic permutations, so your first step is incorrect.


*The operator $V^\dagger V\rho$ isn't square (it is an operator from $\mathcal{N}\rightarrow\mathcal{N\otimes\mathcal{F}}$) so it doesn't make sense to take its partial trace.


*While the trace of an operator $X\in \operatorname{L} (\mathcal{F})$ is computed as $\operatorname{Tr}(X)=\sum_{i}\langle \phi_i, X\phi_i\rangle$, the partial trace is not computed this way. Indeed, if $X\in \operatorname{L} (\mathcal{N\otimes F})$, the expression "$\langle \phi_i, X\phi_i\rangle$" doesn't even make sense.
In general, the expression "$\operatorname{Tr}_\mathcal{F}(V\rho V^\dagger)$" may not be simplified further.
