How do I show that $\text{tr}(A^{\dagger} B) = \langle m | A \otimes B |m\rangle $ for general matrices $A$ and $B$? I am desperately trying to solve the following problem, and would really appreciate help!
Suppose $R$ and $Q$ are two quantum systems with the same Hilbert space $\mathcal{H}$ with $\dim(\mathcal{H})=N$. Let $|i_R\rangle$ and $|i_Q\rangle$ be orthonormal basis sets for $R$ and $Q$. Let $A$ be an operator on $R$ and $B$ an operator on $Q$. Define $|m\rangle := \sum_{i=1}^N |i_R\rangle\otimes |i_Q\rangle$. Show that
$$
tr(A^\dagger B)=\langle m|A\otimes B| m\rangle
$$
where the multiplication on the left hand side is of matrices, and it is understood that the matrix elements of $A$ are taken with respect to the basis $|i_R\rangle$ and those for $B$ with respect to the basis $|i_Q\rangle$.
What I have is the following:
LHS:
$$
tr(A^\dagger B)=\sum_{n=1}^N \langle n_Q|\left( \sum_{i,j,k=1}^N a^*_{ji}b_{jk} |i_Q\rangle \langle k_R| \right)|n_R\rangle
$$
Since $Q$ and $R$ are over the same Hilbert space, we can disregard the fact that one set of basis vectors is in system $Q$ and the other one in system $R$. 
This yields
$$
tr(A^\dagger B)=\sum_{i,j=1}^N a^*_{ji} b_{ji}
$$
RHS:
$$
\begin{align*}
\langle m'|(A\otimes B)|m\rangle &= \sum_{e,f=1}^N \Bigg( \langle e_R| \left( \sum_{k,l=1}^N a_{kl} |k_R\rangle \langle l_R\right) |f_R\rangle \cdot \langle e_Q|\left(\sum_{i,j=1}^N b_{ij}|i_Q\rangle\langle j_Q| \right)|f_Q\rangle\Bigg)\\
&=\sum_{e,f,k,l,i,j=1}^N a_{kl} b_{ij}\langle e_R |k_R \rangle \langle l_R|f_R \rangle \langle e_Q|i_Q \rangle\langle j_Q|f_Q \rangle\\
&=\sum_{i,j=1}^N a_{ij} b_{ij}
\end{align*}
$$
As you can see, the two expressions are almost the same. Only the first number on the LHS evaluation is a complex conjugate and on the RHS it's not. 
Any ideas where I went wrong?
Thanks!
P.S: Full disclosure: I posted this question on PhysicsForum.com as well.
 A: The correct formula is 
$$
\mathrm{tr}[A^TB]=\langle m \vert A\otimes B\vert m\rangle\ ,
$$
so your proof is correct, you're just trying to proof an erroneous formula. (You can easily verify this because with a $\dagger$ the l.h.s. is sesquilinear while the r.h.s. is bilinear.)
But I have the feeling this has been asked before. If you have this from Nielsen-Chuang, did you check the erratum? --- Voila, I found it: Proof of Uhlmann Theorem, see also http://www.michaelnielsen.org/qcqi/errata/errata/errata.html.
A: The other answer already points out the typo, but just in case it may help someone else stumbling upon this: this formula (the corrected version in the other answer) becomes almost trivial using diagrammatic notation.
It essentially amounts to the following statement (which will make sense if you know a bit of diagrammatic notation):
$$
\langle m \vert A\otimes B\vert m\rangle\ =
\mathrm{tr}[A^TB]
$$
is equivalent to:

Where the vertical lines on the LHS represent the maximally mixed states $|m\rangle$ and $\langle m|$, while those on the right arise from the trace.
In this notation, the relation between taking expectation values between maximally mixed states and the trace operation is made more explicit.
