How to find the partial transpose of bipartite states from their matrix representation? Suppose we have a density operator given by $\rho=\mid \Psi \rangle \langle \Psi \mid$ with $\mid \Psi \rangle = \frac{1}{\sqrt{2}}(\mid 1 \rangle\mid \ 0 \rangle-\mid \ 0 \rangle \mid 1 \rangle)=\frac{1}{\sqrt2}(\mid 10 \rangle-\mid01\rangle)$ living in a space like $\mathcal{H_s}\otimes\mathcal{H_a}$, where 1 and 0 are the normal up and down states for example. If we apply the partial transposition $(T_s\otimes I_a)(\rho)$ which transposes the system S and leaves A unchanged, we get something like:
$$\rho=\frac{1}{2}(\mid 10\rangle \langle 10\mid + \mid 01 \rangle\langle01\mid - \mid10\rangle\langle01\mid-\mid01\rangle\langle10\mid)
\\(T_s\otimes I_a)(\rho)=\frac{1}{2}(\mid 10\rangle \langle 10\mid + \mid 01 \rangle\langle01\mid - \mid11\rangle\langle00\mid-\mid00\rangle\langle11\mid)
$$
which can be written in matrix form as:
$$ (T_s\otimes I_a)(\rho)=\frac{1}{2}\left( \begin{matrix} 0 & 0 & 0 & -1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ -1 & 0 & 0 & 0 \end{matrix} \right) $$
If we look at the problem in matrix form from the beginning we see that:
$$ \rho = \frac{1}{2}\left( \begin{matrix} 0 & 0 & 0 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & -1 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{matrix} \right) $$
such that the transposition seems to be transponing the upper right and the lower left quadrants and leaving the others unchanged. This seems to work for every density operator I have tried. My question is: Does is really work and if it does, why are these quadrants specifically changed, namely why are they part of system S and the others of system A?
I appreciate your cooperation.
 A: Given a bipartite state $\rho$ in a space $\mathcal H_1\otimes\mathcal H_2$ with $\dim(\mathcal H_1)=n$ and $\dim(\mathcal H_2)=m$, its (standard) matrix representation has the block form
$$
\rho = \begin{pmatrix}\rho_{1\bullet,1\bullet} & \rho_{1\bullet,2\bullet} & \cdots & \rho_{1\bullet,n\bullet}
 \\ \rho_{2\bullet,1\bullet} & \rho_{2\bullet,2\bullet} & \cdots & \rho_{2\bullet,n\bullet}\\ \vdots & \vdots 
& \ddots & \vdots\\ \rho_{n\bullet,1\bullet} & \rho_{n\bullet,2\bullet} & \cdots & \rho_{n\bullet,n\bullet}\end{pmatrix},\tag1
$$
where $\rho_{i\bullet,j\bullet}$ denotes here the $m\times m$ matrix with components $(\rho_{i k,j\ell})_{k,\ell}$.
The partial transpose is easily computed from (1): $(T\otimes I)\rho$ amounts to taking the transpose of (1) as if it was a regular matrix (thus e.g. $\rho_{1\bullet,2\bullet}$ and $\rho_{2\bullet,1\bullet}$ would switch places).
Moreover, the partial transpose in the second space, $(I\otimes T)\rho$, is obtained taking the transpose of each block in (1), that is, replacing $\rho_{i\bullet,j\bullet}\to\rho_{i\bullet,j\bullet}^T$ for each $i,j$.
In your example, notice how the partial transpose wrt the first space amounts to having the upper-right and bottom-left $2\times2$ blocks switch places.
