How to construct matrix to apply U(2) beam splitter to a N dimentional Beam So I am trying to read this paper (no paywall here). In this the formal construction of a matrix $T$ is given as

The product of matrices is equivalent to setting up experimental
  devices in sequence. Finding an optical experiment belonging to an
  arbitrary unitary matrix is therefore completely equivalent to
  factorizing the uni- tary matrix into a product of block matrices
  containing only beam splitter matrices with appropriate phase shifts
  as defined in Eq. (1). We define a matrix $T_{pq}$ which is an
  $N$-dimensional identity matrix with the elements $|pp$, $|pq$, $|qp$, and $|qq$
  replaced by the corresponding beam splitter matrix elements. This
  matrix performs a unitary transformation on a two-dimensional subspace
  of the $N$ dimensional Hilbert space leaving an $(N — 2)$-dimensional
  subspace unchanged [17]. It can be used to successively make all
  off-diagonal elements of the given $N \times N$ unitary matrix zero, a method
  similar to Gaussian elimination.

I just for the life of me cannot figure out how to make $T_{pq}$. I would be very thankful if someone can please explain it to me.
 A: Without seeing the full paper, I'm almost certain that the following is what is being done. 
Let $\{\hat{X}_j\}_{j=1}^N$ be the chosen orthonormal basis vectors for our state space $\mathbb{C}^N$ with respect to which we wish to write the matrices representing our linear, unitary transformations.
Any homogeneous linear transformation $T:\mathbb{C}^N\to\mathbb{C}^N$ is defined wholly by the $N$ values $T(\hat{X}_j)$.
Now let's think about the authors' $T_{p,q}$. What they mean is $T_{p,q}(\hat{X}_j) = \hat{X}_j$ unless $j=p$ or $j=q$. In this special case, $T_{p,q}$ restricted to the vector space spanned by $\hat{X}_p,\,\hat{X}_q$ has an image in that same vector space, i.e. $T_{p,q}$ maps the space $\operatorname{span}(\{\hat{X}_p,\,\hat{X}_q\})$ to itself:
$$T_{p,q}:\operatorname{span}(\{\hat{X}_p,\,\hat{X}_q\})\to \operatorname{span}(\{\hat{X}_p,\,\hat{X}_q\})$$
and this restricted linear operator is wholly defined by 
$$T_{p,q}(\hat{X}_p) = \ell_{p,p} \hat{X}_p+\ell_{q,p} \hat{X}_q$$
$$T_{p,q}(\hat{X}_q) = \ell_{p,q} \hat{X}_p+\ell_{q,q} \hat{X}_q$$
and, moreover, the $2\times 2$ matrix $\Lambda_{p,q}=\left(\begin{array}{cc}\ell_{p,p}&\ell_{q,p}\\\ell_{p,q}&\ell_{q,q}\end{array}\right)\in U(2)$ and is unitary. So the images of $\hat{X}_p$ and $\hat{X}_q$ define the matrix $T_{p,q}$. (The weird ordering of the indices in the off diagonal terms will become clearer below).
So, how do we write the matrix of $T_{p,q}$? Well, we write the image of $\hat{X}_j$ in the $j^{th}$ column. So all the columns of $T_{p,q}$ aside from the $p^{th}$ and $q^{th}$ column are simply what they would be in the identity matrix, since the image of $\hat{X}_j$ is $\hat{X}_j$. This leaves the last two columns, the columns $p$ and $q$. 
For column $p$:
These are all noughts aside from the element at position $(p,p)$, which is the element $\ell_{p,p}$ in our restricted $2\times 2$ matrix $\Lambda_{p,q}$. Likewise, the element at position $(q,p)$ in the $p^{th}$ column is $\ell_{q,p}$ in our restricted $2\times 2$ matrix $\Lambda_{p,q}$.
For column $q$, the analogous thing happens. All noughts aside from $\ell_{p,q}$ and $\ell_{q,q}$
Now, I am not sure whether the authors allow the restricted $2\times 2$ matrix $\Lambda_{p,q}$ to be any unitary $2\times 2$ matrix. They mention beamsplitter matrices, which could be construed as any unitary $2\times2$ matrix, or they could have the matrices I describe in this answer here in mind. I would guess the former, i.e. any unitary matrix.
