# Generalized Schur Decompostion - Reordering Unitary Matrices

My problem concens two matrices of the following form$$V=\begin{pmatrix}\,0&I\,\\\,A&B\,\end{pmatrix}\quad\text{and}\quad W=\begin{pmatrix}\,I&0\,\\\,0&C\,\end{pmatrix}$$ Here, $$A,B$$ are $$m\times m$$ and may or may not be of full rank. $$C$$ also $$m\times m$$ is not of full rank. So neither $$V$$ nor $$W$$ ($$2m\times 2m$$) are non-singular.

A generalized Schur decomposition yields matrices $$Q,Z,T,S$$ such that $$Q V Z =T$$ and $$QWZ=S$$ where $$T,S$$ are upper triangular and $$Q,Z$$ are unitary.

Given the context of my problem, I'd like to partition $$Q,Z,S,T$$ into $$m\times m$$ blocks such that $$Q=\begin{pmatrix}\,Q_{11}&Q_{12}\,\\\,Q_{21}&Q_{22}\,\end{pmatrix}\quad\text{and}\quad Z=\begin{pmatrix}\,Z_{11}&Z_{12}\,\\\,Z_{21}&Z_{22}\,\end{pmatrix}$$ while $$S=\begin{pmatrix}\,S_{11}&S_{12}\,\\\,0&S_{22}\,\end{pmatrix}\quad\text{and}\quad T=\begin{pmatrix}\,T_{11}&T_{12}\,\\\,0&T_{22}\,\end{pmatrix}$$ such that the terms associated with the smallest generalized eigenvalues are collected in the upper left block.

I do know, that the generalized eigenvalues are given by the diagonal elements of $$TS^{-1}.$$ This is where my problem begins.

Question 1: How do I ascertain the generalized eigenvalues if $$S$$ is not invertible?

Moreover, when I have a plain-vanilla eigenvalue decomposition such that, say, $$A=V \Lambda V^{-1}$$, I can see that the first eigenvalue, i.e. the (1,1) element of $$\Lambda$$, corresponds to the first column vector in $$V.$$ So, when I switch the (1,1) and (2,2) element in $$\Lambda$$ and the first and second column vector in $$V$$ that does not change anything. So I could order my eigenvalues from smallest to largest and order the column vectors in $$V$$ accordingly.

Question 2: Seeing as $$S$$ and $$T$$ are upper triangular, I am afraid that I do not quite see how I would do the same in the context of a generalized Schur decomposition.

I would be grateful for any pointers.

Many thanks.