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.


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