I am working on a problem from the book "Differential-Algebraic Equations: Analysis and Numerical Solution" by Kunkel, Mehrmann. It is the Exercise 3 from Page 53 concerning matrix pairs $(E,A)$ and the properties of a specific block of the blockmatrix A:

Given the matrix pair $$ (E,A)= \left( \begin{bmatrix}I_r&0\\\ 0&0\end{bmatrix},\begin{bmatrix}A_{11}&A_{21}\\\ A_{12}&A_{22}\end{bmatrix} \right) $$

where $E,A \in C^{m \times n}$ and $I_r$ is the identity matrix of dimension $r < \min (m,n)$ one have to proof that the given matrix pair is regular and of index one if and only if $A_{22}$ is a square block and nonsingular.

I have some ideas, but don´t know how to proceed from there. For the $ \Rightarrow $ direction I know, that from regularity of $(E,A)$ it follows:

  • $E,A$ must be square matrices
  • $(E,A)$ is equivalent to the canonical form $ \left( \begin{bmatrix}I&0\\\ 0&N\end{bmatrix},\begin{bmatrix}J&0\\\ 0&I\end{bmatrix} \right) $ , i.e. there are nonsingular Matrices $P,Q$ such that $PEQ = \begin{bmatrix}I&0\\\ 0&N\end{bmatrix}$ and $PAQ = \begin{bmatrix}J&0\\\ 0&I\end{bmatrix}$ where $J$ is a matrix in jordan normal form and $N$ is nilpotent in Jordan normal form
  • $p(s) = \det( sE-A )$ is not $0$ for every $s$

Also, because the index of $(E,A)$ is $\nu=1$ it is that $N^{0} \neq 0,\;N^{1}=0$ the nilpotent block should be of size $1\times1$ : $$N = [0]$$ since $N^0=[0]^0=1$ and $N^1=[0]$. But from here I dont know how to continue. Thank you in advance.

  • $\begingroup$ One has $\det (sE-A) = s^r \det(A_{22})$ plus lower order terms. $\endgroup$ – daw Nov 19 '19 at 10:51
  • $\begingroup$ Also $E$,A$ have to be square. $\endgroup$ – daw Nov 19 '19 at 10:52
  • $\begingroup$ Sorry, but could you clarify, which rule gets me $det(sE−A)=s^r det(A_{22}) + ...$ I have trouble calculating it, since $ A_{22}$ must not be the same size as the zero block in $E$. $\endgroup$ – bambuk Nov 20 '19 at 11:19
  • $\begingroup$ The exercise only makes sense if the sizes of the blocks of $E$ and $A$ are the same. $\endgroup$ – daw Nov 20 '19 at 12:34
  • $\begingroup$ i dont see why, could you explain? $\endgroup$ – bambuk Nov 21 '19 at 10:31

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