# Properties of blocks in blockmatrix A if the matrix pair (E,A) is regular

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.

• One has $\det (sE-A) = s^r \det(A_{22})$ plus lower order terms. – daw Nov 19 '19 at 10:51
• Also $E$,A$have to be square. – daw Nov 19 '19 at 10:52 • Sorry, but could you clarify, which rule gets me$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$. – bambuk Nov 20 '19 at 11:19 • The exercise only makes sense if the sizes of the blocks of$E$and$A\$ are the same. – daw Nov 20 '19 at 12:34
• i dont see why, could you explain? – bambuk Nov 21 '19 at 10:31