Drazin inverse

A Drazin inverse of an operator $A$ is an operator, $B$, such that

$$AB=BA,$$

$$BAB=B,$$

$$ABA=A-U,$$

where the spectral radius $r(U)=0$. The Drazin inverse ($B$) is denoted by $A^D$. It exists, if $0$ is not an accumulation point of $\sigma (A)$.

For example, a projection operator is its own Drazin inverse, $P^D=P$, as$PPP=PP=P$; for a Shift operator $S^D=0$ holds.

The following are some other useful properties of the Drazin inverse:

1. ${(A^D)}^{*}={(A^{*})}^D$;

2. $A^D={(A+\alpha P^{(A)})}^{-1}(I-P^{(A)})$, where $P^{(A)}:=I-A^{D}A$ is the spectral projection of $A$ at $0$ and $\alpha \ne 0$;

3. $A^{†}={(A^{*}A)}^D{A}^{*}=A^{*}{(A{A}^{*})}^D$, where $A^{†}$ is the Moore-Penrose pseudoinverse of $A$;

4. $A^D=A^m{(A^{2m+1})}^{†}{A}^m$ for $m\ge \text{ind}(A)$, if $\text{ind}(A):=\min \{k:\mathrm{Im}{A}^k=\mathrm{Im}{A}^{k+1}\}$ is finite;

5. If the matrix is represented explicitly by its Jordan canonical form, ($\mathrm{\Lambda }$ regular and $N$ nilpotent), then

6. Let $e_{\lambda }^A$ denote an eigenvector of $A$ to the eigenvalue $\lambda$. Then $e_{\lambda }^A+t{(\lambda I-A)}^{D}h{e}_{\lambda }^A+O(t^2)$ is an eigenvector of $A+th$.