Drazin inverse

Remarks

Example: if $A$ is invertible, its index is $0$, and $A^D = A^{-1}$; indeed $A \cdot A^{-1} = I$, $A^{-1} A A^{-1} = A^{-1}$, and $A^{-1}$ commutes with $A$. Example: if $A$ is nilpotent with $A^k = 0$ for its index $k$, then $A^D = 0$; each side of each condition is then $0$, because $A^k = 0$. Non-example: for $A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$, of index $2$, the matrix $X = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$ is not $A^D$, because $AX = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ and $XA = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$ differ, so clause 3 fails. If $A = P \begin{pmatrix} C & 0 \\ 0 & N \end{pmatrix} P^{-1}$ with $C$ invertible and $N$ nilpotent, then $A^D = P \begin{pmatrix} C^{-1} & 0 \\ 0 & 0 \end{pmatrix} P^{-1}$. When the index is at most $1$, the Drazin inverse is called the group inverse and written $A^{\#}$.