Motivation: Consider a matrix $A=\begin{pmatrix}0 &1\\0 &0\end{pmatrix}$, then $A^2=O$. So, $Rank(A)=1\neq 0=Rank(A^2)$ and thus the Group inverse of the matrix $A$ does not exist because the index of $A\neq 1$. Thus, the theory developed in the Group inverse fails when the index of the given matrix is not 1. Hence, we require concept of another generalized inverse which not only preserve spectral properties of the matrix like the group inverse but also exists for any square matrix $A$ of any index and whose inverse coincides with the usual inverse whenever $A$ is nonsingular.
Definition: Let $A\in \mathbb{R}^{n\times n}$, then there exists a unique matrix $X$ satisfying following three equations:
(i) $A^kXA=A^k$, where $k$ is the index of $A$
(ii) $XAX=X$
(iii) $AX=XA$.
The matrix $X$ is called the Drazin inverse of $A$ and is denoted as $A^D$. The Drazin inverse exists for every square matrix and is always unique for a matrix. When $A$ is nonsingular, in that case $A^D=A^{-1}.$
Properties of the Drazin inverse:
(i) $(A^T)^D=(A^D)^T$.
(ii) $A^D)^D=A$ if and only if $A$ has index 1.
(iii) $((A^D)^D)^D=A^D$.
(iv) The Drazin inverse preserves similarity, i.e., $A=ZBZ^{-1}\Rightarrow A^D=ZB^DZ^{-1}$.
(v) $AA^D (=A^DA)$ is a projection on $R(A^D)$ along $N(A^D)$.
(vi) $A^D$ has index 1, and $(A^D)^{\#}=A^2A^D$.
(vii) $A^D(A^D)^{\#}=AA^D$.
(viii) If $A$ is nilpotent, then $A^D=O$.
proofs. (i) Let $A^D$ be the Drazin inverse of $A$ with index $k$, so it satisfies the three conditions: $A^kA^DA=A^k$, $A^DAA^D=A^D$ and $AA^D=A^DA$. Let us denote $B=A^T$ and $X=(A^D)^T$. Then, we have to show that $X=B^D$. Now, $BX=A^T(A^D)^T=(A^DA)^T=(AA^D)^T=(A^D)^TA^T=XB$ and using this we get, $XBX=(A^D)^T(A^T(A^D)^T)=(A^DAA^D)^T=(A^D)^T=X$. Finally,
$\begin{align}B^kXB&=(A^T)^k(A^T(A^D)^T)\\&=(A^k)^TA^T(A^D)^T\\&=(A^kAA^D)^T\\&=(A^k)^T\\&=(A^T)^k\\&=B^k.\end{align}$
(ii) If part: Let $A$ be of index 1. Then, $(A^D)^D=(A^{\#})^{\#}=A$ (since $A^D=A^{\#}$ and $(A^{\#})^{\#}=A$ proved in the Group inverse post).
Only if: Assume that $A$ is singular, i.e., $0$ is an eigenvalue of $A$. Then,
$A=Z\begin{bmatrix}J_{(\lambda\neq 0)} &O\\O &J_{(\lambda=0)}\end{bmatrix}Z^{-1},$ where $J_{(\lambda\neq 0)}$ and $J_{(\lambda=0)}$ ($k\times k$) are the Jordan block corresponding to the non-zero and zero eigenvalues of $A$, respectively. Clearly, $J_{(\lambda\neq 0)}$ is nonsingular and $J^{k}_{(\lambda=0)}=O$ with $J^{k-1}_{(\lambda=0)} \neq O$. Now, $A^D=Z\begin{bmatrix}J^{-1}_{(\lambda\neq 0)} &O\\O &O\end{bmatrix}Z^{-1}$ (see Theorem 2 for the proof). It is easy to check that
$(A^D)^D=Z\begin{bmatrix}J_{(\lambda\neq 0)} &O\\O &O\end{bmatrix}Z^{-1}$. Now, if $(A^D)^D=A$, then $J_{(\lambda=0)}=O$, a zero matrix and thus, $A=Z\begin{bmatrix}J_{(\lambda\neq 0)} &O\\O &O\end{bmatrix}Z^{-1}$. Since $J_{(\lambda\neq 0)}$ is nonsingular, we have $rank(A)=rank(A^2)$ and hence, index of $A$ is 1.
(iii) From (ii), $A=Z\begin{bmatrix}J_{(\lambda\neq 0)} &O\\O &J_{(\lambda=0)}\end{bmatrix}Z^{-1}$, then we can easily check that $A^D=Z\begin{bmatrix}J_{(\lambda\neq 0)} &O\\O &O\end{bmatrix}Z^{-1}$ (see Theorem 2) and $(A^D)^D=Z\begin{bmatrix}J^{-1}_{(\lambda\neq 0)} &O\\O &O\end{bmatrix}Z^{-1}$. Applying Theorem 2 again to find the Drazin inverse of $(A^D)^D$, we get $((A^D)^D)^D=A^D$.
(iv) Let $A=ZBZ^{-1}$ and $X=ZB^DZ^{-1}$, then
$\begin{align}AX&=ZBB^DZ^{-1}\\&=ZB^DBZ^{-1}\\&=ZB^DZ^{-1}ZBZ^{-1}\\&=XA;\end{align}$
$\begin{align}XAX&=ZB^DZ^{-1}ZB^DBZ^{-1}\\&=ZB^DB^DBZ^{-1}\\&=Z(B^DBB^D)Z^{-1}\\&=ZB^DZ^{-1}\\&=X;\end{align}$
$\begin{align}B^kXB&=(ZB^DZ^{-1}ZB^DBZ^{-1})\\&=ZB^DB^DBZ^{-1}\\&=ZB^DBB^DZ^{-1}\\&=ZB^DZ^{-1}.\end{align}$
(v) Let $P=AA^D$. Clearly, $P^2=P$ (since $AA^D=A^DA$). So, $P$ is a projection. Now,
$\begin{align}R(AA^D)&=R(A^DA)\\&\subseteq R(A^D)\\&=R(A^DAA^D)\\&\subseteq R(A^DA)\\&=R(AA^D);\end{align}$
$N(AA^D)\subseteq N(A^D)=N(A^DAA^D)=N(AA^D)$. Thus, $R(AA^D)=R(A^D)$ and $N(AA^D)=N(A^D)$. Hence, $AA^D (=A^DA)$ is a projection on $R(A^D)$ along $N(A^D)$.
(vi) The Drazin inverse of any singular matrix $A$ is of the form, $A^D=Z\begin{bmatrix}J_{(\lambda\neq 0)} &O\\O &O\end{bmatrix}Z^{-1}$. Clearly, $A^D$ has index 1 since $rank(A^D)=rank(A^D)^2$ and thus its Group inverse exists. Let $X=A^2A^D$, then
$\begin{align}A^DXA^D &=A^DAAA^DA^D\\ &=A^DAA^DAA^D\\ &=A^DAA^D\\ &=A^D;\end{align}$
$\begin{align}XA^DX &=A^2A^DA^DA^2A^D\\ &=AA^DAA^DA^2AA^D\\ &=AA^DAA^DAA^D\\ &=A^2A^D;\end{align}$
$\begin{align}AX &=AA^2A^D\\ &=A^3A^D\\ &=A^2A^DA\\ &=XA.\end{align}$
Hence, $A^2A^D=X=(A^D)^{\#}$.
(vii) Using (vi), $A^D(A^D)^{\#}=A^D(A^2A^D)=A^DA =A^DA$.
(viii) Obvious from Theorem 2 (see the proof below).
Theorem 1. Let $A$ be square singular matrix whose index is $k$. If the system
$Ax=b,~~x\in R(A^k)$,
is consistent, then it is uniquely given by $x=A^Db$.
proof. Since $x\in R(A^k)$, then $x=A^ky$ for $y$. Therefore, $A^{k+1}y=A(A^ky)=b$. Using $AA^DA=A$ and $AA^D=A^DA$, we get
$\begin{align}x &=A^ky\\ &=A^{k-1}(AA^DA)y\\ &=A^{k+1}A^Dy\\ &=A^DA^{k+1}y\\ &=A^Db.\end{align}$
Now, we will prove the uniqueness. Let $z_1$ and $z_1$ be two solutions of $Ax=b$ such that $z_1, z_2\in R(A^k)$. Then, $Az_1=b$ and $Az_2=b$ imply that $z_1-z_2\in N(A)\subseteq N(A^k)$. Since $k$ is the index of $A$, therefore $R(A^k)\cap N(A^k)=\{0\}$. Thus, $z_1-z_2\in R(A^k)\cap N(A^k)=\{0\}$. Hence, $z_1=z_2$.
Theorem 2. Let $A\in \mathbb{R}^{n\times n}$ whose Jordan form is given as
$A=Z\begin{bmatrix}J_{(\lambda\neq 0)} &O\\O &J_{(\lambda=0)}\end{bmatrix}Z^{-1},$
where $J_{(\lambda\neq 0)}$ (k\times k) and $J_{(\lambda=0)}$ are the Jordan block corresponding to the non-zero and zero eigenvalues of $A$. Then
$A^D=Z\begin{bmatrix}J_{(\lambda\neq 0)}^{-1} &O\\O &O\end{bmatrix}Z^{-1}.$
proof. We will show that $X=Z\begin{bmatrix}J_{(\lambda\neq 0)}^{-1} &O\\O &O\end{bmatrix}Z^{-1}$ satisfies all three matrix equations: $A^{k}XA=A^{k}$, $XAX=X$ and $AX=XA$. It is clear that $A^k=Z\begin{bmatrix}J_{(\lambda\neq 0)}^{k} &O\\O &O\end{bmatrix}Z^{-1}$ (since the index of $A$ is $k$) and thus by block matrix multiplication $A^{k}XA=A^{k}$. Conditions $XAX=X$ and $AX=XA$ satisfy trivially. Hence the proof.
Example. Let $A=\begin{bmatrix}0 &-2 &0 &-5 &2\\0 &-1 &0 &-2 &0\\0 &0 &0 &2 &0\\0 &1 &0 &2 &0\\-2 &-2 &1 &-8 &4\end{bmatrix}$. Then, the Jordan form of $A$ is given as
$A=Z\begin{bmatrix}1 &0 &0 &0 &0\\0 &2 &1 &0 &0\\0 &0 &2 &0 &0\\0 &0 &0 &1 &0\\0 &0 &0 &0 &0\\\end{bmatrix}Z^{-1}$, where $Z=\begin{bmatrix}1 &-2 &1 &1 &0\\-1 &0 &0 &0 &-2\\2 &0 &0 &2 &0\\1 &0 &0 &0 &1\\2 &-2 &0 &0 &1\\\end{bmatrix}$. The matrix $J=\begin{bmatrix}1 &0 &0 &0 &0\\0 &2 &1 &0 &0\\0 &0 &2 &0 &0\\0 &0 &0 &1 &0\\0 &0 &0 &0 &0\\\end{bmatrix}$ has two jordan blocks, $J_{(\lambda\neq 0)}=\begin{bmatrix}1 &0 &0 \\0 &2 &1 \\0 &0 &2\end{bmatrix}$ and $J_{(\lambda=0)}=\begin{bmatrix}0 &1 \\0 &0 \end{bmatrix}$.
By Theorem 2, the Drazin inverse of $A$ is given as
$A^D=Z\begin{bmatrix} J_{(\lambda\neq 0)}^{-1} &O \\O &O\end{bmatrix}Z^{-1}=Z\begin{bmatrix}1 &0 &0 &0 &0\\0 &\frac{1}{2} &\frac{-1}{4} &0 &0\\0 &0 &\frac{1}{2} &0 &0\\0 &0 &0 &0 &0\\0 &0 &0 &0 &0\\\end{bmatrix}Z^{-1}$, i.e.,
$A^D=\begin{bmatrix}1 &\frac{3}{2} &\frac{-1}{2} &\frac{7}{2} &\frac{-1}{2}\\0 &-1 &0 &-2 &0\\0 &2 &0 &4 &0\\0 &1 &0 &2 &0\\\frac{1}{2} &2 &\frac{-1}{4} &4 &0\\\end{bmatrix}$.
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