Prove an identity which uses Drazin inverses and Moore-Penrose inverses. Let $A$ be $m\times n$ complex matrix. Then how can we prove the following. 
$$A^+=(A^*A)^DA^*=A^*(AA^*)^D$$ where $D$ denotes the Drazin inverse and $A^+$ is the Moore-Penrose Pseudoinverse of $A$. I found this question at http://planetmath.org/drazininverse.
Thanks.
 A: Core-Nilpotent Decomposition
Let $\mathbf{A}$ be a square matrix of rank $\rho<n$, with index $k$ such that $\text{rank} \left( \mathbf{A}^{k} \right) = \rho$:
$$
  \mathbf{A} \in \mathbb{C}^{n\times n}
$$
then there exists a nonsingular matrix $\mathbf{Q}$ such that
$$
  \mathbf{A} = 
  \mathbf{Q}
  \left[ \begin{array}{cc}
    \mathbf{C} & \mathbf{0} \\ \mathbf{0} & \mathbf{N}
  \end{array} \right]
  \mathbf{Q}^{-1}
$$
where the nonsingular core matrix
$$
  \mathbf{C} \in \mathbb{C}^{\rho \times \rho}
$$
and the nilpotent matrix $\mathbf{N}$ has index $k$. We may think of this as a dilute form of diagonalization.
Drazin inverse
The Drazin inverse is defined in terms of the core-nilpotent decomposition
$$
  \mathbf{A}^{D} = 
  \mathbf{Q}
  \left[ \begin{array}{cc}
    \mathbf{C}^{-1} & \mathbf{0} \\ \mathbf{0} & \mathbf{0}
  \end{array} \right]
  \mathbf{Q}^{-1}
$$
Singular value decomposition
The SVD can be viewed as a tool which gets matrices as close to diagonalization as possible. For the target matrix we can lift the requirement that the matrix has as many rows as columns $m=n$:
$$
  \mathbf{A} \in \mathbb{C}^{m\times n}_{\rho}
$$
The SVD provides an orthonormal basis for both domain $\mathbb{C}^{n}$ and codomain $\mathbb{C}^{m}$. 
$$
\begin{align}
%
  \mathbf{C}^{n} = 
    \color{blue}{\mathcal{R} \left( \mathbf{A} \right)} \oplus
    \color{red}{\mathcal{N} \left( \mathbf{A}^{*} \right)} \\
%
  \mathbf{C}^{m} = 
    \color{blue}{\mathcal{R} \left( \mathbf{A}^{*} \right)} \oplus
    \color{red} {\mathcal{N} \left( \mathbf{A} \right)}
%
\end{align}
$$
The $\color{red}{nullspace}$ terms will be silent in the pseudoinverse, just as the nilpotent matrix vanishes in the core-nilpotent decomposition. 
$$
\begin{align}
  \mathbf{A} &=
  \mathbf{U} \, \Sigma \, \mathbf{V}^{*} \\
%
 &=
% U 
  \left[ \begin{array}{cc}
     \color{blue}{\mathbf{U}_{\mathcal{R}}} & \color{red}{\mathbf{U}_{\mathcal{N}}}
  \end{array} \right]  
% Sigma
  \left[ \begin{array}{cccccc}
     \sigma_{1} & 0 & \dots &  &   & \dots &  0 \\
     0 & \sigma_{2}  \\
     \vdots && \ddots \\
       & & & \sigma_{\rho} \\
       & & & & 0 & \\
     \vdots &&&&&\ddots \\
     0 & & &   &   &  & 0 \\
  \end{array} \right]
% V 
  \left[ \begin{array}{c}
     \color{blue}{\mathbf{V}_{\mathcal{R}}}^{*} \\ 
     \color{red}{\mathbf{V}_{\mathcal{N}}}^{*}
  \end{array} \right]  \\
%
  & =
% U
   \left[ \begin{array}{cccccccc}
    \color{blue}{u_{1}} & \dots & \color{blue}{u_{\rho}} & \color{red}{u_{\rho+1}} & \dots & \color{red}{u_{n}}
  \end{array} \right]
% Sigma
  \left[ \begin{array}{cc}
     \mathbf{S}_{\rho\times \rho} & \mathbf{0} \\
     \mathbf{0} & \mathbf{0} 
  \end{array} \right]
% V
   \left[ \begin{array}{c}
    \color{blue}{v_{1}^{*}} \\ 
    \vdots \\
    \color{blue}{v_{\rho}^{*}} \\
    \color{red}{v_{\rho+1}^{*}} \\
    \vdots \\ 
    \color{red}{v_{n}^{*}}
  \end{array} \right]
%
\end{align}
$$
Let's rewrite the SVD in a form similar to the $\mathbf{CN}$ decomposition:
$$
\begin{align}
  \mathbf{A} &=
  \mathbf{U} \, \Sigma \, \mathbf{V}^{*} \\
%
 &=
% U 
  \left[ \begin{array}{cc}
     \color{blue}{\mathbf{U}_{\mathcal{R}}} & \color{red}{\mathbf{U}_{\mathcal{N}}}
  \end{array} \right]  
% Sigma
  \left[ \begin{array}{cc}
     \mathbf{S} & \mathbf{0} \\
     \mathbf{0} & \mathbf{0} 
  \end{array} \right]
% V 
  \left[ \begin{array}{c}
     \color{blue}{\mathbf{V}_{\mathcal{R}}}^{*} \\ 
     \color{red}{\mathbf{V}_{\mathcal{N}}}^{*}
  \end{array} \right]  
\end{align}
$$
While the matrix of singular values $\mathbf{S}$ is diagonal, the core matrix $\mathbf{C}$ is simply nonsingular.
Manipulating the singular value decomposition
The forms needed are the Hermitian conjugate and the Moore-Penrose pseudoinverse. Recall that $\mathbf{S}^{\mathrm{T}} = \mathbf{S}$:
$$
\begin{align}
  \mathbf{A} &=
  \mathbf{U} \, \Sigma \, \mathbf{V}^{*} \\
%
 &=
% U 
  \left[ \begin{array}{cc}
     \color{blue}{\mathbf{U}_{\mathcal{R}}} & \color{red}{\mathbf{U}_{\mathcal{N}}}
  \end{array} \right]  
% Sigma
  \left[ \begin{array}{cc}
     \mathbf{S} & \mathbf{0} \\
     \mathbf{0} & \mathbf{0} 
  \end{array} \right]
% V 
  \left[ \begin{array}{c}
     \color{blue}{\mathbf{V}_{\mathcal{R}}}^{*} \\ 
     \color{red}{\mathbf{V}_{\mathcal{N}}}^{*}
  \end{array} \right] \\[5pt]
%% hc
  \mathbf{A}^{*} &=
  \mathbf{V} \, \Sigma^{\mathrm{T}} \mathbf{U}^{*} \\
%
 &=
% U 
  \left[ \begin{array}{cc}
     \color{blue}{\mathbf{V}_{\mathcal{R}}} & \color{red}{\mathbf{V}_{\mathcal{N}}}
  \end{array} \right]  
% Sigma
  \left[ \begin{array}{cc}
     \mathbf{S} & \mathbf{0} \\
     \mathbf{0} & \mathbf{0} 
  \end{array} \right]
% V 
  \left[ \begin{array}{c}
     \color{blue}{\mathbf{U}_{\mathcal{R}}}^{*} \\ 
     \color{red}{\mathbf{U}_{\mathcal{N}}}^{*}
  \end{array} \right] \\[5pt]
%% mp
  \mathbf{A}^{\dagger} &=
  \mathbf{V} \, \Sigma^{\dagger} \mathbf{U}^{*} \\
%
 &=
% U 
  \left[ \begin{array}{cc}
     \color{blue}{\mathbf{V}_{\mathcal{R}}} & \color{red}{\mathbf{V}_{\mathcal{N}}}
  \end{array} \right]  
% Sigma
  \left[ \begin{array}{cc}
     \mathbf{S}^{-1} & \mathbf{0} \\
     \mathbf{0} & \mathbf{0} 
  \end{array} \right]
% V 
  \left[ \begin{array}{c}
     \color{blue}{\mathbf{U}_{\mathcal{R}}}^{*} \\ 
     \color{red}{\mathbf{U}_{\mathcal{N}}}^{*}
  \end{array} \right] 
%
\end{align}
$$
Assemble the product matrices
$$
\begin{align}
  \mathbf{A}^{*} \mathbf{A} &= 
\left(
  \mathbf{V} \, \Sigma^{\mathrm{T}} \mathbf{U}^{*}
\right)
\left(
  \mathbf{U} \, \Sigma \, \mathbf{V}^{*}
\right) \\
  &= \mathbf{V} \, \Sigma^{\mathrm{T}} \Sigma \, \mathbf{V}^{*} 
   =   \left[ \begin{array}{cc}
     \color{blue}{\mathbf{V}_{\mathcal{R}}} & \color{red}{\mathbf{V}_{\mathcal{N}}}
  \end{array} \right]  
 \, 
     \left[ \begin{array}{cc}
       \mathbf{S}^{2} & \mathbf{0} \\ \mathbf{0} & \mathbf{0}
     \end{array} \right]
       \left[ \begin{array}{cc}
     \color{blue}{\mathbf{V}_{\mathcal{R}}} & \color{red}{\mathbf{V}_{\mathcal{N}}}
  \end{array} \right]  
^{*} \\
%% Wy
  \mathbf{A} \mathbf{A}^{*} 
  &= \mathbf{U} \, \Sigma \Sigma^{\mathrm{T}} \, \mathbf{U}^{*} 
  = \left[ \begin{array}{cc}
     \color{blue}{\mathbf{U}_{\mathcal{R}}} & \color{red}{\mathbf{U}_{\mathcal{N}}}
  \end{array} \right]  
 \, 
     \left[ \begin{array}{cc}
       \mathbf{S}^{2} & \mathbf{0} \\ \mathbf{0} & \mathbf{0}
     \end{array} \right]
       \left[ \begin{array}{cc}
     \color{blue}{\mathbf{U}_{\mathcal{R}}} & \color{red}{\mathbf{U}_{\mathcal{N}}}
  \end{array} \right]  
^{*}
% 
\end{align}
$$
The Moore-Penrose inverses of the product matrices are
$$
\begin{align}
%% Wx
  \left( \mathbf{A}^{*} \mathbf{A} \right)^{\dagger}
  &= \left( \mathbf{V} \, \Sigma^{\mathrm{T}} \Sigma \, \mathbf{V}^{*} \right)^{\dagger} 
%
  = \left[ \begin{array}{cc}
     \color{blue}{\mathbf{V}_{\mathcal{R}}} & \color{red}{\mathbf{V}_{\mathcal{N}}}
  \end{array} \right]  
 \, 
     \left[ \begin{array}{cc}
       \mathbf{S}^{-2} & \mathbf{0} \\ \mathbf{0} & \mathbf{0}
     \end{array} \right]
       \left[ \begin{array}{cc}
     \color{blue}{\mathbf{V}_{\mathcal{R}}} & \color{red}{\mathbf{V}_{\mathcal{N}}}
  \end{array} \right]  
^{*} \\
%% Wy
  \left( \mathbf{A} \mathbf{A}^{*} \right)^{\dagger}
  &= \left( \mathbf{U} \, \Sigma \Sigma^{\mathrm{T}} \, \mathbf{U}^{*} \right)^{\dagger} 
%
  = \left[ \begin{array}{cc}
     \color{blue}{\mathbf{U}_{\mathcal{R}}} & \color{red}{\mathbf{U}_{\mathcal{N}}}
  \end{array} \right]  
 \, 
     \left[ \begin{array}{cc}
       \mathbf{S}^{-2} & \mathbf{0} \\ \mathbf{0} & \mathbf{0}
     \end{array} \right]
       \left[ \begin{array}{cc}
     \color{blue}{\mathbf{U}_{\mathcal{R}}} & \color{red}{\mathbf{U}_{\mathcal{N}}}
  \end{array} \right]  
^{*}
% 
\end{align}
$$
Connecting to the Drazin inverses
After the associations:


*

*$\mathbf{Q} \to \mathbf{V}$

*$\mathbf{Q}^{-1} \to \mathbf{V}^{*}$

*$\mathbf{C} \to \mathbf{S}^{2}$


the Drazin inverse is
$$
\begin{align}
  \color{green}{\left( \mathbf{A}^{*} \mathbf{A} \right)^{D}}
  = \left( \mathbf{V} \, \Sigma^{\mathrm{T}} \Sigma \, \mathbf{V}^{*} \right)^{D} 
%
   = \left(
     \mathbf{V}
     \left[ \begin{array}{cc}
       \mathbf{S}^{2} & \mathbf{0} \\ \mathbf{0} & \mathbf{0}
     \end{array} \right]
     \mathbf{V}^{*}
\right)^{D}
   = 
     \mathbf{V}
     \left[ \begin{array}{cc}
       \mathbf{S}^{-2} & \mathbf{0} \\ \mathbf{0} & \mathbf{0}
     \end{array} \right]
     \mathbf{V}^{*}
%
 &= \color{green}{\left( \mathbf{A}^{*} \mathbf{A} \right)^{\dagger}} \\
\end{align}
$$
The other Drazin inverse should be straightforward:
$$
\begin{align}
  \color{green}{\left( \mathbf{A} \mathbf{A}^{*} \right)^{D}}
  = \left( \mathbf{V} \, \Sigma^{\mathrm{T}} \Sigma \, \mathbf{V}^{*} \right)^{D} 
  =   \mathbf{U}
     \left[ \begin{array}{cc}
       \mathbf{S}^{-2} & \mathbf{0} \\ \mathbf{0} & \mathbf{0}
     \end{array} \right]
     \mathbf{U}^{*}
%
 &= \color{green}{\left( \mathbf{A} \mathbf{A}^{*} \right)^{\dagger}} \\
\end{align}
$$
Conclusion
Another post derives the different forms of the Moore-Penrose pseudoscience.
What forms does the Moore-Penrose inverse take under systems with full rank, full column rank, and full row rank? Two specific cases are of interest in this post.
Case (1):
The only nontrivial nullspace is $\color{red}{\mathcal{N}_{\mathbf{A}^{*}}}$: the target matrix is overdetermined, more rows than columns, full column rank. The normal equations solution is equivalent to the pseudoinverse:
$$
  \mathbf{A}^{\dagger} = 
\left( \mathbf{A}^{*} \mathbf{A} 
\right)^{-1} \mathbf{A}^{*}
$$
Using an identity from the previous section leaves us with
$$
\mathbf{A}^{\dagger} = 
\left( \mathbf{A}^{*} \mathbf{A} 
\right)^{-1} \mathbf{A}^{*} =
\left( \mathbf{A}^{*} \mathbf{A} 
\right)^{D} \mathbf{A}^{*}
$$
Case (2):
The only nontrivial nullspace is $\color{red}{\mathcal{N}_{\mathbf{A}}}$: the target matrix is underdetermined, more columns than rows, full row rank. The normal equations solution is equivalent to the pseudoinverse:
$$
\mathbf{A}^{\dagger} = 
\mathbf{A}^{*} 
\left( \mathbf{A} \mathbf{A}^{*}
\right)^{-1} =
 \mathbf{A}^{*} 
 \left( \mathbf{A} \mathbf{A}^{*} 
\right)^{D}
$$
Case (3)
Neither nullspace is trivial. To prove the conjecture with the Drazin inverse, start with the Moore-Penrose identity
$$
\mathbf{A}^{\dagger} = 
%
\color{green}{\left(
\mathbf{A}^{*} \mathbf{A}
\right)^{\dagger}} \mathbf{A}^{*} =
%
\mathbf{A}^{*}
\color{green}{\left(
 \mathbf{A} \mathbf{A}^{*}
\right)^{\dagger}}
$$
(The proof of this statement in on p. 27 Regression and the Moore-Penrose pseudoinverse)
The block form manipulations of the SVD verify this statement quickly. Notice there is no assumption about the invertability of the product matrices. Using the $\color{green}{equivalence}$ relationships provides
$$
\mathbf{A}^{\dagger} = 
%
\color{green}{\left(
\mathbf{A}^{*} \mathbf{A}
\right)^{D}} \mathbf{A}^{*} =
%
\mathbf{A}^{*}
\color{green}{\left(
 \mathbf{A} \mathbf{A}^{*}
\right)^{D}}
$$
