Moore-Penrose generalized inverse

Let $A$ be an $m\times n$ matrix with entries in $\mathbb{C}$ . The Moore-Penrose generalized inverse, denoted by $A^{\dagger}$ , is an $n\times m$ matrix with entries in $\mathbb{C}$ , such that

1.

$AA^{\dagger}A=A$
2.

$A^{\dagger}AA^{\dagger}=A^{\dagger}$
3.

$AA^{\dagger}$ and $A^{\dagger}A$ are both Hermitian

Remarks

•

The Moore-Penrose generalized inverse of a given matrix is unique.
•

If $A^{\dagger}$ is the Moore-Penrose generalized inverse of $A$ , then $(A^{\dagger})^{\operatorname{T}}$ is the Moore-Penrose generalized inverse of $A^{\operatorname{T}}$ .
•
If $A=BC$ such that
1. (a)
  
  $A\in\mathbb{C}^{m\times n}$ , $B\in\mathbb{C}^{m\times r}$ , and $C\in\mathbb{C}^{r\times n}$ ,
2. (b)
  
  $r=\operatorname{rank}(A)=\operatorname{rank}(B)=\operatorname{rank}(C)$ , then
  
  $A^{\dagger}=C^{\ast}(CC^{\ast})^{-1}(B^{\ast}B)^{-1}B^{\ast}.$

For example, let

A=\begin{pmatrix}1&1&i\\ 0&1&0\end{pmatrix}.

Transform $A$ to its row echelon form to get a decomposition of $A=BC$ , where

B=\begin{pmatrix}1&1\\ 0&1\end{pmatrix}\mbox{ and }C=\begin{pmatrix}1&0&i\\ 0&1&0\end{pmatrix}.

It is readily verified that $2=\operatorname{rank}(A)=\operatorname{rank}(B)=\operatorname{rank}(C)$ . So

A^{\dagger}=\frac{1}{2}\begin{pmatrix}1&-1\\ 0&2\\ -i&i\end{pmatrix}.

We check that

AA^{\dagger}=I\mbox{ and }A^{\dagger}A=\frac{1}{2}\begin{pmatrix}1&0&i\\ 0&2&0\\ -i&0&1\end{pmatrix}

are both Hermitian. Furthermore, $AA^{\dagger}A=A$ and $A^{\dagger}AA^{\dagger}=A^{\dagger}$ . So, $A^{\dagger}$ is the Moore-Penrose generalized inverse of $A$ .

Title	Moore-Penrose generalized inverse
Canonical name	MoorePenroseGeneralizedInverse
Date of creation	2013-03-22 14:31:31
Last modified on	2013-03-22 14:31:31
Owner	CWoo (3771)
Last modified by	CWoo (3771)
Numerical id	8
Author	CWoo (3771)
Entry type	Definition
Classification	msc 15A09
Classification	msc 60J10
Synonym	Moore-Penrose pseudoinverse
Related topic	DrazinInverse
Related topic	Pseudoinverse