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 $A$ .