# 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. 1.

$AA^{\dagger}A=A$

2. 2.

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

3. 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 MoorePenroseGeneralizedInverse 2013-03-22 14:31:31 2013-03-22 14:31:31 CWoo (3771) CWoo (3771) 8 CWoo (3771) Definition msc 15A09 msc 60J10 Moore-Penrose pseudoinverse DrazinInverse Pseudoinverse