# pseudo inverse of a matrix

• Jan 21st 2011, 09:43 PM
pseudo inverse of a matrix
Hi
Let A be an $m \times n$ matrix. An $n\times m$ matrix $A{^\oplus}$ is a pseudoinverse of A if there are matrices $U \ and\ V$ such that

$AA^{\oplus} A= A,$
$A^{\oplus}=UA^{\oplus}=A^{T}V.$
Show that $A^{\oplus}$ exists and is unique.

Whereas i know pseudo inverse is obtained from singular value decomposition (SVD), as any $m \times n$ matrix can be decomposed as

$A=UDV^T$
where $D$ is diagonal matrix and $U, V$ are orthogonal matrices.

So we can find $A^{\oplus}=UD^{\oplus}V^T,$ this stratifies the first property $AA^{\oplus}A = A$.
Where $D^{\oplus}$ is matrix of reciprocal of eigenvalues of $D$.
My Question is how can

1. How can I prove other properties $A^{\oplus}=UA^{\oplus}=A^{T}V$ using this SVD decomposition.

2. Here in the stated problem matrices are not orthogonal, is it OK.

• Jan 21st 2011, 10:01 PM
Drexel28
Quote:

Hi
Let A be an $m \times n$ matrix. An $n\times m$ matrix $A{^\oplus}$ is a pseudoinverse of A if there are matrices $U \ and\ V$ such that

$AA^{\oplus} A= A,$
$A^{\oplus}=UA^{\oplus}=A^{T}V.$
Show that $A^{\oplus}$ exists and is unique.

Whereas i know pseudo inverse is obtained from singular value decomposition (SVD), as any $m \times n$ matrix can be decomposed as

$A=UDV^T$
where $D$ is diagonal matrix and $U, V$ are orthogonal matrices.

So we can find $A^{\oplus}=UD^{\oplus}V^T,$ this stratifies the first property $AA^{\oplus}A = A$.
Where $D^{\oplus}$ is matrix of reciprocal of eigenvalues of $D$.
My Question is how can

1. How can I prove other properties $A^{\oplus}=UA^{\oplus}=A^{T}V$ using this SVD decomposition.

2. Here in the stated problem matrices are not orthogonal, is it OK.