# pseudo inverse of a matrix

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• January 21st 2011, 08:43 PM
nadia321
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.

Thanks in advance
• January 21st 2011, 09:01 PM
Drexel28
Quote:

Originally Posted by nadia321
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.

Thanks in advance

Do you have to use the singular value decomposition (I hate it! Bad experience)? Just in case you want to google information this is often called the Moore-Penrose pseudoinverse.
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