Pseudoinverse of pseudoinverse of matrix A equals A: $(A^+)^+=A$Must a pseudoinverse of a von Neumann regular element be regular?What is the Moore-Penrose pseudoinverse for scaled linear regression?Nilpotent Pseudoinversepseudoinverse of vec-transpose operatorPseudoinverse and orthogonal projectionRepresentation of the Orthogonal ProjectionPseudoinverse of $I-MM^+$pseudo inverse with the minimum $l_2$ norm for each columnUniqueness of matrix pseudoinverseDo fewer axioms suffice to define the Moore-Penrose pseudoinverse? (motivated by least squares method and group theory)
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Pseudoinverse of pseudoinverse of matrix A equals A: $(A^+)^+=A$
Must a pseudoinverse of a von Neumann regular element be regular?What is the Moore-Penrose pseudoinverse for scaled linear regression?Nilpotent Pseudoinversepseudoinverse of vec-transpose operatorPseudoinverse and orthogonal projectionRepresentation of the Orthogonal ProjectionPseudoinverse of $I-MM^+$pseudo inverse with the minimum $l_2$ norm for each columnUniqueness of matrix pseudoinverseDo fewer axioms suffice to define the Moore-Penrose pseudoinverse? (motivated by least squares method and group theory)
$begingroup$
As you know, a Matrix $A^+in mathbbR^mtimes n$ is called a a pseudoinverse of $Ain mathbbR^ntimes m$ if $Vert b-A A^+ b Vert_2leqVert b- A yVert_2 forall bin mathbbR^n forall yin mathbbR^m$ and $A^+b$ is the smallest vector by norm to do so, which is equally characterized by $langle b- A A^+b,Ayrangle=0$, since $A A^+$ is the orthogonal projection of $mathbbR^n$ onto $R(A)$.
Because of it being a orthogonal projection the penrose axioms $AA^+^T=AA^+$ and $A A^+A=A$ follow. What I would like to do now is proof the other two axioms $A^+A A^+=A^+$ and $A^+ A^T= A^+A$ given a Matrix and its pseudoinverse, by showing that $(A^+)^+=A$ or in other words, that $A^+A$ is a projection onto $R(A^+)=N(A)^perp$. On the other hand, if those two penrose-axioms could be proven another way $(A^+)^+=A$ would follow by the penrose-axioms.
I would appreciate any proof - especially one that is not using SVD.
pseudoinverse numerical-calculus
asked Mar 11 at 10:04
Martin ErhardtMartin Erhardt
21019
$endgroup$
add a comment |
$begingroup$
As you know, a Matrix $A^+in mathbbR^mtimes n$ is called a a pseudoinverse of $Ain mathbbR^ntimes m$ if $Vert b-A A^+ b Vert_2leqVert b- A yVert_2 forall bin mathbbR^n forall yin mathbbR^m$ and $A^+b$ is the smallest vector by norm to do so, which is equally characterized by $langle b- A A^+b,Ayrangle=0$, since $A A^+$ is the orthogonal projection of $mathbbR^n$ onto $R(A)$.
Because of it being a orthogonal projection the penrose axioms $AA^+^T=AA^+$ and $A A^+A=A$ follow. What I would like to do now is proof the other two axioms $A^+A A^+=A^+$ and $A^+ A^T= A^+A$ given a Matrix and its pseudoinverse, by showing that $(A^+)^+=A$ or in other words, that $A^+A$ is a projection onto $R(A^+)=N(A)^perp$. On the other hand, if those two penrose-axioms could be proven another way $(A^+)^+=A$ would follow by the penrose-axioms.
I would appreciate any proof - especially one that is not using SVD.
pseudoinverse numerical-calculus
asked Mar 11 at 10:04
Martin ErhardtMartin Erhardt
21019
$endgroup$
add a comment |
$begingroup$
As you know, a Matrix $A^+in mathbbR^mtimes n$ is called a a pseudoinverse of $Ain mathbbR^ntimes m$ if $Vert b-A A^+ b Vert_2leqVert b- A yVert_2 forall bin mathbbR^n forall yin mathbbR^m$ and $A^+b$ is the smallest vector by norm to do so, which is equally characterized by $langle b- A A^+b,Ayrangle=0$, since $A A^+$ is the orthogonal projection of $mathbbR^n$ onto $R(A)$.
Because of it being a orthogonal projection the penrose axioms $AA^+^T=AA^+$ and $A A^+A=A$ follow. What I would like to do now is proof the other two axioms $A^+A A^+=A^+$ and $A^+ A^T= A^+A$ given a Matrix and its pseudoinverse, by showing that $(A^+)^+=A$ or in other words, that $A^+A$ is a projection onto $R(A^+)=N(A)^perp$. On the other hand, if those two penrose-axioms could be proven another way $(A^+)^+=A$ would follow by the penrose-axioms.
I would appreciate any proof - especially one that is not using SVD.
pseudoinverse numerical-calculus
asked Mar 11 at 10:04
Martin ErhardtMartin Erhardt
21019
$endgroup$
As you know, a Matrix $A^+in mathbbR^mtimes n$ is called a a pseudoinverse of $Ain mathbbR^ntimes m$ if $Vert b-A A^+ b Vert_2leqVert b- A yVert_2 forall bin mathbbR^n forall yin mathbbR^m$ and $A^+b$ is the smallest vector by norm to do so, which is equally characterized by $langle b- A A^+b,Ayrangle=0$, since $A A^+$ is the orthogonal projection of $mathbbR^n$ onto $R(A)$.
Because of it being a orthogonal projection the penrose axioms $AA^+^T=AA^+$ and $A A^+A=A$ follow. What I would like to do now is proof the other two axioms $A^+A A^+=A^+$ and $A^+ A^T= A^+A$ given a Matrix and its pseudoinverse, by showing that $(A^+)^+=A$ or in other words, that $A^+A$ is a projection onto $R(A^+)=N(A)^perp$. On the other hand, if those two penrose-axioms could be proven another way $(A^+)^+=A$ would follow by the penrose-axioms.
I would appreciate any proof - especially one that is not using SVD.
pseudoinverse numerical-calculus
pseudoinverse numerical-calculus
asked Mar 11 at 10:04
Martin ErhardtMartin Erhardt
21019
asked Mar 11 at 10:04
Martin ErhardtMartin Erhardt
21019
edited Mar 11 at 10:49
Martin Erhardt
asked Mar 11 at 10:04
Martin ErhardtMartin Erhardt
21019
asked Mar 11 at 10:04
Martin ErhardtMartin Erhardt
21019
asked Mar 11 at 10:04
Martin ErhardtMartin Erhardt
21019
21019
add a comment |
add a comment |
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$begingroup$
$A(b-A^+Ab)=0Rightarrow b-A^+Abin $N(A)$Rightarrow langle b-A^+Ab,A^+yrangle=0$ since $A^+y in R(A^+)=N(A)^perp$
answered Mar 11 at 10:16
Martin ErhardtMartin Erhardt
21019
$endgroup$
add a comment |
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$begingroup$
$A(b-A^+Ab)=0Rightarrow b-A^+Abin $N(A)$Rightarrow langle b-A^+Ab,A^+yrangle=0$ since $A^+y in R(A^+)=N(A)^perp$
answered Mar 11 at 10:16
Martin ErhardtMartin Erhardt
21019
$endgroup$
add a comment |
$begingroup$
$A(b-A^+Ab)=0Rightarrow b-A^+Abin $N(A)$Rightarrow langle b-A^+Ab,A^+yrangle=0$ since $A^+y in R(A^+)=N(A)^perp$
answered Mar 11 at 10:16
Martin ErhardtMartin Erhardt
21019
$endgroup$
add a comment |
$begingroup$
$A(b-A^+Ab)=0Rightarrow b-A^+Abin $N(A)$Rightarrow langle b-A^+Ab,A^+yrangle=0$ since $A^+y in R(A^+)=N(A)^perp$
answered Mar 11 at 10:16
Martin ErhardtMartin Erhardt
21019
$endgroup$
$A(b-A^+Ab)=0Rightarrow b-A^+Abin $N(A)$Rightarrow langle b-A^+Ab,A^+yrangle=0$ since $A^+y in R(A^+)=N(A)^perp$
answered Mar 11 at 10:16
Martin ErhardtMartin Erhardt
21019
answered Mar 11 at 10:16
Martin ErhardtMartin Erhardt
21019
answered Mar 11 at 10:16
Martin ErhardtMartin Erhardt
21019
answered Mar 11 at 10:16
Martin ErhardtMartin Erhardt
21019
21019
add a comment |
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