Inverting $B + TCT'$ if $T$ is of full rank
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In Leeuw & Meijer (2008) 'Handbook of multilevel analysis', I found the following theorem (p. 72):
If $A = B + TCT'$ with A and B positive definite, then
$A^{-1} = B^{-1} - B^{-1}T(C^{-1}+T'B^{-1}T)^{-1}T'B^{-1}$
(note: In the handbook, the bracket term is written as $(C^{-1}+T'C^{-1}T)^{-1}$ but think this is a typo. See: Henderson et al. 1980: On Deriving The Inverse Of a Sum of Matrices, equation 17).
Anyway, the theorem 1.2 is clear. But the authors proceed and state that if, in addition, $T$ is of full rank, then:
$A^{-1} = T(T'T)^{-1}(C+(T'B^{-1}T)^{-1})^{-1}(T'T)^{-1}T'+[B^{-1}-B^{-1}T(T'B^{-1}T)^{-1}T'B^{-1}]$
How to derive this identity?
linear-algebra matrices
asked Dec 12 '18 at 16:40
DomBDomB
205
$endgroup$
|
show 4 more comments
$begingroup$
In Leeuw & Meijer (2008) 'Handbook of multilevel analysis', I found the following theorem (p. 72):
If $A = B + TCT'$ with A and B positive definite, then
$A^{-1} = B^{-1} - B^{-1}T(C^{-1}+T'B^{-1}T)^{-1}T'B^{-1}$
(note: In the handbook, the bracket term is written as $(C^{-1}+T'C^{-1}T)^{-1}$ but think this is a typo. See: Henderson et al. 1980: On Deriving The Inverse Of a Sum of Matrices, equation 17).
Anyway, the theorem 1.2 is clear. But the authors proceed and state that if, in addition, $T$ is of full rank, then:
$A^{-1} = T(T'T)^{-1}(C+(T'B^{-1}T)^{-1})^{-1}(T'T)^{-1}T'+[B^{-1}-B^{-1}T(T'B^{-1}T)^{-1}T'B^{-1}]$
How to derive this identity?
linear-algebra matrices
asked Dec 12 '18 at 16:40
DomBDomB
205
$endgroup$
$begingroup$
I fixed your first statement, since it differed from Theorem 1.2 of your quoted source
$endgroup$
– LinAlg
Dec 12 '18 at 17:49
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My version of Leeuw & Meijer has a reference with a proof for this statement btw.
$endgroup$
– LinAlg
Dec 12 '18 at 17:53
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It is known as the Woodbury Matrix Identity en.wikipedia.org/wiki/Woodbury_matrix_identity
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– Jean Marie
Dec 12 '18 at 18:59
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@LinAlg, it does mention a reference. A report from 1993 that has not been published in any journal. Despite some internet research I was not able to find this reference. Have you?
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– DomB
Dec 16 '18 at 8:57
$begingroup$
@LinAlg, the way theorem 1.2 is presented in the handbook is wrong. I am sure it is a typo.See en.wikipedia.org/wiki/Woodbury_matrix_identity. This is why I presented it the way I did.
$endgroup$
– DomB
Dec 16 '18 at 9:04
|
show 4 more comments
$begingroup$
In Leeuw & Meijer (2008) 'Handbook of multilevel analysis', I found the following theorem (p. 72):
If $A = B + TCT'$ with A and B positive definite, then
$A^{-1} = B^{-1} - B^{-1}T(C^{-1}+T'B^{-1}T)^{-1}T'B^{-1}$
(note: In the handbook, the bracket term is written as $(C^{-1}+T'C^{-1}T)^{-1}$ but think this is a typo. See: Henderson et al. 1980: On Deriving The Inverse Of a Sum of Matrices, equation 17).
Anyway, the theorem 1.2 is clear. But the authors proceed and state that if, in addition, $T$ is of full rank, then:
$A^{-1} = T(T'T)^{-1}(C+(T'B^{-1}T)^{-1})^{-1}(T'T)^{-1}T'+[B^{-1}-B^{-1}T(T'B^{-1}T)^{-1}T'B^{-1}]$
How to derive this identity?
linear-algebra matrices
asked Dec 12 '18 at 16:40
DomBDomB
205
$endgroup$
In Leeuw & Meijer (2008) 'Handbook of multilevel analysis', I found the following theorem (p. 72):
If $A = B + TCT'$ with A and B positive definite, then
$A^{-1} = B^{-1} - B^{-1}T(C^{-1}+T'B^{-1}T)^{-1}T'B^{-1}$
(note: In the handbook, the bracket term is written as $(C^{-1}+T'C^{-1}T)^{-1}$ but think this is a typo. See: Henderson et al. 1980: On Deriving The Inverse Of a Sum of Matrices, equation 17).
Anyway, the theorem 1.2 is clear. But the authors proceed and state that if, in addition, $T$ is of full rank, then:
$A^{-1} = T(T'T)^{-1}(C+(T'B^{-1}T)^{-1})^{-1}(T'T)^{-1}T'+[B^{-1}-B^{-1}T(T'B^{-1}T)^{-1}T'B^{-1}]$
How to derive this identity?
linear-algebra matrices
linear-algebra matrices
asked Dec 12 '18 at 16:40
DomBDomB
205
asked Dec 12 '18 at 16:40
DomBDomB
205
edited Dec 16 '18 at 10:40
DomB
asked Dec 12 '18 at 16:40
DomBDomB
205
asked Dec 12 '18 at 16:40
DomBDomB
205
asked Dec 12 '18 at 16:40
DomBDomB
205
205
$begingroup$
I fixed your first statement, since it differed from Theorem 1.2 of your quoted source
$endgroup$
– LinAlg
Dec 12 '18 at 17:49
$begingroup$
My version of Leeuw & Meijer has a reference with a proof for this statement btw.
$endgroup$
– LinAlg
Dec 12 '18 at 17:53
$begingroup$
It is known as the Woodbury Matrix Identity en.wikipedia.org/wiki/Woodbury_matrix_identity
$endgroup$
– Jean Marie
Dec 12 '18 at 18:59
$begingroup$
@LinAlg, it does mention a reference. A report from 1993 that has not been published in any journal. Despite some internet research I was not able to find this reference. Have you?
$endgroup$
– DomB
Dec 16 '18 at 8:57
$begingroup$
@LinAlg, the way theorem 1.2 is presented in the handbook is wrong. I am sure it is a typo.See en.wikipedia.org/wiki/Woodbury_matrix_identity. This is why I presented it the way I did.
$endgroup$
– DomB
Dec 16 '18 at 9:04
|
show 4 more comments
$begingroup$
I fixed your first statement, since it differed from Theorem 1.2 of your quoted source
$endgroup$
– LinAlg
Dec 12 '18 at 17:49
$begingroup$
My version of Leeuw & Meijer has a reference with a proof for this statement btw.
$endgroup$
– LinAlg
Dec 12 '18 at 17:53
$begingroup$
It is known as the Woodbury Matrix Identity en.wikipedia.org/wiki/Woodbury_matrix_identity
$endgroup$
– Jean Marie
Dec 12 '18 at 18:59
$begingroup$
@LinAlg, it does mention a reference. A report from 1993 that has not been published in any journal. Despite some internet research I was not able to find this reference. Have you?
$endgroup$
– DomB
Dec 16 '18 at 8:57
$begingroup$
@LinAlg, the way theorem 1.2 is presented in the handbook is wrong. I am sure it is a typo.See en.wikipedia.org/wiki/Woodbury_matrix_identity. This is why I presented it the way I did.
$endgroup$
– DomB
Dec 16 '18 at 9:04
$begingroup$
I fixed your first statement, since it differed from Theorem 1.2 of your quoted source
$endgroup$
– LinAlg
Dec 12 '18 at 17:49
$begingroup$
I fixed your first statement, since it differed from Theorem 1.2 of your quoted source
$endgroup$
– LinAlg
Dec 12 '18 at 17:49
$begingroup$
My version of Leeuw & Meijer has a reference with a proof for this statement btw.
$endgroup$
– LinAlg
Dec 12 '18 at 17:53
$begingroup$
My version of Leeuw & Meijer has a reference with a proof for this statement btw.
$endgroup$
– LinAlg
Dec 12 '18 at 17:53
$begingroup$
It is known as the Woodbury Matrix Identity en.wikipedia.org/wiki/Woodbury_matrix_identity
$endgroup$
– Jean Marie
Dec 12 '18 at 18:59
$begingroup$
It is known as the Woodbury Matrix Identity en.wikipedia.org/wiki/Woodbury_matrix_identity
$endgroup$
– Jean Marie
Dec 12 '18 at 18:59
$begingroup$
@LinAlg, it does mention a reference. A report from 1993 that has not been published in any journal. Despite some internet research I was not able to find this reference. Have you?
$endgroup$
– DomB
Dec 16 '18 at 8:57
$begingroup$
@LinAlg, it does mention a reference. A report from 1993 that has not been published in any journal. Despite some internet research I was not able to find this reference. Have you?
$endgroup$
– DomB
Dec 16 '18 at 8:57
$begingroup$
@LinAlg, the way theorem 1.2 is presented in the handbook is wrong. I am sure it is a typo.See en.wikipedia.org/wiki/Woodbury_matrix_identity. This is why I presented it the way I did.
$endgroup$
– DomB
Dec 16 '18 at 9:04
$begingroup$
@LinAlg, the way theorem 1.2 is presented in the handbook is wrong. I am sure it is a typo.See en.wikipedia.org/wiki/Woodbury_matrix_identity. This is why I presented it the way I did.
$endgroup$
– DomB
Dec 16 '18 at 9:04
|
show 4 more comments
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$begingroup$
I fixed your first statement, since it differed from Theorem 1.2 of your quoted source
$endgroup$
– LinAlg
Dec 12 '18 at 17:49
$begingroup$
My version of Leeuw & Meijer has a reference with a proof for this statement btw.
$endgroup$
– LinAlg
Dec 12 '18 at 17:53
$begingroup$
It is known as the Woodbury Matrix Identity en.wikipedia.org/wiki/Woodbury_matrix_identity
$endgroup$
– Jean Marie
Dec 12 '18 at 18:59
$begingroup$
@LinAlg, it does mention a reference. A report from 1993 that has not been published in any journal. Despite some internet research I was not able to find this reference. Have you?
$endgroup$
– DomB
Dec 16 '18 at 8:57
$begingroup$
@LinAlg, the way theorem 1.2 is presented in the handbook is wrong. I am sure it is a typo.See en.wikipedia.org/wiki/Woodbury_matrix_identity. This is why I presented it the way I did.
$endgroup$
– DomB
Dec 16 '18 at 9:04