Properties of matrix inverse
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Suppose $A$ is a $l times l$ matrix and $Gamma$ is a $l times k$ matrix. Is it true that $Gamma[Gamma'AGamma]^{-1}Gamma' = A^{-1}$, assuming $A^{-1}$ exists? If yes, how to show it?
matrices inverse
asked Nov 15 at 15:39
Canine360
229115
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up vote
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Suppose $A$ is a $l times l$ matrix and $Gamma$ is a $l times k$ matrix. Is it true that $Gamma[Gamma'AGamma]^{-1}Gamma' = A^{-1}$, assuming $A^{-1}$ exists? If yes, how to show it?
matrices inverse
asked Nov 15 at 15:39
Canine360
229115
You seem to assume both $A$ and $ Gamma' A Gamma$ are invertible, which requires $ell = k$.
– hardmath
Nov 15 at 15:49
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Suppose $A$ is a $l times l$ matrix and $Gamma$ is a $l times k$ matrix. Is it true that $Gamma[Gamma'AGamma]^{-1}Gamma' = A^{-1}$, assuming $A^{-1}$ exists? If yes, how to show it?
matrices inverse
asked Nov 15 at 15:39
Canine360
229115
Suppose $A$ is a $l times l$ matrix and $Gamma$ is a $l times k$ matrix. Is it true that $Gamma[Gamma'AGamma]^{-1}Gamma' = A^{-1}$, assuming $A^{-1}$ exists? If yes, how to show it?
matrices inverse
matrices inverse
asked Nov 15 at 15:39
Canine360
229115
asked Nov 15 at 15:39
Canine360
229115
asked Nov 15 at 15:39
Canine360
229115
asked Nov 15 at 15:39
Canine360
229115
asked Nov 15 at 15:39
Canine360
229115
229115
You seem to assume both $A$ and $ Gamma' A Gamma$ are invertible, which requires $ell = k$.
– hardmath
Nov 15 at 15:49
add a comment |
You seem to assume both $A$ and $ Gamma' A Gamma$ are invertible, which requires $ell = k$.
– hardmath
Nov 15 at 15:49
You seem to assume both $A$ and $ Gamma' A Gamma$ are invertible, which requires $ell = k$.
– hardmath
Nov 15 at 15:49
You seem to assume both $A$ and $ Gamma' A Gamma$ are invertible, which requires $ell = k$.
– hardmath
Nov 15 at 15:49
add a comment |
1 Answer
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There are clearly counterexamples: for example take $Gamma$ to be the zero matrix. Then $Gamma'AGamma$ is zero, so is not invertible, so the left hand side of your equation doesn't exist.
If we add an assumption that $Gamma'AGamma$ is invertible, then, in particular, we must have $k geq l$ (the dimension of the image is at most $k$), and $k leq l$ (the dimension of the kernel is at least $k - l$), so $l = k$, and $Gamma$ is an invertible $ltimes l$ matrix, so we have $Gamma[Gamma'AGamma]^{-1}Gamma' = Gamma[Gamma^{-1}A^{-1}Gamma'^{-1}]Gamma' = (GammaGamma^{-1})A^{-1}(Gamma'^{-1}Gamma') = A^{-1}$.
answered Nov 15 at 15:52
user3482749
1,709411
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
There are clearly counterexamples: for example take $Gamma$ to be the zero matrix. Then $Gamma'AGamma$ is zero, so is not invertible, so the left hand side of your equation doesn't exist.
If we add an assumption that $Gamma'AGamma$ is invertible, then, in particular, we must have $k geq l$ (the dimension of the image is at most $k$), and $k leq l$ (the dimension of the kernel is at least $k - l$), so $l = k$, and $Gamma$ is an invertible $ltimes l$ matrix, so we have $Gamma[Gamma'AGamma]^{-1}Gamma' = Gamma[Gamma^{-1}A^{-1}Gamma'^{-1}]Gamma' = (GammaGamma^{-1})A^{-1}(Gamma'^{-1}Gamma') = A^{-1}$.
answered Nov 15 at 15:52
user3482749
1,709411
add a comment |
up vote
1
down vote
accepted
There are clearly counterexamples: for example take $Gamma$ to be the zero matrix. Then $Gamma'AGamma$ is zero, so is not invertible, so the left hand side of your equation doesn't exist.
If we add an assumption that $Gamma'AGamma$ is invertible, then, in particular, we must have $k geq l$ (the dimension of the image is at most $k$), and $k leq l$ (the dimension of the kernel is at least $k - l$), so $l = k$, and $Gamma$ is an invertible $ltimes l$ matrix, so we have $Gamma[Gamma'AGamma]^{-1}Gamma' = Gamma[Gamma^{-1}A^{-1}Gamma'^{-1}]Gamma' = (GammaGamma^{-1})A^{-1}(Gamma'^{-1}Gamma') = A^{-1}$.
answered Nov 15 at 15:52
user3482749
1,709411
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
There are clearly counterexamples: for example take $Gamma$ to be the zero matrix. Then $Gamma'AGamma$ is zero, so is not invertible, so the left hand side of your equation doesn't exist.
If we add an assumption that $Gamma'AGamma$ is invertible, then, in particular, we must have $k geq l$ (the dimension of the image is at most $k$), and $k leq l$ (the dimension of the kernel is at least $k - l$), so $l = k$, and $Gamma$ is an invertible $ltimes l$ matrix, so we have $Gamma[Gamma'AGamma]^{-1}Gamma' = Gamma[Gamma^{-1}A^{-1}Gamma'^{-1}]Gamma' = (GammaGamma^{-1})A^{-1}(Gamma'^{-1}Gamma') = A^{-1}$.
answered Nov 15 at 15:52
user3482749
1,709411
There are clearly counterexamples: for example take $Gamma$ to be the zero matrix. Then $Gamma'AGamma$ is zero, so is not invertible, so the left hand side of your equation doesn't exist.
If we add an assumption that $Gamma'AGamma$ is invertible, then, in particular, we must have $k geq l$ (the dimension of the image is at most $k$), and $k leq l$ (the dimension of the kernel is at least $k - l$), so $l = k$, and $Gamma$ is an invertible $ltimes l$ matrix, so we have $Gamma[Gamma'AGamma]^{-1}Gamma' = Gamma[Gamma^{-1}A^{-1}Gamma'^{-1}]Gamma' = (GammaGamma^{-1})A^{-1}(Gamma'^{-1}Gamma') = A^{-1}$.
answered Nov 15 at 15:52
user3482749
1,709411
answered Nov 15 at 15:52
user3482749
1,709411
answered Nov 15 at 15:52
user3482749
1,709411
answered Nov 15 at 15:52
user3482749
1,709411
1,709411
add a comment |
add a comment |
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You seem to assume both $A$ and $ Gamma' A Gamma$ are invertible, which requires $ell = k$.
– hardmath
Nov 15 at 15:49