What are all the $A$-invariant subspaces of $mathbb{C}^n$
$begingroup$
Let $ngeq 1$.
Let $A$ be an invertible linear map $mathbb{C}^nrightarrowmathbb{C}^n$.
I would like to recognize all the $A$-invariant subspaces.
So, I consider the Jordan normal form of $A$. Each Jordan block of dimensions $dtimes d$ gives a chain of $d$ nested $A$-invariant subspaces. I call these the "basic $A$-invariant subspaces".
Every direct sum of basic $A$-invariant subspaces is $A$-invariant.
Is every $A$-invariant subspace a direct sum of basic $A$-invariant subspaces?
linear-algebra
asked Jun 15 '16 at 11:34
TerryTerry
1167
$endgroup$
add a comment |
$begingroup$
Let $ngeq 1$.
Let $A$ be an invertible linear map $mathbb{C}^nrightarrowmathbb{C}^n$.
I would like to recognize all the $A$-invariant subspaces.
So, I consider the Jordan normal form of $A$. Each Jordan block of dimensions $dtimes d$ gives a chain of $d$ nested $A$-invariant subspaces. I call these the "basic $A$-invariant subspaces".
Every direct sum of basic $A$-invariant subspaces is $A$-invariant.
Is every $A$-invariant subspace a direct sum of basic $A$-invariant subspaces?
linear-algebra
asked Jun 15 '16 at 11:34
TerryTerry
1167
$endgroup$
add a comment |
$begingroup$
Let $ngeq 1$.
Let $A$ be an invertible linear map $mathbb{C}^nrightarrowmathbb{C}^n$.
I would like to recognize all the $A$-invariant subspaces.
So, I consider the Jordan normal form of $A$. Each Jordan block of dimensions $dtimes d$ gives a chain of $d$ nested $A$-invariant subspaces. I call these the "basic $A$-invariant subspaces".
Every direct sum of basic $A$-invariant subspaces is $A$-invariant.
Is every $A$-invariant subspace a direct sum of basic $A$-invariant subspaces?
linear-algebra
asked Jun 15 '16 at 11:34
TerryTerry
1167
$endgroup$
Let $ngeq 1$.
Let $A$ be an invertible linear map $mathbb{C}^nrightarrowmathbb{C}^n$.
I would like to recognize all the $A$-invariant subspaces.
So, I consider the Jordan normal form of $A$. Each Jordan block of dimensions $dtimes d$ gives a chain of $d$ nested $A$-invariant subspaces. I call these the "basic $A$-invariant subspaces".
Every direct sum of basic $A$-invariant subspaces is $A$-invariant.
Is every $A$-invariant subspace a direct sum of basic $A$-invariant subspaces?
linear-algebra
linear-algebra
asked Jun 15 '16 at 11:34
TerryTerry
1167
asked Jun 15 '16 at 11:34
TerryTerry
1167
asked Jun 15 '16 at 11:34
TerryTerry
1167
asked Jun 15 '16 at 11:34
TerryTerry
1167
asked Jun 15 '16 at 11:34
TerryTerry
1167
1167
add a comment |
add a comment |
2 Answers
2
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oldest
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$begingroup$
Yes, up to the change of coordinates giving the Jordan form (which of course need not be unique) and up to the space ${0}$.
And why do you want $A$ to be invertible? It is not needed.
answered Jun 15 '16 at 14:01
John BJohn B
12.2k51840
$endgroup$
$begingroup$
Can you please provide or sketch a proof?
$endgroup$
– Terry
Jun 16 '16 at 6:47
add a comment |
$begingroup$
The answer is NO. In other words, the answer of John B is false.
Let $Ain M_n(mathbb{C})$ s.t. every $A$-invariant subspace is a direct sum of basic A-invariant subspaces. Since a Jordan block has a finite number of invariant subspaces, $A$ has a finite number of invariant subspaces; this property is equivalent to "$A$ is a cyclic matrix" or equivalent to "$A$ admits exactly one Jordan block associated to each of its eigenvalues".
For example: i) $lambda I_2$ has two Jordan blocks associated to $lambda$ and admits an infinity of invariant subspaces.
ii) $A=diag(J_2,J_2)$ where $J_2$ is the nilpotent Jordan block of dimension $2$. Then $span(ae_1+be_3)$ is an invariant line for each $(a,b)in mathbb{C}^2setminus {(0,0)}$.
answered Dec 28 '18 at 23:25
loup blancloup blanc
23.3k21851
$endgroup$
$begingroup$
Who is the ignorant who has downvoted my post ?
$endgroup$
– loup blanc
Feb 8 at 9:27
add a comment |
Your Answer
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes, up to the change of coordinates giving the Jordan form (which of course need not be unique) and up to the space ${0}$.
And why do you want $A$ to be invertible? It is not needed.
answered Jun 15 '16 at 14:01
John BJohn B
12.2k51840
$endgroup$
$begingroup$
Can you please provide or sketch a proof?
$endgroup$
– Terry
Jun 16 '16 at 6:47
add a comment |
$begingroup$
Yes, up to the change of coordinates giving the Jordan form (which of course need not be unique) and up to the space ${0}$.
And why do you want $A$ to be invertible? It is not needed.
answered Jun 15 '16 at 14:01
John BJohn B
12.2k51840
$endgroup$
$begingroup$
Can you please provide or sketch a proof?
$endgroup$
– Terry
Jun 16 '16 at 6:47
add a comment |
$begingroup$
Yes, up to the change of coordinates giving the Jordan form (which of course need not be unique) and up to the space ${0}$.
And why do you want $A$ to be invertible? It is not needed.
answered Jun 15 '16 at 14:01
John BJohn B
12.2k51840
$endgroup$
Yes, up to the change of coordinates giving the Jordan form (which of course need not be unique) and up to the space ${0}$.
And why do you want $A$ to be invertible? It is not needed.
answered Jun 15 '16 at 14:01
John BJohn B
12.2k51840
answered Jun 15 '16 at 14:01
John BJohn B
12.2k51840
answered Jun 15 '16 at 14:01
John BJohn B
12.2k51840
answered Jun 15 '16 at 14:01
John BJohn B
12.2k51840
12.2k51840
$begingroup$
Can you please provide or sketch a proof?
$endgroup$
– Terry
Jun 16 '16 at 6:47
add a comment |
$begingroup$
Can you please provide or sketch a proof?
$endgroup$
– Terry
Jun 16 '16 at 6:47
$begingroup$
Can you please provide or sketch a proof?
$endgroup$
– Terry
Jun 16 '16 at 6:47
$begingroup$
Can you please provide or sketch a proof?
$endgroup$
– Terry
Jun 16 '16 at 6:47
add a comment |
$begingroup$
The answer is NO. In other words, the answer of John B is false.
Let $Ain M_n(mathbb{C})$ s.t. every $A$-invariant subspace is a direct sum of basic A-invariant subspaces. Since a Jordan block has a finite number of invariant subspaces, $A$ has a finite number of invariant subspaces; this property is equivalent to "$A$ is a cyclic matrix" or equivalent to "$A$ admits exactly one Jordan block associated to each of its eigenvalues".
For example: i) $lambda I_2$ has two Jordan blocks associated to $lambda$ and admits an infinity of invariant subspaces.
ii) $A=diag(J_2,J_2)$ where $J_2$ is the nilpotent Jordan block of dimension $2$. Then $span(ae_1+be_3)$ is an invariant line for each $(a,b)in mathbb{C}^2setminus {(0,0)}$.
answered Dec 28 '18 at 23:25
loup blancloup blanc
23.3k21851
$endgroup$
$begingroup$
Who is the ignorant who has downvoted my post ?
$endgroup$
– loup blanc
Feb 8 at 9:27
add a comment |
$begingroup$
The answer is NO. In other words, the answer of John B is false.
Let $Ain M_n(mathbb{C})$ s.t. every $A$-invariant subspace is a direct sum of basic A-invariant subspaces. Since a Jordan block has a finite number of invariant subspaces, $A$ has a finite number of invariant subspaces; this property is equivalent to "$A$ is a cyclic matrix" or equivalent to "$A$ admits exactly one Jordan block associated to each of its eigenvalues".
For example: i) $lambda I_2$ has two Jordan blocks associated to $lambda$ and admits an infinity of invariant subspaces.
ii) $A=diag(J_2,J_2)$ where $J_2$ is the nilpotent Jordan block of dimension $2$. Then $span(ae_1+be_3)$ is an invariant line for each $(a,b)in mathbb{C}^2setminus {(0,0)}$.
answered Dec 28 '18 at 23:25
loup blancloup blanc
23.3k21851
$endgroup$
$begingroup$
Who is the ignorant who has downvoted my post ?
$endgroup$
– loup blanc
Feb 8 at 9:27
add a comment |
$begingroup$
The answer is NO. In other words, the answer of John B is false.
Let $Ain M_n(mathbb{C})$ s.t. every $A$-invariant subspace is a direct sum of basic A-invariant subspaces. Since a Jordan block has a finite number of invariant subspaces, $A$ has a finite number of invariant subspaces; this property is equivalent to "$A$ is a cyclic matrix" or equivalent to "$A$ admits exactly one Jordan block associated to each of its eigenvalues".
For example: i) $lambda I_2$ has two Jordan blocks associated to $lambda$ and admits an infinity of invariant subspaces.
ii) $A=diag(J_2,J_2)$ where $J_2$ is the nilpotent Jordan block of dimension $2$. Then $span(ae_1+be_3)$ is an invariant line for each $(a,b)in mathbb{C}^2setminus {(0,0)}$.
answered Dec 28 '18 at 23:25
loup blancloup blanc
23.3k21851
$endgroup$
The answer is NO. In other words, the answer of John B is false.
Let $Ain M_n(mathbb{C})$ s.t. every $A$-invariant subspace is a direct sum of basic A-invariant subspaces. Since a Jordan block has a finite number of invariant subspaces, $A$ has a finite number of invariant subspaces; this property is equivalent to "$A$ is a cyclic matrix" or equivalent to "$A$ admits exactly one Jordan block associated to each of its eigenvalues".
For example: i) $lambda I_2$ has two Jordan blocks associated to $lambda$ and admits an infinity of invariant subspaces.
ii) $A=diag(J_2,J_2)$ where $J_2$ is the nilpotent Jordan block of dimension $2$. Then $span(ae_1+be_3)$ is an invariant line for each $(a,b)in mathbb{C}^2setminus {(0,0)}$.
answered Dec 28 '18 at 23:25
loup blancloup blanc
23.3k21851
edited Dec 28 '18 at 23:31
answered Dec 28 '18 at 23:25
loup blancloup blanc
23.3k21851
answered Dec 28 '18 at 23:25
loup blancloup blanc
23.3k21851
answered Dec 28 '18 at 23:25
loup blancloup blanc
23.3k21851
23.3k21851
$begingroup$
Who is the ignorant who has downvoted my post ?
$endgroup$
– loup blanc
Feb 8 at 9:27
add a comment |
$begingroup$
Who is the ignorant who has downvoted my post ?
$endgroup$
– loup blanc
Feb 8 at 9:27
$begingroup$
Who is the ignorant who has downvoted my post ?
$endgroup$
– loup blanc
Feb 8 at 9:27
$begingroup$
Who is the ignorant who has downvoted my post ?
$endgroup$
– loup blanc
Feb 8 at 9:27
add a comment |
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