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An $n times n$ matrix $A$ naturally induces a mapping on $Grass(p,n)$
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I start reading about matrix manifolds and in an introduction section it is mentioned the statement below. Is it possible to provide an intuition in terms of basic (e.g. fundamental sub-spaces) linear algebra terms?
An $n times n$ matrix $A$ naturally induces a mapping on $Grass(p,n)$ defined by $$mathcalSin Grass(p,n) mapsto AmathcalS:=Ay : y in mathcalS $$ where $Grass(p,n)$ admits a structure of manifold called the Grassmann manifold.
Thank you in advance!
linear-algebra general-topology geometric-topology
asked Jul 30 at 11:25
darkmoor
263210
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up vote
1
down vote
favorite
I start reading about matrix manifolds and in an introduction section it is mentioned the statement below. Is it possible to provide an intuition in terms of basic (e.g. fundamental sub-spaces) linear algebra terms?
An $n times n$ matrix $A$ naturally induces a mapping on $Grass(p,n)$ defined by $$mathcalSin Grass(p,n) mapsto AmathcalS:=Ay : y in mathcalS $$ where $Grass(p,n)$ admits a structure of manifold called the Grassmann manifold.
Thank you in advance!
linear-algebra general-topology geometric-topology
asked Jul 30 at 11:25
darkmoor
263210
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I start reading about matrix manifolds and in an introduction section it is mentioned the statement below. Is it possible to provide an intuition in terms of basic (e.g. fundamental sub-spaces) linear algebra terms?
An $n times n$ matrix $A$ naturally induces a mapping on $Grass(p,n)$ defined by $$mathcalSin Grass(p,n) mapsto AmathcalS:=Ay : y in mathcalS $$ where $Grass(p,n)$ admits a structure of manifold called the Grassmann manifold.
Thank you in advance!
linear-algebra general-topology geometric-topology
asked Jul 30 at 11:25
darkmoor
263210
I start reading about matrix manifolds and in an introduction section it is mentioned the statement below. Is it possible to provide an intuition in terms of basic (e.g. fundamental sub-spaces) linear algebra terms?
An $n times n$ matrix $A$ naturally induces a mapping on $Grass(p,n)$ defined by $$mathcalSin Grass(p,n) mapsto AmathcalS:=Ay : y in mathcalS $$ where $Grass(p,n)$ admits a structure of manifold called the Grassmann manifold.
Thank you in advance!
linear-algebra general-topology geometric-topology
asked Jul 30 at 11:25
darkmoor
263210
asked Jul 30 at 11:25
darkmoor
263210
asked Jul 30 at 11:25
darkmoor
263210
asked Jul 30 at 11:25
darkmoor
263210
263210
add a comment |Â
add a comment |Â
1 Answer
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active
oldest
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The statement is actually incorrect. Unless $det Ane0$, the image of $S$ under $A$ may have smaller dimension, and therefore "fall out" of the Grassmannian manifold being considered. If $A$ is invertible, then it induces a linear isomorphism of the ambient space and the image of $S$ will have the same dimension as $S$.
answered Jul 30 at 11:29
Mikhail Katz
30k13996
Because I am not familiar with topology terminology could you please provide a little more explanations, if it is possible of course, to understand the two last statements? More specifically, statements in the first statement like "smaller dimension", "fall out", and in the second statement, like "linear isomorphism of the ambient space" and "ave the same dimension as S" can be explained with examples?
– darkmoor
Jul 30 at 12:13
@darkmoor These are statements of linear algebra, not topology. The only thing not standard is "fall out", meaning that the dimension of the vector space $AS$ may be strictly smaller than the dimension of $S$, and so certainly is not an element of the Grassmannian of p-dimensional subspaces if $S$ was. As an example, consider the 0 matrix.
– Mike Miller
Jul 30 at 15:57
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The statement is actually incorrect. Unless $det Ane0$, the image of $S$ under $A$ may have smaller dimension, and therefore "fall out" of the Grassmannian manifold being considered. If $A$ is invertible, then it induces a linear isomorphism of the ambient space and the image of $S$ will have the same dimension as $S$.
answered Jul 30 at 11:29
Mikhail Katz
30k13996
Because I am not familiar with topology terminology could you please provide a little more explanations, if it is possible of course, to understand the two last statements? More specifically, statements in the first statement like "smaller dimension", "fall out", and in the second statement, like "linear isomorphism of the ambient space" and "ave the same dimension as S" can be explained with examples?
– darkmoor
Jul 30 at 12:13
@darkmoor These are statements of linear algebra, not topology. The only thing not standard is "fall out", meaning that the dimension of the vector space $AS$ may be strictly smaller than the dimension of $S$, and so certainly is not an element of the Grassmannian of p-dimensional subspaces if $S$ was. As an example, consider the 0 matrix.
– Mike Miller
Jul 30 at 15:57
add a comment |Â
up vote
0
down vote
The statement is actually incorrect. Unless $det Ane0$, the image of $S$ under $A$ may have smaller dimension, and therefore "fall out" of the Grassmannian manifold being considered. If $A$ is invertible, then it induces a linear isomorphism of the ambient space and the image of $S$ will have the same dimension as $S$.
answered Jul 30 at 11:29
Mikhail Katz
30k13996
Because I am not familiar with topology terminology could you please provide a little more explanations, if it is possible of course, to understand the two last statements? More specifically, statements in the first statement like "smaller dimension", "fall out", and in the second statement, like "linear isomorphism of the ambient space" and "ave the same dimension as S" can be explained with examples?
– darkmoor
Jul 30 at 12:13
@darkmoor These are statements of linear algebra, not topology. The only thing not standard is "fall out", meaning that the dimension of the vector space $AS$ may be strictly smaller than the dimension of $S$, and so certainly is not an element of the Grassmannian of p-dimensional subspaces if $S$ was. As an example, consider the 0 matrix.
– Mike Miller
Jul 30 at 15:57
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The statement is actually incorrect. Unless $det Ane0$, the image of $S$ under $A$ may have smaller dimension, and therefore "fall out" of the Grassmannian manifold being considered. If $A$ is invertible, then it induces a linear isomorphism of the ambient space and the image of $S$ will have the same dimension as $S$.
answered Jul 30 at 11:29
Mikhail Katz
30k13996
The statement is actually incorrect. Unless $det Ane0$, the image of $S$ under $A$ may have smaller dimension, and therefore "fall out" of the Grassmannian manifold being considered. If $A$ is invertible, then it induces a linear isomorphism of the ambient space and the image of $S$ will have the same dimension as $S$.
answered Jul 30 at 11:29
Mikhail Katz
30k13996
answered Jul 30 at 11:29
Mikhail Katz
30k13996
answered Jul 30 at 11:29
Mikhail Katz
30k13996
answered Jul 30 at 11:29
Mikhail Katz
30k13996
30k13996
Because I am not familiar with topology terminology could you please provide a little more explanations, if it is possible of course, to understand the two last statements? More specifically, statements in the first statement like "smaller dimension", "fall out", and in the second statement, like "linear isomorphism of the ambient space" and "ave the same dimension as S" can be explained with examples?
– darkmoor
Jul 30 at 12:13
@darkmoor These are statements of linear algebra, not topology. The only thing not standard is "fall out", meaning that the dimension of the vector space $AS$ may be strictly smaller than the dimension of $S$, and so certainly is not an element of the Grassmannian of p-dimensional subspaces if $S$ was. As an example, consider the 0 matrix.
– Mike Miller
Jul 30 at 15:57
add a comment |Â
Because I am not familiar with topology terminology could you please provide a little more explanations, if it is possible of course, to understand the two last statements? More specifically, statements in the first statement like "smaller dimension", "fall out", and in the second statement, like "linear isomorphism of the ambient space" and "ave the same dimension as S" can be explained with examples?
– darkmoor
Jul 30 at 12:13
@darkmoor These are statements of linear algebra, not topology. The only thing not standard is "fall out", meaning that the dimension of the vector space $AS$ may be strictly smaller than the dimension of $S$, and so certainly is not an element of the Grassmannian of p-dimensional subspaces if $S$ was. As an example, consider the 0 matrix.
– Mike Miller
Jul 30 at 15:57
Because I am not familiar with topology terminology could you please provide a little more explanations, if it is possible of course, to understand the two last statements? More specifically, statements in the first statement like "smaller dimension", "fall out", and in the second statement, like "linear isomorphism of the ambient space" and "ave the same dimension as S" can be explained with examples?
– darkmoor
Jul 30 at 12:13
Because I am not familiar with topology terminology could you please provide a little more explanations, if it is possible of course, to understand the two last statements? More specifically, statements in the first statement like "smaller dimension", "fall out", and in the second statement, like "linear isomorphism of the ambient space" and "ave the same dimension as S" can be explained with examples?
– darkmoor
Jul 30 at 12:13
@darkmoor These are statements of linear algebra, not topology. The only thing not standard is "fall out", meaning that the dimension of the vector space $AS$ may be strictly smaller than the dimension of $S$, and so certainly is not an element of the Grassmannian of p-dimensional subspaces if $S$ was. As an example, consider the 0 matrix.
– Mike Miller
Jul 30 at 15:57
@darkmoor These are statements of linear algebra, not topology. The only thing not standard is "fall out", meaning that the dimension of the vector space $AS$ may be strictly smaller than the dimension of $S$, and so certainly is not an element of the Grassmannian of p-dimensional subspaces if $S$ was. As an example, consider the 0 matrix.
– Mike Miller
Jul 30 at 15:57
add a comment |Â
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