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Invariants between two isomorphic vector spaces

Complex vector spacesSimple questions about isomorphisms between vector spacesWhat is needed to make Euclidean spaces isomorphic as groups?Vector Spaces and GroupsWhat do mathematicians mean by if two vector spaces are isomorphic, then “they are the same”What are the implications of the fact that vector spaces are isomorphic?Assume that $V$ and $W$ are isomorphic vector spaces. What does it imply if $V$ and $W$ are not finite dimensional.If there is a bijective linear transformation between two vector spaces, do they bear same properties?Are $C[0,1]$ and $C[0,1)$ isomorphic?Subspaces and elements of isomorphic vector spaces

2

$begingroup$

I have a general question about isomorphisms between vector spaces.

From a general point of view, there are common properties (invariants) between two isomorphic structures (e.g., properties about compactness or connectedness between two isomorphic topological spaces, commutativity between two isomorphic groups). Can you give me some examples of invariants between two isomorphic vector spaces ?

Thank you for your help !

vector-spaces vector-space-isomorphism invariance

share|cite|improve this question

edited Mar 26 at 15:01

José Carlos Santos

172k22132239

asked Mar 26 at 14:13

deeppinkwaterdeeppinkwater

858

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Dimension certainly.
  $endgroup$
  – Wuestenfux
  Mar 26 at 14:15
  

  
  
  
  

add a comment | 

2

$begingroup$

I have a general question about isomorphisms between vector spaces.

From a general point of view, there are common properties (invariants) between two isomorphic structures (e.g., properties about compactness or connectedness between two isomorphic topological spaces, commutativity between two isomorphic groups). Can you give me some examples of invariants between two isomorphic vector spaces ?

Thank you for your help !

vector-spaces vector-space-isomorphism invariance

share|cite|improve this question

edited Mar 26 at 15:01

José Carlos Santos

172k22132239

asked Mar 26 at 14:13

deeppinkwaterdeeppinkwater

858

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Dimension certainly.
  $endgroup$
  – Wuestenfux
  Mar 26 at 14:15
  

  
  
  
  

add a comment | 

$begingroup$

I have a general question about isomorphisms between vector spaces.

From a general point of view, there are common properties (invariants) between two isomorphic structures (e.g., properties about compactness or connectedness between two isomorphic topological spaces, commutativity between two isomorphic groups). Can you give me some examples of invariants between two isomorphic vector spaces ?

Thank you for your help !

vector-spaces vector-space-isomorphism invariance

share|cite|improve this question

edited Mar 26 at 15:01

José Carlos Santos

172k22132239

asked Mar 26 at 14:13

deeppinkwaterdeeppinkwater

858

$endgroup$

share|cite|improve this question

edited Mar 26 at 15:01

José Carlos Santos

172k22132239

asked Mar 26 at 14:13

deeppinkwaterdeeppinkwater

858

share|cite|improve this question

edited Mar 26 at 15:01

José Carlos Santos

172k22132239

asked Mar 26 at 14:13

deeppinkwaterdeeppinkwater

858

edited Mar 26 at 15:01

José Carlos Santos

172k22132239

edited Mar 26 at 15:01

José Carlos Santos

172k22132239

asked Mar 26 at 14:13

deeppinkwaterdeeppinkwater

858

asked Mar 26 at 14:13

deeppinkwaterdeeppinkwater

858

2 Answers
2

active

oldest

votes

6

$begingroup$

One of the fundamental theorems of linear algebra is that there is only one invariant you really need to care about: the dimension. Two vector spaces are isomorphic if and only if they have the same dimension. The reason is that any map that bijectively takes a basis of one vector space to a basis of another vector space must extend linearly (in a unique way) to a vector space isomorphism.

Another example is the cardinality of the underlying set (of course, the isomorphism is a bijection).

This observation is actually useful in field theory. Suppose you have a field $K$ which you know is finite. Then the prime subfield of $K$ is $mathbbF_p$ for some prime $p.$ Now $K$ must have finite dimension as a vector space over $mathbbF_p$ (since $K$ is finite), so, again as a vector space, $K=(mathbbF_p)^oplus r$ for some $r.$ It follows immediately that $lvertKrvert=p^r$ for the same $r.$ Moreover, if two finite fields are of the same cardinality, then that cardinality must be $p^r$ for some prime $p$ and some integer $r,$ and it follows that those two fields are isomorphic as vector spaces over the prime subfield $mathbbF_p.$ (With a bit more work, one can show that any two finite fields of the same cardinality are isomorphic as fields, and that a finite field of cardinality $p^r$ does exist for every prime $p$ and every integer $r>0.$)

share|cite|improve this answer

edited Mar 26 at 20:27

answered Mar 26 at 14:29

Will RWill R

6,78231429

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for this answer.
  $endgroup$
  – deeppinkwater
  Mar 26 at 14:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @deeppinkwater: It is perhaps also worth noting that although (or rather, because!) the dimension is the only thing that determines the isomorphism class of a vector space, most linear algebra courses actually spend more time focusing on the linear maps between vector spaces. Then there are lots of "invariants", depending upon your notion of isomorphism, the primary examples being eigenvalues (with their respective multiplicities), and therefore also the determinant and the trace, and the singular values.
  $endgroup$
  – Will R
  Mar 26 at 17:13
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I take it by "underlying set", you mean "scalar field"? I think that the latter is more precise.
  $endgroup$
  – Acccumulation
  Mar 26 at 21:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Acccumulation: No, I mean the set of vectors. If a vector space over a field $K$ is an abelian group $V$ with a linear action of $K$ on $V$ (basically, you have some operations and a load of axioms are satisfied), then by "underlying set" I mean the set $V.$
  $endgroup$
  – Will R
  Mar 26 at 21:41
  

  
  
  
  

add a comment | 

2

$begingroup$

The number one is the dimension, of course, since two vector spaces over the same field are isomorphic if and only if they have the same dimension.

share|cite|improve this answer

answered Mar 26 at 14:15

José Carlos SantosJosé Carlos Santos

172k22132239

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, that's the only one that came to my mind actually !
  $endgroup$
  – deeppinkwater
  Mar 26 at 14:16
  

  
  
  
  

add a comment | 

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6

$begingroup$

One of the fundamental theorems of linear algebra is that there is only one invariant you really need to care about: the dimension. Two vector spaces are isomorphic if and only if they have the same dimension. The reason is that any map that bijectively takes a basis of one vector space to a basis of another vector space must extend linearly (in a unique way) to a vector space isomorphism.

Another example is the cardinality of the underlying set (of course, the isomorphism is a bijection).

This observation is actually useful in field theory. Suppose you have a field $K$ which you know is finite. Then the prime subfield of $K$ is $mathbbF_p$ for some prime $p.$ Now $K$ must have finite dimension as a vector space over $mathbbF_p$ (since $K$ is finite), so, again as a vector space, $K=(mathbbF_p)^oplus r$ for some $r.$ It follows immediately that $lvertKrvert=p^r$ for the same $r.$ Moreover, if two finite fields are of the same cardinality, then that cardinality must be $p^r$ for some prime $p$ and some integer $r,$ and it follows that those two fields are isomorphic as vector spaces over the prime subfield $mathbbF_p.$ (With a bit more work, one can show that any two finite fields of the same cardinality are isomorphic as fields, and that a finite field of cardinality $p^r$ does exist for every prime $p$ and every integer $r>0.$)

share|cite|improve this answer

edited Mar 26 at 20:27

answered Mar 26 at 14:29

Will RWill R

6,78231429

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for this answer.
  $endgroup$
  – deeppinkwater
  Mar 26 at 14:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @deeppinkwater: It is perhaps also worth noting that although (or rather, because!) the dimension is the only thing that determines the isomorphism class of a vector space, most linear algebra courses actually spend more time focusing on the linear maps between vector spaces. Then there are lots of "invariants", depending upon your notion of isomorphism, the primary examples being eigenvalues (with their respective multiplicities), and therefore also the determinant and the trace, and the singular values.
  $endgroup$
  – Will R
  Mar 26 at 17:13
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I take it by "underlying set", you mean "scalar field"? I think that the latter is more precise.
  $endgroup$
  – Acccumulation
  Mar 26 at 21:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Acccumulation: No, I mean the set of vectors. If a vector space over a field $K$ is an abelian group $V$ with a linear action of $K$ on $V$ (basically, you have some operations and a load of axioms are satisfied), then by "underlying set" I mean the set $V.$
  $endgroup$
  – Will R
  Mar 26 at 21:41
  

  
  
  
  

add a comment | 

6

$begingroup$

One of the fundamental theorems of linear algebra is that there is only one invariant you really need to care about: the dimension. Two vector spaces are isomorphic if and only if they have the same dimension. The reason is that any map that bijectively takes a basis of one vector space to a basis of another vector space must extend linearly (in a unique way) to a vector space isomorphism.

Another example is the cardinality of the underlying set (of course, the isomorphism is a bijection).

This observation is actually useful in field theory. Suppose you have a field $K$ which you know is finite. Then the prime subfield of $K$ is $mathbbF_p$ for some prime $p.$ Now $K$ must have finite dimension as a vector space over $mathbbF_p$ (since $K$ is finite), so, again as a vector space, $K=(mathbbF_p)^oplus r$ for some $r.$ It follows immediately that $lvertKrvert=p^r$ for the same $r.$ Moreover, if two finite fields are of the same cardinality, then that cardinality must be $p^r$ for some prime $p$ and some integer $r,$ and it follows that those two fields are isomorphic as vector spaces over the prime subfield $mathbbF_p.$ (With a bit more work, one can show that any two finite fields of the same cardinality are isomorphic as fields, and that a finite field of cardinality $p^r$ does exist for every prime $p$ and every integer $r>0.$)

share|cite|improve this answer

edited Mar 26 at 20:27

answered Mar 26 at 14:29

Will RWill R

6,78231429

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for this answer.
  $endgroup$
  – deeppinkwater
  Mar 26 at 14:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @deeppinkwater: It is perhaps also worth noting that although (or rather, because!) the dimension is the only thing that determines the isomorphism class of a vector space, most linear algebra courses actually spend more time focusing on the linear maps between vector spaces. Then there are lots of "invariants", depending upon your notion of isomorphism, the primary examples being eigenvalues (with their respective multiplicities), and therefore also the determinant and the trace, and the singular values.
  $endgroup$
  – Will R
  Mar 26 at 17:13
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I take it by "underlying set", you mean "scalar field"? I think that the latter is more precise.
  $endgroup$
  – Acccumulation
  Mar 26 at 21:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Acccumulation: No, I mean the set of vectors. If a vector space over a field $K$ is an abelian group $V$ with a linear action of $K$ on $V$ (basically, you have some operations and a load of axioms are satisfied), then by "underlying set" I mean the set $V.$
  $endgroup$
  – Will R
  Mar 26 at 21:41
  

  
  
  
  

add a comment | 

$begingroup$

One of the fundamental theorems of linear algebra is that there is only one invariant you really need to care about: the dimension. Two vector spaces are isomorphic if and only if they have the same dimension. The reason is that any map that bijectively takes a basis of one vector space to a basis of another vector space must extend linearly (in a unique way) to a vector space isomorphism.

Another example is the cardinality of the underlying set (of course, the isomorphism is a bijection).

This observation is actually useful in field theory. Suppose you have a field $K$ which you know is finite. Then the prime subfield of $K$ is $mathbbF_p$ for some prime $p.$ Now $K$ must have finite dimension as a vector space over $mathbbF_p$ (since $K$ is finite), so, again as a vector space, $K=(mathbbF_p)^oplus r$ for some $r.$ It follows immediately that $lvertKrvert=p^r$ for the same $r.$ Moreover, if two finite fields are of the same cardinality, then that cardinality must be $p^r$ for some prime $p$ and some integer $r,$ and it follows that those two fields are isomorphic as vector spaces over the prime subfield $mathbbF_p.$ (With a bit more work, one can show that any two finite fields of the same cardinality are isomorphic as fields, and that a finite field of cardinality $p^r$ does exist for every prime $p$ and every integer $r>0.$)

share|cite|improve this answer

edited Mar 26 at 20:27

answered Mar 26 at 14:29

Will RWill R

6,78231429

$endgroup$

share|cite|improve this answer

edited Mar 26 at 20:27

answered Mar 26 at 14:29

Will RWill R

6,78231429

answered Mar 26 at 14:29

Will RWill R

6,78231429

answered Mar 26 at 14:29

Will RWill R

6,78231429

2

$begingroup$

The number one is the dimension, of course, since two vector spaces over the same field are isomorphic if and only if they have the same dimension.

share|cite|improve this answer

answered Mar 26 at 14:15

José Carlos SantosJosé Carlos Santos

172k22132239

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, that's the only one that came to my mind actually !
  $endgroup$
  – deeppinkwater
  Mar 26 at 14:16
  

  
  
  
  

add a comment | 

2

$begingroup$

The number one is the dimension, of course, since two vector spaces over the same field are isomorphic if and only if they have the same dimension.

share|cite|improve this answer

answered Mar 26 at 14:15

José Carlos SantosJosé Carlos Santos

172k22132239

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, that's the only one that came to my mind actually !
  $endgroup$
  – deeppinkwater
  Mar 26 at 14:16
  

  
  
  
  

add a comment | 

$begingroup$

The number one is the dimension, of course, since two vector spaces over the same field are isomorphic if and only if they have the same dimension.

share|cite|improve this answer

answered Mar 26 at 14:15

José Carlos SantosJosé Carlos Santos

172k22132239

$endgroup$

share|cite|improve this answer

answered Mar 26 at 14:15

José Carlos SantosJosé Carlos Santos

172k22132239

answered Mar 26 at 14:15

José Carlos SantosJosé Carlos Santos

172k22132239

answered Mar 26 at 14:15

José Carlos SantosJosé Carlos Santos

172k22132239