Invariants between two isomorphic vector spacesComplex 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
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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
$begingroup$
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
edited 4 hours ago
José Carlos Santos
170k23132238
asked 5 hours ago
deeppinkwaterdeeppinkwater
858
$endgroup$
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
edited 4 hours ago
José Carlos Santos
170k23132238
asked 5 hours ago
deeppinkwaterdeeppinkwater
858
$endgroup$
1
$begingroup$
Dimension certainly.
$endgroup$
– Wuestenfux
4 hours ago
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
edited 4 hours ago
José Carlos Santos
170k23132238
asked 5 hours ago
deeppinkwaterdeeppinkwater
858
$endgroup$
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
vector-spaces vector-space-isomorphism invariance
edited 4 hours ago
José Carlos Santos
170k23132238
asked 5 hours ago
deeppinkwaterdeeppinkwater
858
edited 4 hours ago
José Carlos Santos
170k23132238
asked 5 hours ago
deeppinkwaterdeeppinkwater
858
edited 4 hours ago
José Carlos Santos
170k23132238
edited 4 hours ago
José Carlos Santos
170k23132238
edited 4 hours ago
José Carlos Santos
170k23132238
170k23132238
asked 5 hours ago
deeppinkwaterdeeppinkwater
858
asked 5 hours ago
deeppinkwaterdeeppinkwater
858
asked 5 hours ago
deeppinkwaterdeeppinkwater
858
858
1
$begingroup$
Dimension certainly.
$endgroup$
– Wuestenfux
4 hours ago
add a comment |
1
$begingroup$
Dimension certainly.
$endgroup$
– Wuestenfux
4 hours ago
1
1
$begingroup$
Dimension certainly.
$endgroup$
– Wuestenfux
4 hours ago
$begingroup$
Dimension certainly.
$endgroup$
– Wuestenfux
4 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 (since $K$ is finite), so $K=(mathbbF_p)^r$ for some $r.$ It follows immediately that $lvertKrvert=p^r$ for some $r.$ 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 $r>0.$)
answered 4 hours ago
Will RWill R
6,75231429
$endgroup$
$begingroup$
Thank you for this answer.
$endgroup$
– deeppinkwater
4 hours ago
$begingroup$
@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
2 hours ago
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.
answered 4 hours ago
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
$begingroup$
Thank you, that's the only one that came to my mind actually !
$endgroup$
– deeppinkwater
4 hours ago
add a comment |
Your Answer
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2 Answers
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2 Answers
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$begingroup$
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 (since $K$ is finite), so $K=(mathbbF_p)^r$ for some $r.$ It follows immediately that $lvertKrvert=p^r$ for some $r.$ 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 $r>0.$)
answered 4 hours ago
Will RWill R
6,75231429
$endgroup$
$begingroup$
Thank you for this answer.
$endgroup$
– deeppinkwater
4 hours ago
$begingroup$
@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
2 hours ago
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 (since $K$ is finite), so $K=(mathbbF_p)^r$ for some $r.$ It follows immediately that $lvertKrvert=p^r$ for some $r.$ 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 $r>0.$)
answered 4 hours ago
Will RWill R
6,75231429
$endgroup$
$begingroup$
Thank you for this answer.
$endgroup$
– deeppinkwater
4 hours ago
$begingroup$
@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
2 hours ago
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 (since $K$ is finite), so $K=(mathbbF_p)^r$ for some $r.$ It follows immediately that $lvertKrvert=p^r$ for some $r.$ 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 $r>0.$)
answered 4 hours ago
Will RWill R
6,75231429
$endgroup$
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 (since $K$ is finite), so $K=(mathbbF_p)^r$ for some $r.$ It follows immediately that $lvertKrvert=p^r$ for some $r.$ 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 $r>0.$)
answered 4 hours ago
Will RWill R
6,75231429
edited 4 hours ago
answered 4 hours ago
Will RWill R
6,75231429
answered 4 hours ago
Will RWill R
6,75231429
answered 4 hours ago
Will RWill R
6,75231429
6,75231429
$begingroup$
Thank you for this answer.
$endgroup$
– deeppinkwater
4 hours ago
$begingroup$
@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
2 hours ago
add a comment |
$begingroup$
Thank you for this answer.
$endgroup$
– deeppinkwater
4 hours ago
$begingroup$
@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
2 hours ago
$begingroup$
Thank you for this answer.
$endgroup$
– deeppinkwater
4 hours ago
$begingroup$
Thank you for this answer.
$endgroup$
– deeppinkwater
4 hours ago
$begingroup$
@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
2 hours ago
$begingroup$
@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
2 hours ago
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.
answered 4 hours ago
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
$begingroup$
Thank you, that's the only one that came to my mind actually !
$endgroup$
– deeppinkwater
4 hours ago
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.
answered 4 hours ago
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
$begingroup$
Thank you, that's the only one that came to my mind actually !
$endgroup$
– deeppinkwater
4 hours ago
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.
answered 4 hours ago
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
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.
answered 4 hours ago
José Carlos SantosJosé Carlos Santos
170k23132238
answered 4 hours ago
José Carlos SantosJosé Carlos Santos
170k23132238
answered 4 hours ago
José Carlos SantosJosé Carlos Santos
170k23132238
answered 4 hours ago
José Carlos SantosJosé Carlos Santos
170k23132238
170k23132238
$begingroup$
Thank you, that's the only one that came to my mind actually !
$endgroup$
– deeppinkwater
4 hours ago
add a comment |
$begingroup$
Thank you, that's the only one that came to my mind actually !
$endgroup$
– deeppinkwater
4 hours ago
$begingroup$
Thank you, that's the only one that came to my mind actually !
$endgroup$
– deeppinkwater
4 hours ago
$begingroup$
Thank you, that's the only one that came to my mind actually !
$endgroup$
– deeppinkwater
4 hours ago
add a comment |
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1
$begingroup$
Dimension certainly.
$endgroup$
– Wuestenfux
4 hours ago