Invariants between two 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







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  • 1




    $begingroup$
    Dimension certainly.
    $endgroup$
    – Wuestenfux
    15 hours ago
















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











$endgroup$








  • 1




    $begingroup$
    Dimension certainly.
    $endgroup$
    – Wuestenfux
    15 hours ago














2












2








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











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 14 hours ago









José Carlos Santos

170k23132238




170k23132238










asked 15 hours ago









deeppinkwaterdeeppinkwater

858




858








  • 1




    $begingroup$
    Dimension certainly.
    $endgroup$
    – Wuestenfux
    15 hours ago














  • 1




    $begingroup$
    Dimension certainly.
    $endgroup$
    – Wuestenfux
    15 hours ago








1




1




$begingroup$
Dimension certainly.
$endgroup$
– Wuestenfux
15 hours ago




$begingroup$
Dimension certainly.
$endgroup$
– Wuestenfux
15 hours ago










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 $mathbb{F}_{p}$ for some prime $p.$ Now $K$ must have finite dimension as a vector space over $mathbb{F}_{p}$ (since $K$ is finite), so, again as a vector space, $K=(mathbb{F}_{p})^{oplus r}$ for some $r.$ It follows immediately that $lvert{K}rvert=p^{r}$ 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 $mathbb{F}_{p}.$ (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











$endgroup$













  • $begingroup$
    Thank you for this answer.
    $endgroup$
    – deeppinkwater
    14 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
    12 hours ago












  • $begingroup$
    I take it by "underlying set", you mean "scalar field"? I think that the latter is more precise.
    $endgroup$
    – Acccumulation
    7 hours ago










  • $begingroup$
    @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
    7 hours ago



















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









$endgroup$













  • $begingroup$
    Thank you, that's the only one that came to my mind actually !
    $endgroup$
    – deeppinkwater
    15 hours ago











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






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 $mathbb{F}_{p}$ for some prime $p.$ Now $K$ must have finite dimension as a vector space over $mathbb{F}_{p}$ (since $K$ is finite), so, again as a vector space, $K=(mathbb{F}_{p})^{oplus r}$ for some $r.$ It follows immediately that $lvert{K}rvert=p^{r}$ 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 $mathbb{F}_{p}.$ (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











$endgroup$













  • $begingroup$
    Thank you for this answer.
    $endgroup$
    – deeppinkwater
    14 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
    12 hours ago












  • $begingroup$
    I take it by "underlying set", you mean "scalar field"? I think that the latter is more precise.
    $endgroup$
    – Acccumulation
    7 hours ago










  • $begingroup$
    @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
    7 hours ago
















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 $mathbb{F}_{p}$ for some prime $p.$ Now $K$ must have finite dimension as a vector space over $mathbb{F}_{p}$ (since $K$ is finite), so, again as a vector space, $K=(mathbb{F}_{p})^{oplus r}$ for some $r.$ It follows immediately that $lvert{K}rvert=p^{r}$ 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 $mathbb{F}_{p}.$ (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











$endgroup$













  • $begingroup$
    Thank you for this answer.
    $endgroup$
    – deeppinkwater
    14 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
    12 hours ago












  • $begingroup$
    I take it by "underlying set", you mean "scalar field"? I think that the latter is more precise.
    $endgroup$
    – Acccumulation
    7 hours ago










  • $begingroup$
    @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
    7 hours ago














6












6








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 $mathbb{F}_{p}$ for some prime $p.$ Now $K$ must have finite dimension as a vector space over $mathbb{F}_{p}$ (since $K$ is finite), so, again as a vector space, $K=(mathbb{F}_{p})^{oplus r}$ for some $r.$ It follows immediately that $lvert{K}rvert=p^{r}$ 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 $mathbb{F}_{p}.$ (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











$endgroup$



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 $mathbb{F}_{p}$ for some prime $p.$ Now $K$ must have finite dimension as a vector space over $mathbb{F}_{p}$ (since $K$ is finite), so, again as a vector space, $K=(mathbb{F}_{p})^{oplus r}$ for some $r.$ It follows immediately that $lvert{K}rvert=p^{r}$ 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 $mathbb{F}_{p}.$ (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














share|cite|improve this answer



share|cite|improve this answer








edited 9 hours ago

























answered 15 hours ago









Will RWill R

6,78231429




6,78231429












  • $begingroup$
    Thank you for this answer.
    $endgroup$
    – deeppinkwater
    14 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
    12 hours ago












  • $begingroup$
    I take it by "underlying set", you mean "scalar field"? I think that the latter is more precise.
    $endgroup$
    – Acccumulation
    7 hours ago










  • $begingroup$
    @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
    7 hours ago


















  • $begingroup$
    Thank you for this answer.
    $endgroup$
    – deeppinkwater
    14 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
    12 hours ago












  • $begingroup$
    I take it by "underlying set", you mean "scalar field"? I think that the latter is more precise.
    $endgroup$
    – Acccumulation
    7 hours ago










  • $begingroup$
    @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
    7 hours ago
















$begingroup$
Thank you for this answer.
$endgroup$
– deeppinkwater
14 hours ago




$begingroup$
Thank you for this answer.
$endgroup$
– deeppinkwater
14 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
12 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
12 hours ago














$begingroup$
I take it by "underlying set", you mean "scalar field"? I think that the latter is more precise.
$endgroup$
– Acccumulation
7 hours ago




$begingroup$
I take it by "underlying set", you mean "scalar field"? I think that the latter is more precise.
$endgroup$
– Acccumulation
7 hours ago












$begingroup$
@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
7 hours ago




$begingroup$
@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
7 hours ago











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









$endgroup$













  • $begingroup$
    Thank you, that's the only one that came to my mind actually !
    $endgroup$
    – deeppinkwater
    15 hours ago
















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









$endgroup$













  • $begingroup$
    Thank you, that's the only one that came to my mind actually !
    $endgroup$
    – deeppinkwater
    15 hours ago














2












2








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









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 15 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
    15 hours ago


















  • $begingroup$
    Thank you, that's the only one that came to my mind actually !
    $endgroup$
    – deeppinkwater
    15 hours ago
















$begingroup$
Thank you, that's the only one that came to my mind actually !
$endgroup$
– deeppinkwater
15 hours ago




$begingroup$
Thank you, that's the only one that came to my mind actually !
$endgroup$
– deeppinkwater
15 hours ago


















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