Invariants between two 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 17 hours ago
José Carlos Santos
170k23132238
asked 18 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 17 hours ago
José Carlos Santos
170k23132238
asked 18 hours ago
deeppinkwaterdeeppinkwater
858
$endgroup$
1
$begingroup$
Dimension certainly.
$endgroup$
– Wuestenfux
18 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 17 hours ago
José Carlos Santos
170k23132238
asked 18 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 17 hours ago
José Carlos Santos
170k23132238
asked 18 hours ago
deeppinkwaterdeeppinkwater
858
edited 17 hours ago
José Carlos Santos
170k23132238
asked 18 hours ago
deeppinkwaterdeeppinkwater
858
edited 17 hours ago
José Carlos Santos
170k23132238
edited 17 hours ago
José Carlos Santos
170k23132238
edited 17 hours ago
José Carlos Santos
170k23132238
170k23132238
asked 18 hours ago
deeppinkwaterdeeppinkwater
858
asked 18 hours ago
deeppinkwaterdeeppinkwater
858
asked 18 hours ago
deeppinkwaterdeeppinkwater
858
858
1
$begingroup$
Dimension certainly.
$endgroup$
– Wuestenfux
18 hours ago
add a comment |
1
$begingroup$
Dimension certainly.
$endgroup$
– Wuestenfux
18 hours ago
1
1
$begingroup$
Dimension certainly.
$endgroup$
– Wuestenfux
18 hours ago
$begingroup$
Dimension certainly.
$endgroup$
– Wuestenfux
18 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 $mathbb{F}_{p}$ 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 $r>0.$)
answered 18 hours ago
Will RWill R
6,78231429
$endgroup$
$begingroup$
Thank you for this answer.
$endgroup$
– deeppinkwater
17 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
15 hours ago
$begingroup$
I take it by "underlying set", you mean "scalar field"? I think that the latter is more precise.
$endgroup$
– Acccumulation
10 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
10 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 18 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
18 hours ago
add a comment |
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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 $mathbb{F}_{p}$ 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 $r>0.$)
answered 18 hours ago
Will RWill R
6,78231429
$endgroup$
$begingroup$
Thank you for this answer.
$endgroup$
– deeppinkwater
17 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
15 hours ago
$begingroup$
I take it by "underlying set", you mean "scalar field"? I think that the latter is more precise.
$endgroup$
– Acccumulation
10 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
10 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 $mathbb{F}_{p}$ 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 $r>0.$)
answered 18 hours ago
Will RWill R
6,78231429
$endgroup$
$begingroup$
Thank you for this answer.
$endgroup$
– deeppinkwater
17 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
15 hours ago
$begingroup$
I take it by "underlying set", you mean "scalar field"? I think that the latter is more precise.
$endgroup$
– Acccumulation
10 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
10 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 $mathbb{F}_{p}$ 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 $r>0.$)
answered 18 hours ago
Will RWill R
6,78231429
$endgroup$
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 $r>0.$)
answered 18 hours ago
Will RWill R
6,78231429
edited 12 hours ago
answered 18 hours ago
Will RWill R
6,78231429
answered 18 hours ago
Will RWill R
6,78231429
answered 18 hours ago
Will RWill R
6,78231429
6,78231429
$begingroup$
Thank you for this answer.
$endgroup$
– deeppinkwater
17 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
15 hours ago
$begingroup$
I take it by "underlying set", you mean "scalar field"? I think that the latter is more precise.
$endgroup$
– Acccumulation
10 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
10 hours ago
add a comment |
$begingroup$
Thank you for this answer.
$endgroup$
– deeppinkwater
17 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
15 hours ago
$begingroup$
I take it by "underlying set", you mean "scalar field"? I think that the latter is more precise.
$endgroup$
– Acccumulation
10 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
10 hours ago
$begingroup$
Thank you for this answer.
$endgroup$
– deeppinkwater
17 hours ago
$begingroup$
Thank you for this answer.
$endgroup$
– deeppinkwater
17 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
15 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
15 hours ago
$begingroup$
I take it by "underlying set", you mean "scalar field"? I think that the latter is more precise.
$endgroup$
– Acccumulation
10 hours ago
$begingroup$
I take it by "underlying set", you mean "scalar field"? I think that the latter is more precise.
$endgroup$
– Acccumulation
10 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
10 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
10 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 18 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
18 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 18 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
18 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 18 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 18 hours ago
José Carlos SantosJosé Carlos Santos
170k23132238
answered 18 hours ago
José Carlos SantosJosé Carlos Santos
170k23132238
answered 18 hours ago
José Carlos SantosJosé Carlos Santos
170k23132238
answered 18 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
18 hours ago
add a comment |
$begingroup$
Thank you, that's the only one that came to my mind actually !
$endgroup$
– deeppinkwater
18 hours ago
$begingroup$
Thank you, that's the only one that came to my mind actually !
$endgroup$
– deeppinkwater
18 hours ago
$begingroup$
Thank you, that's the only one that came to my mind actually !
$endgroup$
– deeppinkwater
18 hours ago
add a comment |
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$begingroup$
Dimension certainly.
$endgroup$
– Wuestenfux
18 hours ago