Example of a certain vector space [closed]

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Let $k$ be a field. I am looking for an example of infinite dimensional $k$-vector space $V$ such that $Vcong Voplus V$. Also kindly let me know in brief how you've thought about it . Thanks in advance!

linear-algebra vector-spaces

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edited Aug 6 at 11:19

Babelfish

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asked Aug 6 at 11:11

solgaleo

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closed as off-topic by Jendrik Stelzner, Shailesh, Clayton, José Carlos Santos, Xander Henderson Aug 7 at 0:02

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jendrik Stelzner, Shailesh, Clayton, José Carlos Santos, Xander Henderson

If this question can be reworded to fit the rules in the help center, please edit the question.

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  Have you tried anything? Because if by "$cong$" you just mean isomorphic as vector spaces then absolutely any infinite dimensional $k$-vector space works.
  – Rhys Steele
  Aug 6 at 11:15
  

  
  
  
  

add a comment | 

up vote
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Let $k$ be a field. I am looking for an example of infinite dimensional $k$-vector space $V$ such that $Vcong Voplus V$. Also kindly let me know in brief how you've thought about it . Thanks in advance!

linear-algebra vector-spaces

share|cite|improve this question

edited Aug 6 at 11:19

Babelfish

629114

asked Aug 6 at 11:11

solgaleo

638

closed as off-topic by Jendrik Stelzner, Shailesh, Clayton, José Carlos Santos, Xander Henderson Aug 7 at 0:02

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jendrik Stelzner, Shailesh, Clayton, José Carlos Santos, Xander Henderson

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Have you tried anything? Because if by "$cong$" you just mean isomorphic as vector spaces then absolutely any infinite dimensional $k$-vector space works.
  – Rhys Steele
  Aug 6 at 11:15
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

1

up vote
0
down vote

favorite

1

1

Let $k$ be a field. I am looking for an example of infinite dimensional $k$-vector space $V$ such that $Vcong Voplus V$. Also kindly let me know in brief how you've thought about it . Thanks in advance!

linear-algebra vector-spaces

share|cite|improve this question

edited Aug 6 at 11:19

Babelfish

629114

asked Aug 6 at 11:11

solgaleo

638

Let $k$ be a field. I am looking for an example of infinite dimensional $k$-vector space $V$ such that $Vcong Voplus V$. Also kindly let me know in brief how you've thought about it . Thanks in advance!

linear-algebra vector-spaces

share|cite|improve this question

edited Aug 6 at 11:19

Babelfish

629114

asked Aug 6 at 11:11

solgaleo

638

share|cite|improve this question

share|cite|improve this question

edited Aug 6 at 11:19

Babelfish

629114

edited Aug 6 at 11:19

Babelfish

629114

edited Aug 6 at 11:19

Babelfish

629114

629114

asked Aug 6 at 11:11

solgaleo

638

asked Aug 6 at 11:11

solgaleo

638

asked Aug 6 at 11:11

solgaleo

638

638

closed as off-topic by Jendrik Stelzner, Shailesh, Clayton, José Carlos Santos, Xander Henderson Aug 7 at 0:02

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jendrik Stelzner, Shailesh, Clayton, José Carlos Santos, Xander Henderson

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by Jendrik Stelzner, Shailesh, Clayton, José Carlos Santos, Xander Henderson Aug 7 at 0:02

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jendrik Stelzner, Shailesh, Clayton, José Carlos Santos, Xander Henderson

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Have you tried anything? Because if by "$cong$" you just mean isomorphic as vector spaces then absolutely any infinite dimensional $k$-vector space works.
  – Rhys Steele
  Aug 6 at 11:15
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Have you tried anything? Because if by "$cong$" you just mean isomorphic as vector spaces then absolutely any infinite dimensional $k$-vector space works.
  – Rhys Steele
  Aug 6 at 11:15
  

  
  
  
  

3

3

Have you tried anything? Because if by "$cong$" you just mean isomorphic as vector spaces then absolutely any infinite dimensional $k$-vector space works.
– Rhys Steele
Aug 6 at 11:15

Have you tried anything? Because if by "$cong$" you just mean isomorphic as vector spaces then absolutely any infinite dimensional $k$-vector space works.
– Rhys Steele
Aug 6 at 11:15

add a comment | 

1 Answer
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As Rhys Steele mentioned, two vector spaces over $k$ are isomorphic iff they have the same dimension. For an infinite dimensional $k$-vector space $V$, $dim(V) + dim(V)= dim(V)$ holds (see this question). Therefore $V cong V oplus V$.

Let's pick our favourite infinite-dimensional vector space and do this. Mine is $k[X]$, the vector space of polynomials in one variable over $k$.
If you add it with itself, you obtain $k[X] oplus k[Y] cong k[X,Y]$, the vector space of polynomials in two variables. All those vector spaces have dimension $mathbb N$, so they are abstractly isomorphic. But where is the isomorphism? $k[X]$ and $k[X,Y]$ look very different, don't they?

We can choose a basis of $k[X]$ (e.g. the monomial basis) and $k[X,Y]$ (e.g. the basis of all monomial products $1,X,Y,XY,X^2,Y^2,XY^2,X^2Ydots$). Choosing any bijection between those two basis (which exists, since $mathbb N times mathbb N$ and $mathbb N$ have the same cardinality, i.e. are both countably infinite) provides a vector space isomorphism, by linear extension.

You might feel a bit cheated now. Aren't $k[X]$ and $k[X,Y]$ different things? They are the same as vector spaces though. The difference is the additional structure on those two vector spaces, since you can multiply polynomials. The multiplication turns $k[X]$ and $k[X,Y]$ into algebras (even graded algebras), and the isomorphism above won't preserve this structure. If you want to fix this, you obtain an example which feels more natural:

Let $V=k[X_1,dots,X_n,ldots] = bigoplus_i=1^infty k[X_i]$ the vector space of polynomials with coefficients in $k$ and countably infinitely many unknowns. Like before, if you take the direct sum of $V$ with itself, you just "double" the unknowns. So $Voplus V$ is the vector space over twice as many unknowns (which could be called $X_1,X_2,X_3,ldots$ and $Y_1,Y_2,Y_3,ldots$). But that's still the same amount of unknowns. A bijection of the set $,X_1,X_2,X_3,ldots$ and $,X_1,Y_1,X_2,Y_2,X_3,Y_3,ldots,$ gives rise to an algebra-isomorphism (which is in particular a linear isomorphism).

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answered Aug 6 at 15:38

Babelfish

629114

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1 Answer
1

active

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1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote

As Rhys Steele mentioned, two vector spaces over $k$ are isomorphic iff they have the same dimension. For an infinite dimensional $k$-vector space $V$, $dim(V) + dim(V)= dim(V)$ holds (see this question). Therefore $V cong V oplus V$.

Let's pick our favourite infinite-dimensional vector space and do this. Mine is $k[X]$, the vector space of polynomials in one variable over $k$.
If you add it with itself, you obtain $k[X] oplus k[Y] cong k[X,Y]$, the vector space of polynomials in two variables. All those vector spaces have dimension $mathbb N$, so they are abstractly isomorphic. But where is the isomorphism? $k[X]$ and $k[X,Y]$ look very different, don't they?

We can choose a basis of $k[X]$ (e.g. the monomial basis) and $k[X,Y]$ (e.g. the basis of all monomial products $1,X,Y,XY,X^2,Y^2,XY^2,X^2Ydots$). Choosing any bijection between those two basis (which exists, since $mathbb N times mathbb N$ and $mathbb N$ have the same cardinality, i.e. are both countably infinite) provides a vector space isomorphism, by linear extension.

You might feel a bit cheated now. Aren't $k[X]$ and $k[X,Y]$ different things? They are the same as vector spaces though. The difference is the additional structure on those two vector spaces, since you can multiply polynomials. The multiplication turns $k[X]$ and $k[X,Y]$ into algebras (even graded algebras), and the isomorphism above won't preserve this structure. If you want to fix this, you obtain an example which feels more natural:

Let $V=k[X_1,dots,X_n,ldots] = bigoplus_i=1^infty k[X_i]$ the vector space of polynomials with coefficients in $k$ and countably infinitely many unknowns. Like before, if you take the direct sum of $V$ with itself, you just "double" the unknowns. So $Voplus V$ is the vector space over twice as many unknowns (which could be called $X_1,X_2,X_3,ldots$ and $Y_1,Y_2,Y_3,ldots$). But that's still the same amount of unknowns. A bijection of the set $,X_1,X_2,X_3,ldots$ and $,X_1,Y_1,X_2,Y_2,X_3,Y_3,ldots,$ gives rise to an algebra-isomorphism (which is in particular a linear isomorphism).

share|cite|improve this answer

answered Aug 6 at 15:38

Babelfish

629114

add a comment | 

up vote
1
down vote

As Rhys Steele mentioned, two vector spaces over $k$ are isomorphic iff they have the same dimension. For an infinite dimensional $k$-vector space $V$, $dim(V) + dim(V)= dim(V)$ holds (see this question). Therefore $V cong V oplus V$.

Let's pick our favourite infinite-dimensional vector space and do this. Mine is $k[X]$, the vector space of polynomials in one variable over $k$.
If you add it with itself, you obtain $k[X] oplus k[Y] cong k[X,Y]$, the vector space of polynomials in two variables. All those vector spaces have dimension $mathbb N$, so they are abstractly isomorphic. But where is the isomorphism? $k[X]$ and $k[X,Y]$ look very different, don't they?

We can choose a basis of $k[X]$ (e.g. the monomial basis) and $k[X,Y]$ (e.g. the basis of all monomial products $1,X,Y,XY,X^2,Y^2,XY^2,X^2Ydots$). Choosing any bijection between those two basis (which exists, since $mathbb N times mathbb N$ and $mathbb N$ have the same cardinality, i.e. are both countably infinite) provides a vector space isomorphism, by linear extension.

You might feel a bit cheated now. Aren't $k[X]$ and $k[X,Y]$ different things? They are the same as vector spaces though. The difference is the additional structure on those two vector spaces, since you can multiply polynomials. The multiplication turns $k[X]$ and $k[X,Y]$ into algebras (even graded algebras), and the isomorphism above won't preserve this structure. If you want to fix this, you obtain an example which feels more natural:

Let $V=k[X_1,dots,X_n,ldots] = bigoplus_i=1^infty k[X_i]$ the vector space of polynomials with coefficients in $k$ and countably infinitely many unknowns. Like before, if you take the direct sum of $V$ with itself, you just "double" the unknowns. So $Voplus V$ is the vector space over twice as many unknowns (which could be called $X_1,X_2,X_3,ldots$ and $Y_1,Y_2,Y_3,ldots$). But that's still the same amount of unknowns. A bijection of the set $,X_1,X_2,X_3,ldots$ and $,X_1,Y_1,X_2,Y_2,X_3,Y_3,ldots,$ gives rise to an algebra-isomorphism (which is in particular a linear isomorphism).

share|cite|improve this answer

answered Aug 6 at 15:38

Babelfish

629114

add a comment | 

up vote
1
down vote

up vote
1
down vote

As Rhys Steele mentioned, two vector spaces over $k$ are isomorphic iff they have the same dimension. For an infinite dimensional $k$-vector space $V$, $dim(V) + dim(V)= dim(V)$ holds (see this question). Therefore $V cong V oplus V$.

Let's pick our favourite infinite-dimensional vector space and do this. Mine is $k[X]$, the vector space of polynomials in one variable over $k$.
If you add it with itself, you obtain $k[X] oplus k[Y] cong k[X,Y]$, the vector space of polynomials in two variables. All those vector spaces have dimension $mathbb N$, so they are abstractly isomorphic. But where is the isomorphism? $k[X]$ and $k[X,Y]$ look very different, don't they?

We can choose a basis of $k[X]$ (e.g. the monomial basis) and $k[X,Y]$ (e.g. the basis of all monomial products $1,X,Y,XY,X^2,Y^2,XY^2,X^2Ydots$). Choosing any bijection between those two basis (which exists, since $mathbb N times mathbb N$ and $mathbb N$ have the same cardinality, i.e. are both countably infinite) provides a vector space isomorphism, by linear extension.

You might feel a bit cheated now. Aren't $k[X]$ and $k[X,Y]$ different things? They are the same as vector spaces though. The difference is the additional structure on those two vector spaces, since you can multiply polynomials. The multiplication turns $k[X]$ and $k[X,Y]$ into algebras (even graded algebras), and the isomorphism above won't preserve this structure. If you want to fix this, you obtain an example which feels more natural:

Let $V=k[X_1,dots,X_n,ldots] = bigoplus_i=1^infty k[X_i]$ the vector space of polynomials with coefficients in $k$ and countably infinitely many unknowns. Like before, if you take the direct sum of $V$ with itself, you just "double" the unknowns. So $Voplus V$ is the vector space over twice as many unknowns (which could be called $X_1,X_2,X_3,ldots$ and $Y_1,Y_2,Y_3,ldots$). But that's still the same amount of unknowns. A bijection of the set $,X_1,X_2,X_3,ldots$ and $,X_1,Y_1,X_2,Y_2,X_3,Y_3,ldots,$ gives rise to an algebra-isomorphism (which is in particular a linear isomorphism).

share|cite|improve this answer

answered Aug 6 at 15:38

Babelfish

629114

As Rhys Steele mentioned, two vector spaces over $k$ are isomorphic iff they have the same dimension. For an infinite dimensional $k$-vector space $V$, $dim(V) + dim(V)= dim(V)$ holds (see this question). Therefore $V cong V oplus V$.

Let's pick our favourite infinite-dimensional vector space and do this. Mine is $k[X]$, the vector space of polynomials in one variable over $k$.
If you add it with itself, you obtain $k[X] oplus k[Y] cong k[X,Y]$, the vector space of polynomials in two variables. All those vector spaces have dimension $mathbb N$, so they are abstractly isomorphic. But where is the isomorphism? $k[X]$ and $k[X,Y]$ look very different, don't they?

We can choose a basis of $k[X]$ (e.g. the monomial basis) and $k[X,Y]$ (e.g. the basis of all monomial products $1,X,Y,XY,X^2,Y^2,XY^2,X^2Ydots$). Choosing any bijection between those two basis (which exists, since $mathbb N times mathbb N$ and $mathbb N$ have the same cardinality, i.e. are both countably infinite) provides a vector space isomorphism, by linear extension.

You might feel a bit cheated now. Aren't $k[X]$ and $k[X,Y]$ different things? They are the same as vector spaces though. The difference is the additional structure on those two vector spaces, since you can multiply polynomials. The multiplication turns $k[X]$ and $k[X,Y]$ into algebras (even graded algebras), and the isomorphism above won't preserve this structure. If you want to fix this, you obtain an example which feels more natural:

Let $V=k[X_1,dots,X_n,ldots] = bigoplus_i=1^infty k[X_i]$ the vector space of polynomials with coefficients in $k$ and countably infinitely many unknowns. Like before, if you take the direct sum of $V$ with itself, you just "double" the unknowns. So $Voplus V$ is the vector space over twice as many unknowns (which could be called $X_1,X_2,X_3,ldots$ and $Y_1,Y_2,Y_3,ldots$). But that's still the same amount of unknowns. A bijection of the set $,X_1,X_2,X_3,ldots$ and $,X_1,Y_1,X_2,Y_2,X_3,Y_3,ldots,$ gives rise to an algebra-isomorphism (which is in particular a linear isomorphism).

share|cite|improve this answer

answered Aug 6 at 15:38

Babelfish

629114

share|cite|improve this answer

share|cite|improve this answer

answered Aug 6 at 15:38

Babelfish

629114

answered Aug 6 at 15:38

Babelfish

629114

answered Aug 6 at 15:38

Babelfish

629114

629114

add a comment | 

add a comment |