What is a linear isomorphism between $underset{ntimes m}{times} mathbb{R}$ and...
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What is a linear isomorphism between $underset{ntimes m}{times} mathbb{R}$ and $mathbb{R}^notimesmathbb{R}^m$? [closed]
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Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Isomorphism between $mathbb{C} X otimes mathbb{C} X$ and $mathbb{C} (X times X)$Are $mathbb{C} otimes _mathbb{R} mathbb{C}$ and $mathbb{C} otimes _mathbb{C} mathbb{C}$ isomorphic as $mathbb{R}$-vector spaces?Isomorphism between symmetric and upper triangular matricesTensor Algebra and IsomorphismIsomorphism between $U$ and $mathbb R^n$Natural Isomorphism between $V^*otimes W^*$ and $mathcal L^2(V,W; F)$.Commuting of Hom and Tensor Product functors?Oh Times, $otimes$ in linear algebra and tensorsNatural isomorphism between $V otimes V$ and bilinear forms on $V^*$.How to construct a $k[G]$ isomorphism $W^G otimes V cong (W otimes V_H)^G$.
$begingroup$
What is a linear isomorphism between
$underset{ntimes m}{times} mathbb{R}$ and $mathbb{R}^notimesmathbb{R}^m$?
Where $ntimes m :={(i,j):0le n-1,0le j le m-1}$.
Since $underset{ntimes m}{times} mathbb{R}$ consists of $nm$ copies of $mathbb{R}$ it has dimension $nm$. Also $mathbb{R}^notimesmathbb{R}^m$ has dimension $nm$. Hence they must be isomorphic as vector spaces.
Now what is a map $psi:underset{ntimes m}{times} mathbb{R}rightarrow mathbb{R}^notimesmathbb{R}^m$ that gives this isomorphism?
So what is $psi(w_{i,j})$? Where $w_{i,j}:ntimes m rightarrow mathbb{R}$.
linear-algebra vector-spaces tensor-products tensors multilinear-algebra
asked Mar 22 at 20:39
Jens WagemakerJens Wagemaker
593312
$endgroup$
closed as off-topic by Matthew Towers, Alex Provost, Leucippus, Shailesh, Parcly Taxel Mar 26 at 3:26
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Alex Provost, Leucippus, Shailesh, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
What is a linear isomorphism between
$underset{ntimes m}{times} mathbb{R}$ and $mathbb{R}^notimesmathbb{R}^m$?
Where $ntimes m :={(i,j):0le n-1,0le j le m-1}$.
Since $underset{ntimes m}{times} mathbb{R}$ consists of $nm$ copies of $mathbb{R}$ it has dimension $nm$. Also $mathbb{R}^notimesmathbb{R}^m$ has dimension $nm$. Hence they must be isomorphic as vector spaces.
Now what is a map $psi:underset{ntimes m}{times} mathbb{R}rightarrow mathbb{R}^notimesmathbb{R}^m$ that gives this isomorphism?
So what is $psi(w_{i,j})$? Where $w_{i,j}:ntimes m rightarrow mathbb{R}$.
linear-algebra vector-spaces tensor-products tensors multilinear-algebra
asked Mar 22 at 20:39
Jens WagemakerJens Wagemaker
593312
$endgroup$
closed as off-topic by Matthew Towers, Alex Provost, Leucippus, Shailesh, Parcly Taxel Mar 26 at 3:26
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Alex Provost, Leucippus, Shailesh, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
If $v_i$ are a basis of $Bbb{R}^n$ and $w_j$ are a basis of $Bbb{R}^m$, then $v_i otimes w_j$ are a basis of $Bbb{R}^n otimes Bbb{R}^m$. So just order these $v_1 otimes w_1$, $v_1 otimes w_2$, ..., etc, and map these to the basis vectors of $Bbb{R}^{mn}$.
$endgroup$
– Nick
Mar 22 at 20:53
1
$begingroup$
You say "the linear isomorphism". Note that "the" is the singular definite article, and implies uniqueness: there is one linear isomorphism, or there is a prefered linear isomorphism. There is no such thing here; you should ask for " a linear isomorphism".
$endgroup$
– Arturo Magidin
Mar 22 at 21:08
1
$begingroup$
The map is the wrong question. Trivially, there are many linear isomorphisms between both spaces because have the same dimension.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 22 at 21:09
$begingroup$
Your both right; I replaced it by a
$endgroup$
– Jens Wagemaker
Mar 22 at 21:10
1
$begingroup$
Are you thinking of $times_{ntimes m}mathbb{R}$ as the vector space of linear maps $mathbb{R}^m to mathbb{R}^n$? If so, there is a natural isomorphism $text{Hom}_F(V,W) to Wotimes V^*$ for any finite-dimensional $F$-vector spaces $V,W$, which may be more satisfying than choosing two arbitrary bases and identifying them. Of course, you'll need to choose bases to identify linear maps with matrices in the end.
$endgroup$
– csprun
Mar 22 at 21:18
add a comment |
$begingroup$
What is a linear isomorphism between
$underset{ntimes m}{times} mathbb{R}$ and $mathbb{R}^notimesmathbb{R}^m$?
Where $ntimes m :={(i,j):0le n-1,0le j le m-1}$.
Since $underset{ntimes m}{times} mathbb{R}$ consists of $nm$ copies of $mathbb{R}$ it has dimension $nm$. Also $mathbb{R}^notimesmathbb{R}^m$ has dimension $nm$. Hence they must be isomorphic as vector spaces.
Now what is a map $psi:underset{ntimes m}{times} mathbb{R}rightarrow mathbb{R}^notimesmathbb{R}^m$ that gives this isomorphism?
So what is $psi(w_{i,j})$? Where $w_{i,j}:ntimes m rightarrow mathbb{R}$.
linear-algebra vector-spaces tensor-products tensors multilinear-algebra
asked Mar 22 at 20:39
Jens WagemakerJens Wagemaker
593312
$endgroup$
What is a linear isomorphism between
$underset{ntimes m}{times} mathbb{R}$ and $mathbb{R}^notimesmathbb{R}^m$?
Where $ntimes m :={(i,j):0le n-1,0le j le m-1}$.
Since $underset{ntimes m}{times} mathbb{R}$ consists of $nm$ copies of $mathbb{R}$ it has dimension $nm$. Also $mathbb{R}^notimesmathbb{R}^m$ has dimension $nm$. Hence they must be isomorphic as vector spaces.
Now what is a map $psi:underset{ntimes m}{times} mathbb{R}rightarrow mathbb{R}^notimesmathbb{R}^m$ that gives this isomorphism?
So what is $psi(w_{i,j})$? Where $w_{i,j}:ntimes m rightarrow mathbb{R}$.
linear-algebra vector-spaces tensor-products tensors multilinear-algebra
linear-algebra vector-spaces tensor-products tensors multilinear-algebra
asked Mar 22 at 20:39
Jens WagemakerJens Wagemaker
593312
asked Mar 22 at 20:39
Jens WagemakerJens Wagemaker
593312
edited Mar 22 at 21:10
Jens Wagemaker
asked Mar 22 at 20:39
Jens WagemakerJens Wagemaker
593312
asked Mar 22 at 20:39
Jens WagemakerJens Wagemaker
593312
asked Mar 22 at 20:39
Jens WagemakerJens Wagemaker
593312
593312
closed as off-topic by Matthew Towers, Alex Provost, Leucippus, Shailesh, Parcly Taxel Mar 26 at 3:26
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Alex Provost, Leucippus, Shailesh, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Matthew Towers, Alex Provost, Leucippus, Shailesh, Parcly Taxel Mar 26 at 3:26
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Alex Provost, Leucippus, Shailesh, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
If $v_i$ are a basis of $Bbb{R}^n$ and $w_j$ are a basis of $Bbb{R}^m$, then $v_i otimes w_j$ are a basis of $Bbb{R}^n otimes Bbb{R}^m$. So just order these $v_1 otimes w_1$, $v_1 otimes w_2$, ..., etc, and map these to the basis vectors of $Bbb{R}^{mn}$.
$endgroup$
– Nick
Mar 22 at 20:53
1
$begingroup$
You say "the linear isomorphism". Note that "the" is the singular definite article, and implies uniqueness: there is one linear isomorphism, or there is a prefered linear isomorphism. There is no such thing here; you should ask for " a linear isomorphism".
$endgroup$
– Arturo Magidin
Mar 22 at 21:08
1
$begingroup$
The map is the wrong question. Trivially, there are many linear isomorphisms between both spaces because have the same dimension.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 22 at 21:09
$begingroup$
Your both right; I replaced it by a
$endgroup$
– Jens Wagemaker
Mar 22 at 21:10
1
$begingroup$
Are you thinking of $times_{ntimes m}mathbb{R}$ as the vector space of linear maps $mathbb{R}^m to mathbb{R}^n$? If so, there is a natural isomorphism $text{Hom}_F(V,W) to Wotimes V^*$ for any finite-dimensional $F$-vector spaces $V,W$, which may be more satisfying than choosing two arbitrary bases and identifying them. Of course, you'll need to choose bases to identify linear maps with matrices in the end.
$endgroup$
– csprun
Mar 22 at 21:18
add a comment |
2
$begingroup$
If $v_i$ are a basis of $Bbb{R}^n$ and $w_j$ are a basis of $Bbb{R}^m$, then $v_i otimes w_j$ are a basis of $Bbb{R}^n otimes Bbb{R}^m$. So just order these $v_1 otimes w_1$, $v_1 otimes w_2$, ..., etc, and map these to the basis vectors of $Bbb{R}^{mn}$.
$endgroup$
– Nick
Mar 22 at 20:53
1
$begingroup$
You say "the linear isomorphism". Note that "the" is the singular definite article, and implies uniqueness: there is one linear isomorphism, or there is a prefered linear isomorphism. There is no such thing here; you should ask for " a linear isomorphism".
$endgroup$
– Arturo Magidin
Mar 22 at 21:08
1
$begingroup$
The map is the wrong question. Trivially, there are many linear isomorphisms between both spaces because have the same dimension.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 22 at 21:09
$begingroup$
Your both right; I replaced it by a
$endgroup$
– Jens Wagemaker
Mar 22 at 21:10
1
$begingroup$
Are you thinking of $times_{ntimes m}mathbb{R}$ as the vector space of linear maps $mathbb{R}^m to mathbb{R}^n$? If so, there is a natural isomorphism $text{Hom}_F(V,W) to Wotimes V^*$ for any finite-dimensional $F$-vector spaces $V,W$, which may be more satisfying than choosing two arbitrary bases and identifying them. Of course, you'll need to choose bases to identify linear maps with matrices in the end.
$endgroup$
– csprun
Mar 22 at 21:18
2
2
$begingroup$
If $v_i$ are a basis of $Bbb{R}^n$ and $w_j$ are a basis of $Bbb{R}^m$, then $v_i otimes w_j$ are a basis of $Bbb{R}^n otimes Bbb{R}^m$. So just order these $v_1 otimes w_1$, $v_1 otimes w_2$, ..., etc, and map these to the basis vectors of $Bbb{R}^{mn}$.
$endgroup$
– Nick
Mar 22 at 20:53
$begingroup$
If $v_i$ are a basis of $Bbb{R}^n$ and $w_j$ are a basis of $Bbb{R}^m$, then $v_i otimes w_j$ are a basis of $Bbb{R}^n otimes Bbb{R}^m$. So just order these $v_1 otimes w_1$, $v_1 otimes w_2$, ..., etc, and map these to the basis vectors of $Bbb{R}^{mn}$.
$endgroup$
– Nick
Mar 22 at 20:53
1
1
$begingroup$
You say "the linear isomorphism". Note that "the" is the singular definite article, and implies uniqueness: there is one linear isomorphism, or there is a prefered linear isomorphism. There is no such thing here; you should ask for " a linear isomorphism".
$endgroup$
– Arturo Magidin
Mar 22 at 21:08
$begingroup$
You say "the linear isomorphism". Note that "the" is the singular definite article, and implies uniqueness: there is one linear isomorphism, or there is a prefered linear isomorphism. There is no such thing here; you should ask for " a linear isomorphism".
$endgroup$
– Arturo Magidin
Mar 22 at 21:08
1
1
$begingroup$
The map is the wrong question. Trivially, there are many linear isomorphisms between both spaces because have the same dimension.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 22 at 21:09
$begingroup$
The map is the wrong question. Trivially, there are many linear isomorphisms between both spaces because have the same dimension.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 22 at 21:09
$begingroup$
Your both right; I replaced it by a
$endgroup$
– Jens Wagemaker
Mar 22 at 21:10
$begingroup$
Your both right; I replaced it by a
$endgroup$
– Jens Wagemaker
Mar 22 at 21:10
1
1
$begingroup$
Are you thinking of $times_{ntimes m}mathbb{R}$ as the vector space of linear maps $mathbb{R}^m to mathbb{R}^n$? If so, there is a natural isomorphism $text{Hom}_F(V,W) to Wotimes V^*$ for any finite-dimensional $F$-vector spaces $V,W$, which may be more satisfying than choosing two arbitrary bases and identifying them. Of course, you'll need to choose bases to identify linear maps with matrices in the end.
$endgroup$
– csprun
Mar 22 at 21:18
$begingroup$
Are you thinking of $times_{ntimes m}mathbb{R}$ as the vector space of linear maps $mathbb{R}^m to mathbb{R}^n$? If so, there is a natural isomorphism $text{Hom}_F(V,W) to Wotimes V^*$ for any finite-dimensional $F$-vector spaces $V,W$, which may be more satisfying than choosing two arbitrary bases and identifying them. Of course, you'll need to choose bases to identify linear maps with matrices in the end.
$endgroup$
– csprun
Mar 22 at 21:18
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let ${e_1^n, dots, e_n^n}$ be the standard basis on $mathbb{R}^n$ and ${e_1^m, dots, e_m^m}$ be the standard basis on $mathbb{R}^m$.
Then a basis of $mathbb{R}^n otimes mathbb{R}^m$ is given by ${e_i^n otimes e_j^m: 0 le i le n-1, 0 le j le m -1 }$.
We define $psi(w_{i,j}) = sum_{i,j}w_{i,j} ; e_i^n otimes e_j^m$.
answered Mar 22 at 21:23
Jens WagemakerJens Wagemaker
593312
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let ${e_1^n, dots, e_n^n}$ be the standard basis on $mathbb{R}^n$ and ${e_1^m, dots, e_m^m}$ be the standard basis on $mathbb{R}^m$.
Then a basis of $mathbb{R}^n otimes mathbb{R}^m$ is given by ${e_i^n otimes e_j^m: 0 le i le n-1, 0 le j le m -1 }$.
We define $psi(w_{i,j}) = sum_{i,j}w_{i,j} ; e_i^n otimes e_j^m$.
answered Mar 22 at 21:23
Jens WagemakerJens Wagemaker
593312
$endgroup$
add a comment |
$begingroup$
Let ${e_1^n, dots, e_n^n}$ be the standard basis on $mathbb{R}^n$ and ${e_1^m, dots, e_m^m}$ be the standard basis on $mathbb{R}^m$.
Then a basis of $mathbb{R}^n otimes mathbb{R}^m$ is given by ${e_i^n otimes e_j^m: 0 le i le n-1, 0 le j le m -1 }$.
We define $psi(w_{i,j}) = sum_{i,j}w_{i,j} ; e_i^n otimes e_j^m$.
answered Mar 22 at 21:23
Jens WagemakerJens Wagemaker
593312
$endgroup$
add a comment |
$begingroup$
Let ${e_1^n, dots, e_n^n}$ be the standard basis on $mathbb{R}^n$ and ${e_1^m, dots, e_m^m}$ be the standard basis on $mathbb{R}^m$.
Then a basis of $mathbb{R}^n otimes mathbb{R}^m$ is given by ${e_i^n otimes e_j^m: 0 le i le n-1, 0 le j le m -1 }$.
We define $psi(w_{i,j}) = sum_{i,j}w_{i,j} ; e_i^n otimes e_j^m$.
answered Mar 22 at 21:23
Jens WagemakerJens Wagemaker
593312
$endgroup$
Let ${e_1^n, dots, e_n^n}$ be the standard basis on $mathbb{R}^n$ and ${e_1^m, dots, e_m^m}$ be the standard basis on $mathbb{R}^m$.
Then a basis of $mathbb{R}^n otimes mathbb{R}^m$ is given by ${e_i^n otimes e_j^m: 0 le i le n-1, 0 le j le m -1 }$.
We define $psi(w_{i,j}) = sum_{i,j}w_{i,j} ; e_i^n otimes e_j^m$.
answered Mar 22 at 21:23
Jens WagemakerJens Wagemaker
593312
answered Mar 22 at 21:23
Jens WagemakerJens Wagemaker
593312
answered Mar 22 at 21:23
Jens WagemakerJens Wagemaker
593312
answered Mar 22 at 21:23
Jens WagemakerJens Wagemaker
593312
593312
add a comment |
add a comment |
2
$begingroup$
If $v_i$ are a basis of $Bbb{R}^n$ and $w_j$ are a basis of $Bbb{R}^m$, then $v_i otimes w_j$ are a basis of $Bbb{R}^n otimes Bbb{R}^m$. So just order these $v_1 otimes w_1$, $v_1 otimes w_2$, ..., etc, and map these to the basis vectors of $Bbb{R}^{mn}$.
$endgroup$
– Nick
Mar 22 at 20:53
1
$begingroup$
You say "the linear isomorphism". Note that "the" is the singular definite article, and implies uniqueness: there is one linear isomorphism, or there is a prefered linear isomorphism. There is no such thing here; you should ask for " a linear isomorphism".
$endgroup$
– Arturo Magidin
Mar 22 at 21:08
1
$begingroup$
The map is the wrong question. Trivially, there are many linear isomorphisms between both spaces because have the same dimension.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 22 at 21:09
$begingroup$
Your both right; I replaced it by a
$endgroup$
– Jens Wagemaker
Mar 22 at 21:10
1
$begingroup$
Are you thinking of $times_{ntimes m}mathbb{R}$ as the vector space of linear maps $mathbb{R}^m to mathbb{R}^n$? If so, there is a natural isomorphism $text{Hom}_F(V,W) to Wotimes V^*$ for any finite-dimensional $F$-vector spaces $V,W$, which may be more satisfying than choosing two arbitrary bases and identifying them. Of course, you'll need to choose bases to identify linear maps with matrices in the end.
$endgroup$
– csprun
Mar 22 at 21:18