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Tensor Product with Trivial Vector Space

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3

$begingroup$

This question seems obvious and yet I can't seem to find a good answer anywhere.

Let $V$ be a finite dimensional vector space, and let $0$ denote the trivial vector space. Is $V otimes 0 = 0$ or $V otimes 0 = V$? My gut tells me that it is the second case, but in thinking about dimension, tensor product should multiply dimension in which case I think it is the first case.

linear-algebra vector-spaces tensor-products

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asked 8 hours ago

Emilio MinichielloEmilio Minichiello

50411 silver badge1010 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The elements of $Votimes 0$ are sums of pure tensors of the form ${bf v}otimes{bf 0}$. Then, as we can slide scalars across the $otimes$ symbol, we have ${bf v}otimes{bf 0}={bf v}otimes 0{bf 0}=0{bf v}otimes{bf 0}={bf 0}otimes{bf 0}$. Thus, there is only one element in $Votimes 0$, the zero tensor.
  $endgroup$
  – runway44
  7 hours ago
  
  
  

  
  
  
  

add a comment | 

3

$begingroup$

This question seems obvious and yet I can't seem to find a good answer anywhere.

Let $V$ be a finite dimensional vector space, and let $0$ denote the trivial vector space. Is $V otimes 0 = 0$ or $V otimes 0 = V$? My gut tells me that it is the second case, but in thinking about dimension, tensor product should multiply dimension in which case I think it is the first case.

linear-algebra vector-spaces tensor-products

share|cite|improve this question

asked 8 hours ago

Emilio MinichielloEmilio Minichiello

50411 silver badge1010 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The elements of $Votimes 0$ are sums of pure tensors of the form ${bf v}otimes{bf 0}$. Then, as we can slide scalars across the $otimes$ symbol, we have ${bf v}otimes{bf 0}={bf v}otimes 0{bf 0}=0{bf v}otimes{bf 0}={bf 0}otimes{bf 0}$. Thus, there is only one element in $Votimes 0$, the zero tensor.
  $endgroup$
  – runway44
  7 hours ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

This question seems obvious and yet I can't seem to find a good answer anywhere.

Let $V$ be a finite dimensional vector space, and let $0$ denote the trivial vector space. Is $V otimes 0 = 0$ or $V otimes 0 = V$? My gut tells me that it is the second case, but in thinking about dimension, tensor product should multiply dimension in which case I think it is the first case.

linear-algebra vector-spaces tensor-products

share|cite|improve this question

asked 8 hours ago

Emilio MinichielloEmilio Minichiello

50411 silver badge1010 bronze badges

$endgroup$

share|cite|improve this question

asked 8 hours ago

Emilio MinichielloEmilio Minichiello

50411 silver badge1010 bronze badges

share|cite|improve this question

asked 8 hours ago

Emilio MinichielloEmilio Minichiello

50411 silver badge1010 bronze badges

asked 8 hours ago

Emilio MinichielloEmilio Minichiello

50411 silver badge1010 bronze badges

asked 8 hours ago

Emilio MinichielloEmilio Minichiello

50411 silver badge1010 bronze badges

2 Answers
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5

$begingroup$

Notice that the space of bilinear maps $f:Vtimes {0}to k$ consists of exactly the zero map, therefore the constant map $w:Vtimes {0}to{0}$ satisfies the universal property of the tensor product: $w$ is bilinear and any bilinear map $f:Vtimes{0}to k$ is the constant zero map, therefore the linear map $0:{0}to k$ satisfies $f=0circ w$. $0$ is also the only linear map ${0}to k$, so it's a fortiori the only linear map $g$ such that $f=gcirc w$.

share|cite|improve this answer

edited 1 hour ago

answered 7 hours ago

Gae. S.Gae. S.

79544 silver badges1414 bronze badges

$endgroup$

add a comment | 

4

$begingroup$

Recall that the tensor product $Votimes W$ of two finite-dimensional vector spaces $V$ and $W$ satisfy the dimension formula $dim(Votimes W)=dim(V)cdotdim(W)$.

So, tensoring a finite-dimensional vector space $V$ with the trivial vector space $0$ yields a vector space $Votimes 0$ with dimension
$$
dim(Votimes 0)=dim(V)cdotdim(0)=dim(V)cdot 0 = 0
$$
This implies that $Votimes0$ is itself the trivial vector space $Votimes 0=0$.

share|cite|improve this answer

answered 7 hours ago

Brian FitzpatrickBrian Fitzpatrick

22.4k44 gold badges3131 silver badges6262 bronze badges

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add a comment | 

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5

$begingroup$

Notice that the space of bilinear maps $f:Vtimes {0}to k$ consists of exactly the zero map, therefore the constant map $w:Vtimes {0}to{0}$ satisfies the universal property of the tensor product: $w$ is bilinear and any bilinear map $f:Vtimes{0}to k$ is the constant zero map, therefore the linear map $0:{0}to k$ satisfies $f=0circ w$. $0$ is also the only linear map ${0}to k$, so it's a fortiori the only linear map $g$ such that $f=gcirc w$.

share|cite|improve this answer

edited 1 hour ago

answered 7 hours ago

Gae. S.Gae. S.

79544 silver badges1414 bronze badges

$endgroup$

add a comment | 

5

$begingroup$

Notice that the space of bilinear maps $f:Vtimes {0}to k$ consists of exactly the zero map, therefore the constant map $w:Vtimes {0}to{0}$ satisfies the universal property of the tensor product: $w$ is bilinear and any bilinear map $f:Vtimes{0}to k$ is the constant zero map, therefore the linear map $0:{0}to k$ satisfies $f=0circ w$. $0$ is also the only linear map ${0}to k$, so it's a fortiori the only linear map $g$ such that $f=gcirc w$.

share|cite|improve this answer

edited 1 hour ago

answered 7 hours ago

Gae. S.Gae. S.

79544 silver badges1414 bronze badges

$endgroup$

add a comment | 

$begingroup$

Notice that the space of bilinear maps $f:Vtimes {0}to k$ consists of exactly the zero map, therefore the constant map $w:Vtimes {0}to{0}$ satisfies the universal property of the tensor product: $w$ is bilinear and any bilinear map $f:Vtimes{0}to k$ is the constant zero map, therefore the linear map $0:{0}to k$ satisfies $f=0circ w$. $0$ is also the only linear map ${0}to k$, so it's a fortiori the only linear map $g$ such that $f=gcirc w$.

share|cite|improve this answer

edited 1 hour ago

answered 7 hours ago

Gae. S.Gae. S.

79544 silver badges1414 bronze badges

$endgroup$

share|cite|improve this answer

edited 1 hour ago

answered 7 hours ago

Gae. S.Gae. S.

79544 silver badges1414 bronze badges

answered 7 hours ago

Gae. S.Gae. S.

79544 silver badges1414 bronze badges

answered 7 hours ago

Gae. S.Gae. S.

79544 silver badges1414 bronze badges

4

$begingroup$

Recall that the tensor product $Votimes W$ of two finite-dimensional vector spaces $V$ and $W$ satisfy the dimension formula $dim(Votimes W)=dim(V)cdotdim(W)$.

So, tensoring a finite-dimensional vector space $V$ with the trivial vector space $0$ yields a vector space $Votimes 0$ with dimension
$$
dim(Votimes 0)=dim(V)cdotdim(0)=dim(V)cdot 0 = 0
$$
This implies that $Votimes0$ is itself the trivial vector space $Votimes 0=0$.

share|cite|improve this answer

answered 7 hours ago

Brian FitzpatrickBrian Fitzpatrick

22.4k44 gold badges3131 silver badges6262 bronze badges

$endgroup$

add a comment | 

4

$begingroup$

Recall that the tensor product $Votimes W$ of two finite-dimensional vector spaces $V$ and $W$ satisfy the dimension formula $dim(Votimes W)=dim(V)cdotdim(W)$.

So, tensoring a finite-dimensional vector space $V$ with the trivial vector space $0$ yields a vector space $Votimes 0$ with dimension
$$
dim(Votimes 0)=dim(V)cdotdim(0)=dim(V)cdot 0 = 0
$$
This implies that $Votimes0$ is itself the trivial vector space $Votimes 0=0$.

share|cite|improve this answer

answered 7 hours ago

Brian FitzpatrickBrian Fitzpatrick

22.4k44 gold badges3131 silver badges6262 bronze badges

$endgroup$

add a comment | 

$begingroup$

Recall that the tensor product $Votimes W$ of two finite-dimensional vector spaces $V$ and $W$ satisfy the dimension formula $dim(Votimes W)=dim(V)cdotdim(W)$.

So, tensoring a finite-dimensional vector space $V$ with the trivial vector space $0$ yields a vector space $Votimes 0$ with dimension
$$
dim(Votimes 0)=dim(V)cdotdim(0)=dim(V)cdot 0 = 0
$$
This implies that $Votimes0$ is itself the trivial vector space $Votimes 0=0$.

share|cite|improve this answer

answered 7 hours ago

Brian FitzpatrickBrian Fitzpatrick

22.4k44 gold badges3131 silver badges6262 bronze badges

$endgroup$

share|cite|improve this answer

answered 7 hours ago

Brian FitzpatrickBrian Fitzpatrick

22.4k44 gold badges3131 silver badges6262 bronze badges

answered 7 hours ago

Brian FitzpatrickBrian Fitzpatrick

22.4k44 gold badges3131 silver badges6262 bronze badges

answered 7 hours ago

Brian FitzpatrickBrian Fitzpatrick

22.4k44 gold badges3131 silver badges6262 bronze badges