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Tensor Product with Trivial Vector Space

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6

$begingroup$

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 Jul 24 at 22:23

Emilio MinichielloEmilio Minichiello

53111 silver badge1010 bronze badges

$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  The elements of $Votimes 0$ are sums of pure tensors of the form $bf votimesbf 0$. Then, as we can slide scalars across the $otimes$ symbol, we have $bf votimesbf 0=bf votimes 0bf 0=0bf votimesbf 0=bf 0otimesbf 0$. Thus, there is only one element in $Votimes 0$, the zero tensor.
  $endgroup$
  – runway44
  Jul 24 at 22:54
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @runway44 that is IMO a better answer than the two given ones. Why did you post it as a comment?
  $endgroup$
  – leftaroundabout
  Jul 25 at 10:11
  

  
  
  
  

add a comment | 

6

$begingroup$

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 Jul 24 at 22:23

Emilio MinichielloEmilio Minichiello

53111 silver badge1010 bronze badges

$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  The elements of $Votimes 0$ are sums of pure tensors of the form $bf votimesbf 0$. Then, as we can slide scalars across the $otimes$ symbol, we have $bf votimesbf 0=bf votimes 0bf 0=0bf votimesbf 0=bf 0otimesbf 0$. Thus, there is only one element in $Votimes 0$, the zero tensor.
  $endgroup$
  – runway44
  Jul 24 at 22:54
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @runway44 that is IMO a better answer than the two given ones. Why did you post it as a comment?
  $endgroup$
  – leftaroundabout
  Jul 25 at 10:11
  

  
  
  
  

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 Jul 24 at 22:23

Emilio MinichielloEmilio Minichiello

53111 silver badge1010 bronze badges

$endgroup$

share|cite|improve this question

asked Jul 24 at 22:23

Emilio MinichielloEmilio Minichiello

53111 silver badge1010 bronze badges

share|cite|improve this question

asked Jul 24 at 22:23

Emilio MinichielloEmilio Minichiello

53111 silver badge1010 bronze badges

asked Jul 24 at 22:23

Emilio MinichielloEmilio Minichiello

53111 silver badge1010 bronze badges

asked Jul 24 at 22:23

Emilio MinichielloEmilio Minichiello

53111 silver badge1010 bronze badges

3 Answers
3

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oldest

votes

8

$begingroup$

Notice that the space of bilinear maps $f:Vtimes 0to k$ consists of exactly the zero map, therefore the constant map $w:Vtimes 0to0$ satisfies the universal property of the tensor product: $w$ is bilinear and any bilinear map $f:Vtimes0to k$ is the constant zero map, therefore the linear map $0:0to k$ satisfies $f=0circ w$. $0$ is also the only linear map $0to k$, so it's a fortiori the only linear map $g$ such that $f=gcirc w$.

share|cite|improve this answer

edited Jul 25 at 5:24

answered Jul 24 at 22:38

Gae. S.Gae. S.

1,13055 silver badges1414 bronze badges

$endgroup$

add a comment | 

7

$begingroup$

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 Jul 24 at 22:42

Brian FitzpatrickBrian Fitzpatrick

22.8k55 gold badges3131 silver badges6464 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

The elements of $Votimes 0$ are sums of pure tensors of the form $bf v⊗bf 0$. Then, as we can slide scalars across the $otimes$ symbol, we have $bf votimesbf 0=bf votimes (0cdotbf 0)=(0cdotbf v)otimesbf 0=bf 0otimesbf 0$.

Thus, there is only one element in $Votimes 0$, the zero tensor.

share|cite|improve this answer

answered Jul 26 at 5:22

runway44runway44

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add a comment | 

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8

$begingroup$

Notice that the space of bilinear maps $f:Vtimes 0to k$ consists of exactly the zero map, therefore the constant map $w:Vtimes 0to0$ satisfies the universal property of the tensor product: $w$ is bilinear and any bilinear map $f:Vtimes0to k$ is the constant zero map, therefore the linear map $0:0to k$ satisfies $f=0circ w$. $0$ is also the only linear map $0to k$, so it's a fortiori the only linear map $g$ such that $f=gcirc w$.

share|cite|improve this answer

edited Jul 25 at 5:24

answered Jul 24 at 22:38

Gae. S.Gae. S.

1,13055 silver badges1414 bronze badges

$endgroup$

add a comment | 

8

$begingroup$

Notice that the space of bilinear maps $f:Vtimes 0to k$ consists of exactly the zero map, therefore the constant map $w:Vtimes 0to0$ satisfies the universal property of the tensor product: $w$ is bilinear and any bilinear map $f:Vtimes0to k$ is the constant zero map, therefore the linear map $0:0to k$ satisfies $f=0circ w$. $0$ is also the only linear map $0to k$, so it's a fortiori the only linear map $g$ such that $f=gcirc w$.

share|cite|improve this answer

edited Jul 25 at 5:24

answered Jul 24 at 22:38

Gae. S.Gae. S.

1,13055 silver badges1414 bronze badges

$endgroup$

add a comment | 

$begingroup$

Notice that the space of bilinear maps $f:Vtimes 0to k$ consists of exactly the zero map, therefore the constant map $w:Vtimes 0to0$ satisfies the universal property of the tensor product: $w$ is bilinear and any bilinear map $f:Vtimes0to k$ is the constant zero map, therefore the linear map $0:0to k$ satisfies $f=0circ w$. $0$ is also the only linear map $0to k$, so it's a fortiori the only linear map $g$ such that $f=gcirc w$.

share|cite|improve this answer

edited Jul 25 at 5:24

answered Jul 24 at 22:38

Gae. S.Gae. S.

1,13055 silver badges1414 bronze badges

$endgroup$

share|cite|improve this answer

edited Jul 25 at 5:24

answered Jul 24 at 22:38

Gae. S.Gae. S.

1,13055 silver badges1414 bronze badges

answered Jul 24 at 22:38

Gae. S.Gae. S.

1,13055 silver badges1414 bronze badges

answered Jul 24 at 22:38

Gae. S.Gae. S.

1,13055 silver badges1414 bronze badges

7

$begingroup$

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 Jul 24 at 22:42

Brian FitzpatrickBrian Fitzpatrick

22.8k55 gold badges3131 silver badges6464 bronze badges

$endgroup$

add a comment | 

7

$begingroup$

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 Jul 24 at 22:42

Brian FitzpatrickBrian Fitzpatrick

22.8k55 gold badges3131 silver badges6464 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 Jul 24 at 22:42

Brian FitzpatrickBrian Fitzpatrick

22.8k55 gold badges3131 silver badges6464 bronze badges

$endgroup$

share|cite|improve this answer

answered Jul 24 at 22:42

Brian FitzpatrickBrian Fitzpatrick

22.8k55 gold badges3131 silver badges6464 bronze badges

answered Jul 24 at 22:42

Brian FitzpatrickBrian Fitzpatrick

22.8k55 gold badges3131 silver badges6464 bronze badges

answered Jul 24 at 22:42

Brian FitzpatrickBrian Fitzpatrick

22.8k55 gold badges3131 silver badges6464 bronze badges

1

$begingroup$

The elements of $Votimes 0$ are sums of pure tensors of the form $bf v⊗bf 0$. Then, as we can slide scalars across the $otimes$ symbol, we have $bf votimesbf 0=bf votimes (0cdotbf 0)=(0cdotbf v)otimesbf 0=bf 0otimesbf 0$.

Thus, there is only one element in $Votimes 0$, the zero tensor.

share|cite|improve this answer

answered Jul 26 at 5:22

runway44runway44

1,44011 gold badge22 silver badges99 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

The elements of $Votimes 0$ are sums of pure tensors of the form $bf v⊗bf 0$. Then, as we can slide scalars across the $otimes$ symbol, we have $bf votimesbf 0=bf votimes (0cdotbf 0)=(0cdotbf v)otimesbf 0=bf 0otimesbf 0$.

Thus, there is only one element in $Votimes 0$, the zero tensor.

share|cite|improve this answer

answered Jul 26 at 5:22

runway44runway44

1,44011 gold badge22 silver badges99 bronze badges

$endgroup$

add a comment | 

$begingroup$

The elements of $Votimes 0$ are sums of pure tensors of the form $bf v⊗bf 0$. Then, as we can slide scalars across the $otimes$ symbol, we have $bf votimesbf 0=bf votimes (0cdotbf 0)=(0cdotbf v)otimesbf 0=bf 0otimesbf 0$.

Thus, there is only one element in $Votimes 0$, the zero tensor.

share|cite|improve this answer

answered Jul 26 at 5:22

runway44runway44

1,44011 gold badge22 silver badges99 bronze badges

$endgroup$

share|cite|improve this answer

answered Jul 26 at 5:22

runway44runway44

1,44011 gold badge22 silver badges99 bronze badges

answered Jul 26 at 5:22

runway44runway44

1,44011 gold badge22 silver badges99 bronze badges

answered Jul 26 at 5:22

runway44runway44

1,44011 gold badge22 silver badges99 bronze badges