Tensor Product with Trivial Vector SpaceProjection on Tensor Product of Hilbert SpaceTensor product of a vector space and a fieldAlternative introduction to tensor products of vector spacesClarification of definition of tensor productBasis for Tensor Product of Infinite Dimensional Vector SpacesSymmetric kernel of tensor productUnderstanding definition of tensor productTensor Product Vector Space ExampleTensor product, Cartesian product and dualsTwo different definitions of tensor product space?
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Tensor Product with Trivial Vector Space
Projection on Tensor Product of Hilbert SpaceTensor product of a vector space and a fieldAlternative introduction to tensor products of vector spacesClarification of definition of tensor productBasis for Tensor Product of Infinite Dimensional Vector SpacesSymmetric kernel of tensor productUnderstanding definition of tensor productTensor Product Vector Space ExampleTensor product, Cartesian product and dualsTwo different definitions of tensor product space?
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$begingroup$
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
asked 8 hours ago
Emilio MinichielloEmilio Minichiello
5041 silver badge10 bronze badges
$endgroup$
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
asked 8 hours ago
Emilio MinichielloEmilio Minichiello
5041 silver badge10 bronze badges
$endgroup$
$begingroup$
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
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
asked 8 hours ago
Emilio MinichielloEmilio Minichiello
5041 silver badge10 bronze badges
$endgroup$
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
linear-algebra vector-spaces tensor-products
asked 8 hours ago
Emilio MinichielloEmilio Minichiello
5041 silver badge10 bronze badges
asked 8 hours ago
Emilio MinichielloEmilio Minichiello
5041 silver badge10 bronze badges
asked 8 hours ago
Emilio MinichielloEmilio Minichiello
5041 silver badge10 bronze badges
asked 8 hours ago
Emilio MinichielloEmilio Minichiello
5041 silver badge10 bronze badges
asked 8 hours ago
Emilio MinichielloEmilio Minichiello
5041 silver badge10 bronze badges
5041 silver badge10 bronze badges
$begingroup$
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
7 hours ago
add a comment |
$begingroup$
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
7 hours ago
$begingroup$
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
7 hours ago
$begingroup$
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
7 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 $f=gcirc w$.
answered 8 hours ago
Gae. S.Gae. S.
7954 silver badges14 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 $Votimes 0=0$.
answered 8 hours ago
Brian FitzpatrickBrian Fitzpatrick
22.4k4 gold badges31 silver badges62 bronze badges
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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active
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active
oldest
votes
$begingroup$
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 $f=gcirc w$.
answered 8 hours ago
Gae. S.Gae. S.
7954 silver badges14 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 $f=gcirc w$.
answered 8 hours ago
Gae. S.Gae. S.
7954 silver badges14 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 $f=gcirc w$.
answered 8 hours ago
Gae. S.Gae. S.
7954 silver badges14 bronze badges
$endgroup$
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 $f=gcirc w$.
answered 8 hours ago
Gae. S.Gae. S.
7954 silver badges14 bronze badges
edited 1 hour ago
answered 8 hours ago
Gae. S.Gae. S.
7954 silver badges14 bronze badges
answered 8 hours ago
Gae. S.Gae. S.
7954 silver badges14 bronze badges
answered 8 hours ago
Gae. S.Gae. S.
7954 silver badges14 bronze badges
7954 silver badges14 bronze badges
add a comment |
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 $Votimes 0=0$.
answered 8 hours ago
Brian FitzpatrickBrian Fitzpatrick
22.4k4 gold badges31 silver badges62 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 $Votimes 0=0$.
answered 8 hours ago
Brian FitzpatrickBrian Fitzpatrick
22.4k4 gold badges31 silver badges62 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 $Votimes 0=0$.
answered 8 hours ago
Brian FitzpatrickBrian Fitzpatrick
22.4k4 gold badges31 silver badges62 bronze badges
$endgroup$
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 $Votimes 0=0$.
answered 8 hours ago
Brian FitzpatrickBrian Fitzpatrick
22.4k4 gold badges31 silver badges62 bronze badges
answered 8 hours ago
Brian FitzpatrickBrian Fitzpatrick
22.4k4 gold badges31 silver badges62 bronze badges
answered 8 hours ago
Brian FitzpatrickBrian Fitzpatrick
22.4k4 gold badges31 silver badges62 bronze badges
answered 8 hours ago
Brian FitzpatrickBrian Fitzpatrick
22.4k4 gold badges31 silver badges62 bronze badges
22.4k4 gold badges31 silver badges62 bronze badges
add a comment |
add a comment |
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$begingroup$
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
7 hours ago