Why is the tensor product of two vector spaces a vector space?











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We defined the tensor product of $v$ and $w$ to just mean a "symbol" $e_{vw}$, then considered the subspace spanned by all these symbols, and finally we quotient out by relations to make the tensor product between two vectors bilinear. The resulting vector space is the tensor product of $V$ and $W$.



I don't understand why this is a vector space.. sure, we've said it's the span of a bunch of symbols, but what does it even mean to "add" together two symbols? How is equality defined in our vector space?






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    You can "generate" the smallest vector space containing your "symbols" $e_{vw}$, no problem. It will be the linear span, as you said. A linear span is a vector space by definition.
    – Dietrich Burde
    2 days ago












  • If $V$ is $n$-dimensional, and $W$ is $m$-dimensional, then the tensor product is $nm$-dimensional. So as a vector space, $V otimes W$ is just $Bbb{R}^{nm}$. The "symbols" $e_{vw}$ just correspond to the standard basis vectors in $Bbb{R}^{nm}$, after choosing some ordering of the bases of $V$ and $W$.
    – Nick
    2 days ago

















up vote
0
down vote

favorite












We defined the tensor product of $v$ and $w$ to just mean a "symbol" $e_{vw}$, then considered the subspace spanned by all these symbols, and finally we quotient out by relations to make the tensor product between two vectors bilinear. The resulting vector space is the tensor product of $V$ and $W$.



I don't understand why this is a vector space.. sure, we've said it's the span of a bunch of symbols, but what does it even mean to "add" together two symbols? How is equality defined in our vector space?






vector-spaces tensor-products tensors







share|cite|improve this question


















  • 2




    You can "generate" the smallest vector space containing your "symbols" $e_{vw}$, no problem. It will be the linear span, as you said. A linear span is a vector space by definition.
    – Dietrich Burde
    2 days ago












  • If $V$ is $n$-dimensional, and $W$ is $m$-dimensional, then the tensor product is $nm$-dimensional. So as a vector space, $V otimes W$ is just $Bbb{R}^{nm}$. The "symbols" $e_{vw}$ just correspond to the standard basis vectors in $Bbb{R}^{nm}$, after choosing some ordering of the bases of $V$ and $W$.
    – Nick
    2 days ago















up vote
0
down vote

favorite









up vote
0
down vote

favorite











We defined the tensor product of $v$ and $w$ to just mean a "symbol" $e_{vw}$, then considered the subspace spanned by all these symbols, and finally we quotient out by relations to make the tensor product between two vectors bilinear. The resulting vector space is the tensor product of $V$ and $W$.



I don't understand why this is a vector space.. sure, we've said it's the span of a bunch of symbols, but what does it even mean to "add" together two symbols? How is equality defined in our vector space?






vector-spaces tensor-products tensors







share|cite|improve this question













We defined the tensor product of $v$ and $w$ to just mean a "symbol" $e_{vw}$, then considered the subspace spanned by all these symbols, and finally we quotient out by relations to make the tensor product between two vectors bilinear. The resulting vector space is the tensor product of $V$ and $W$.



I don't understand why this is a vector space.. sure, we've said it's the span of a bunch of symbols, but what does it even mean to "add" together two symbols? How is equality defined in our vector space?






vector-spaces tensor-products tensors




vector-spaces tensor-products tensors






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 2 days ago









Saad

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500210








  • 2




    You can "generate" the smallest vector space containing your "symbols" $e_{vw}$, no problem. It will be the linear span, as you said. A linear span is a vector space by definition.
    – Dietrich Burde
    2 days ago












  • If $V$ is $n$-dimensional, and $W$ is $m$-dimensional, then the tensor product is $nm$-dimensional. So as a vector space, $V otimes W$ is just $Bbb{R}^{nm}$. The "symbols" $e_{vw}$ just correspond to the standard basis vectors in $Bbb{R}^{nm}$, after choosing some ordering of the bases of $V$ and $W$.
    – Nick
    2 days ago
















  • 2




    You can "generate" the smallest vector space containing your "symbols" $e_{vw}$, no problem. It will be the linear span, as you said. A linear span is a vector space by definition.
    – Dietrich Burde
    2 days ago












  • If $V$ is $n$-dimensional, and $W$ is $m$-dimensional, then the tensor product is $nm$-dimensional. So as a vector space, $V otimes W$ is just $Bbb{R}^{nm}$. The "symbols" $e_{vw}$ just correspond to the standard basis vectors in $Bbb{R}^{nm}$, after choosing some ordering of the bases of $V$ and $W$.
    – Nick
    2 days ago










2




2




You can "generate" the smallest vector space containing your "symbols" $e_{vw}$, no problem. It will be the linear span, as you said. A linear span is a vector space by definition.
– Dietrich Burde
2 days ago






You can "generate" the smallest vector space containing your "symbols" $e_{vw}$, no problem. It will be the linear span, as you said. A linear span is a vector space by definition.
– Dietrich Burde
2 days ago














If $V$ is $n$-dimensional, and $W$ is $m$-dimensional, then the tensor product is $nm$-dimensional. So as a vector space, $V otimes W$ is just $Bbb{R}^{nm}$. The "symbols" $e_{vw}$ just correspond to the standard basis vectors in $Bbb{R}^{nm}$, after choosing some ordering of the bases of $V$ and $W$.
– Nick
2 days ago






If $V$ is $n$-dimensional, and $W$ is $m$-dimensional, then the tensor product is $nm$-dimensional. So as a vector space, $V otimes W$ is just $Bbb{R}^{nm}$. The "symbols" $e_{vw}$ just correspond to the standard basis vectors in $Bbb{R}^{nm}$, after choosing some ordering of the bases of $V$ and $W$.
– Nick
2 days ago

















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