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?
vector-spaces tensor-products tensors
asked Nov 19 at 19:11
Saad
500210
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up vote
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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
asked Nov 19 at 19:11
Saad
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
Nov 19 at 19:13
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
Nov 19 at 20:57
@DietrichBurde but my question is why is the linear span "well-defined"? For example, suppose we consider the span {$(1,0)$,$ (1,1,1)$}? It doesn't make sense since we have no method of adding $(1,0)$ and $(1,1,1)$, which is the point of confusion for me
– Saad
Nov 26 at 18:24
1
We do not add $e_v=(1,0)$ and $e_w=(1,1,1)$, but we add elements of type $e_{vw}=e_votimes e_w$. The basis for $Votimes W$ consists of vectors $e_{vw}$ and not of $e_v$, $e_w$, which are basis vectors of $V$ and $W$ respectively.
– Dietrich Burde
Nov 26 at 19:14
add a comment |
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
asked Nov 19 at 19:11
Saad
500210
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
asked Nov 19 at 19:11
Saad
500210
asked Nov 19 at 19:11
Saad
500210
asked Nov 19 at 19:11
Saad
500210
asked Nov 19 at 19:11
Saad
500210
asked Nov 19 at 19:11
Saad
500210
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
Nov 19 at 19:13
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
Nov 19 at 20:57
@DietrichBurde but my question is why is the linear span "well-defined"? For example, suppose we consider the span {$(1,0)$,$ (1,1,1)$}? It doesn't make sense since we have no method of adding $(1,0)$ and $(1,1,1)$, which is the point of confusion for me
– Saad
Nov 26 at 18:24
1
We do not add $e_v=(1,0)$ and $e_w=(1,1,1)$, but we add elements of type $e_{vw}=e_votimes e_w$. The basis for $Votimes W$ consists of vectors $e_{vw}$ and not of $e_v$, $e_w$, which are basis vectors of $V$ and $W$ respectively.
– Dietrich Burde
Nov 26 at 19:14
add a comment |
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
Nov 19 at 19:13
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
Nov 19 at 20:57
@DietrichBurde but my question is why is the linear span "well-defined"? For example, suppose we consider the span {$(1,0)$,$ (1,1,1)$}? It doesn't make sense since we have no method of adding $(1,0)$ and $(1,1,1)$, which is the point of confusion for me
– Saad
Nov 26 at 18:24
1
We do not add $e_v=(1,0)$ and $e_w=(1,1,1)$, but we add elements of type $e_{vw}=e_votimes e_w$. The basis for $Votimes W$ consists of vectors $e_{vw}$ and not of $e_v$, $e_w$, which are basis vectors of $V$ and $W$ respectively.
– Dietrich Burde
Nov 26 at 19:14
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
Nov 19 at 19:13
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
Nov 19 at 19:13
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
Nov 19 at 20:57
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
Nov 19 at 20:57
@DietrichBurde but my question is why is the linear span "well-defined"? For example, suppose we consider the span {$(1,0)$,$ (1,1,1)$}? It doesn't make sense since we have no method of adding $(1,0)$ and $(1,1,1)$, which is the point of confusion for me
– Saad
Nov 26 at 18:24
@DietrichBurde but my question is why is the linear span "well-defined"? For example, suppose we consider the span {$(1,0)$,$ (1,1,1)$}? It doesn't make sense since we have no method of adding $(1,0)$ and $(1,1,1)$, which is the point of confusion for me
– Saad
Nov 26 at 18:24
1
1
We do not add $e_v=(1,0)$ and $e_w=(1,1,1)$, but we add elements of type $e_{vw}=e_votimes e_w$. The basis for $Votimes W$ consists of vectors $e_{vw}$ and not of $e_v$, $e_w$, which are basis vectors of $V$ and $W$ respectively.
– Dietrich Burde
Nov 26 at 19:14
We do not add $e_v=(1,0)$ and $e_w=(1,1,1)$, but we add elements of type $e_{vw}=e_votimes e_w$. The basis for $Votimes W$ consists of vectors $e_{vw}$ and not of $e_v$, $e_w$, which are basis vectors of $V$ and $W$ respectively.
– Dietrich Burde
Nov 26 at 19:14
add a comment |
1 Answer
1
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0
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If $v_1,cdots v_n$ is a basis of $V$ and $w_1,ldots w_m$ is a basis of $W$, then
$$
{v_iotimes w_jmid 1le ile n,1le jle m}
$$
is a basis of $Votimes W$. So we write $e_{vw}=e_votimes e_w$.
answered Nov 26 at 19:20
Dietrich Burde
76.7k64286
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
If $v_1,cdots v_n$ is a basis of $V$ and $w_1,ldots w_m$ is a basis of $W$, then
$$
{v_iotimes w_jmid 1le ile n,1le jle m}
$$
is a basis of $Votimes W$. So we write $e_{vw}=e_votimes e_w$.
answered Nov 26 at 19:20
Dietrich Burde
76.7k64286
add a comment |
up vote
0
down vote
If $v_1,cdots v_n$ is a basis of $V$ and $w_1,ldots w_m$ is a basis of $W$, then
$$
{v_iotimes w_jmid 1le ile n,1le jle m}
$$
is a basis of $Votimes W$. So we write $e_{vw}=e_votimes e_w$.
answered Nov 26 at 19:20
Dietrich Burde
76.7k64286
add a comment |
up vote
0
down vote
up vote
0
down vote
If $v_1,cdots v_n$ is a basis of $V$ and $w_1,ldots w_m$ is a basis of $W$, then
$$
{v_iotimes w_jmid 1le ile n,1le jle m}
$$
is a basis of $Votimes W$. So we write $e_{vw}=e_votimes e_w$.
answered Nov 26 at 19:20
Dietrich Burde
76.7k64286
If $v_1,cdots v_n$ is a basis of $V$ and $w_1,ldots w_m$ is a basis of $W$, then
$$
{v_iotimes w_jmid 1le ile n,1le jle m}
$$
is a basis of $Votimes W$. So we write $e_{vw}=e_votimes e_w$.
answered Nov 26 at 19:20
Dietrich Burde
76.7k64286
answered Nov 26 at 19:20
Dietrich Burde
76.7k64286
answered Nov 26 at 19:20
Dietrich Burde
76.7k64286
answered Nov 26 at 19:20
Dietrich Burde
76.7k64286
76.7k64286
add a comment |
add a comment |
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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
Nov 19 at 19:13
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
Nov 19 at 20:57
@DietrichBurde but my question is why is the linear span "well-defined"? For example, suppose we consider the span {$(1,0)$,$ (1,1,1)$}? It doesn't make sense since we have no method of adding $(1,0)$ and $(1,1,1)$, which is the point of confusion for me
– Saad
Nov 26 at 18:24
1
We do not add $e_v=(1,0)$ and $e_w=(1,1,1)$, but we add elements of type $e_{vw}=e_votimes e_w$. The basis for $Votimes W$ consists of vectors $e_{vw}$ and not of $e_v$, $e_w$, which are basis vectors of $V$ and $W$ respectively.
– Dietrich Burde
Nov 26 at 19:14