Mixing
Uniqueness of the basis for a tensor product of vector spaces.
Clash Royale CLAN TAG#URR8PPP
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Having in mind the following answered question:
Tensor Product of Spaces has Basis of Tensor Products
about bases for a tensor product of vector spaces I wonder if that is the only way to get a basis for that tensor other than the obvious. That is, the basis $(e_1)otimes(e_2)$ for $V_1otimesV_2$ where $(e_1)$ is a basis for $V_1$ and $(e_2)$ for $V_2$.
In case it was not unique, would that mean a different tensor product between the corresponding matrices than the usual Kronecker product between matrices?
linear-algebra tensor-products
asked Jul 24 at 6:50
gibarian
5610
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up vote
0
down vote
favorite
Having in mind the following answered question:
Tensor Product of Spaces has Basis of Tensor Products
about bases for a tensor product of vector spaces I wonder if that is the only way to get a basis for that tensor other than the obvious. That is, the basis $(e_1)otimes(e_2)$ for $V_1otimesV_2$ where $(e_1)$ is a basis for $V_1$ and $(e_2)$ for $V_2$.
In case it was not unique, would that mean a different tensor product between the corresponding matrices than the usual Kronecker product between matrices?
linear-algebra tensor-products
asked Jul 24 at 6:50
gibarian
5610
Bases are never unique.
– Paul Frost
Jul 24 at 8:15
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Having in mind the following answered question:
Tensor Product of Spaces has Basis of Tensor Products
about bases for a tensor product of vector spaces I wonder if that is the only way to get a basis for that tensor other than the obvious. That is, the basis $(e_1)otimes(e_2)$ for $V_1otimesV_2$ where $(e_1)$ is a basis for $V_1$ and $(e_2)$ for $V_2$.
In case it was not unique, would that mean a different tensor product between the corresponding matrices than the usual Kronecker product between matrices?
linear-algebra tensor-products
asked Jul 24 at 6:50
gibarian
5610
Having in mind the following answered question:
Tensor Product of Spaces has Basis of Tensor Products
about bases for a tensor product of vector spaces I wonder if that is the only way to get a basis for that tensor other than the obvious. That is, the basis $(e_1)otimes(e_2)$ for $V_1otimesV_2$ where $(e_1)$ is a basis for $V_1$ and $(e_2)$ for $V_2$.
In case it was not unique, would that mean a different tensor product between the corresponding matrices than the usual Kronecker product between matrices?
linear-algebra tensor-products
asked Jul 24 at 6:50
gibarian
5610
asked Jul 24 at 6:50
gibarian
5610
asked Jul 24 at 6:50
gibarian
5610
asked Jul 24 at 6:50
gibarian
5610
5610
Bases are never unique.
– Paul Frost
Jul 24 at 8:15
add a comment |Â
Bases are never unique.
– Paul Frost
Jul 24 at 8:15
Bases are never unique.
– Paul Frost
Jul 24 at 8:15
Bases are never unique.
– Paul Frost
Jul 24 at 8:15
add a comment |Â
1 Answer
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That is certainly not unique.
To begin with, there are many basis for both $V_1$ and $V_2$ so there are several choices of basis on the form $(e_1) otimes (e_2)$ where $(e_1)$ is a basis for $V_1$ and $(e_2)$ for $V_2.$
Then, one can have "cross-bases". For example, assume that $V_1$ and $V_2$ are both 2-dimensional with bases $ e_1^(1), e_1^(2) $ and $ e_2^(1), e_2^(2) $ respectively. Then as a basis for $V_1 otimes V_2$ you can take
$$
e_1^(1) otimes e_2^(1) + e_1^(1) otimes e_2^(1),
e_1^(1) otimes e_2^(2) + e_1^(1) otimes e_2^(2), \
e_1^(2) otimes e_2^(1) + e_1^(2) otimes e_2^(1),
e_1^(2) otimes e_2^(2) + e_1^(2) otimes e_2^(2)
$$
It isn't certain that you can rewrite any of these basis vectors as a simple tensor product.
Maybe you have heard of quantum entanglement. The mathematical description of these consists of this kind of sums of simple tensor products. See for example these states.
answered Jul 24 at 8:17
md2perpe
5,83511022
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
That is certainly not unique.
To begin with, there are many basis for both $V_1$ and $V_2$ so there are several choices of basis on the form $(e_1) otimes (e_2)$ where $(e_1)$ is a basis for $V_1$ and $(e_2)$ for $V_2.$
Then, one can have "cross-bases". For example, assume that $V_1$ and $V_2$ are both 2-dimensional with bases $ e_1^(1), e_1^(2) $ and $ e_2^(1), e_2^(2) $ respectively. Then as a basis for $V_1 otimes V_2$ you can take
$$
e_1^(1) otimes e_2^(1) + e_1^(1) otimes e_2^(1),
e_1^(1) otimes e_2^(2) + e_1^(1) otimes e_2^(2), \
e_1^(2) otimes e_2^(1) + e_1^(2) otimes e_2^(1),
e_1^(2) otimes e_2^(2) + e_1^(2) otimes e_2^(2)
$$
It isn't certain that you can rewrite any of these basis vectors as a simple tensor product.
Maybe you have heard of quantum entanglement. The mathematical description of these consists of this kind of sums of simple tensor products. See for example these states.
answered Jul 24 at 8:17
md2perpe
5,83511022
add a comment |Â
up vote
0
down vote
That is certainly not unique.
To begin with, there are many basis for both $V_1$ and $V_2$ so there are several choices of basis on the form $(e_1) otimes (e_2)$ where $(e_1)$ is a basis for $V_1$ and $(e_2)$ for $V_2.$
Then, one can have "cross-bases". For example, assume that $V_1$ and $V_2$ are both 2-dimensional with bases $ e_1^(1), e_1^(2) $ and $ e_2^(1), e_2^(2) $ respectively. Then as a basis for $V_1 otimes V_2$ you can take
$$
e_1^(1) otimes e_2^(1) + e_1^(1) otimes e_2^(1),
e_1^(1) otimes e_2^(2) + e_1^(1) otimes e_2^(2), \
e_1^(2) otimes e_2^(1) + e_1^(2) otimes e_2^(1),
e_1^(2) otimes e_2^(2) + e_1^(2) otimes e_2^(2)
$$
It isn't certain that you can rewrite any of these basis vectors as a simple tensor product.
Maybe you have heard of quantum entanglement. The mathematical description of these consists of this kind of sums of simple tensor products. See for example these states.
answered Jul 24 at 8:17
md2perpe
5,83511022
add a comment |Â
up vote
0
down vote
up vote
0
down vote
That is certainly not unique.
To begin with, there are many basis for both $V_1$ and $V_2$ so there are several choices of basis on the form $(e_1) otimes (e_2)$ where $(e_1)$ is a basis for $V_1$ and $(e_2)$ for $V_2.$
Then, one can have "cross-bases". For example, assume that $V_1$ and $V_2$ are both 2-dimensional with bases $ e_1^(1), e_1^(2) $ and $ e_2^(1), e_2^(2) $ respectively. Then as a basis for $V_1 otimes V_2$ you can take
$$
e_1^(1) otimes e_2^(1) + e_1^(1) otimes e_2^(1),
e_1^(1) otimes e_2^(2) + e_1^(1) otimes e_2^(2), \
e_1^(2) otimes e_2^(1) + e_1^(2) otimes e_2^(1),
e_1^(2) otimes e_2^(2) + e_1^(2) otimes e_2^(2)
$$
It isn't certain that you can rewrite any of these basis vectors as a simple tensor product.
Maybe you have heard of quantum entanglement. The mathematical description of these consists of this kind of sums of simple tensor products. See for example these states.
answered Jul 24 at 8:17
md2perpe
5,83511022
That is certainly not unique.
To begin with, there are many basis for both $V_1$ and $V_2$ so there are several choices of basis on the form $(e_1) otimes (e_2)$ where $(e_1)$ is a basis for $V_1$ and $(e_2)$ for $V_2.$
Then, one can have "cross-bases". For example, assume that $V_1$ and $V_2$ are both 2-dimensional with bases $ e_1^(1), e_1^(2) $ and $ e_2^(1), e_2^(2) $ respectively. Then as a basis for $V_1 otimes V_2$ you can take
$$
e_1^(1) otimes e_2^(1) + e_1^(1) otimes e_2^(1),
e_1^(1) otimes e_2^(2) + e_1^(1) otimes e_2^(2), \
e_1^(2) otimes e_2^(1) + e_1^(2) otimes e_2^(1),
e_1^(2) otimes e_2^(2) + e_1^(2) otimes e_2^(2)
$$
It isn't certain that you can rewrite any of these basis vectors as a simple tensor product.
Maybe you have heard of quantum entanglement. The mathematical description of these consists of this kind of sums of simple tensor products. See for example these states.
answered Jul 24 at 8:17
md2perpe
5,83511022
answered Jul 24 at 8:17
md2perpe
5,83511022
answered Jul 24 at 8:17
md2perpe
5,83511022
answered Jul 24 at 8:17
md2perpe
5,83511022
5,83511022
add a comment |Â
add a comment |Â
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Bases are never unique.
– Paul Frost
Jul 24 at 8:15