Convert from one tensor canonical form to another
Suppose we have two canonical forms $A, B in mathbb{F}_2^{2 times 2 times 2}$ of a 3-dimensional tensor product space over the finite field with two elements, where
$A = e_1 otimes e_2 otimes e_1 + e_2 otimes e_1 otimes e_1$ and $B = e_1 otimes e_1 otimes e_1 + e_2 otimes e_2 otimes e_2$,
and $e_1, e_2$ are the usual standard basis of a 2-d vector space. Is it possible to "convert" from A to B? I am wondering if there are transformations that convert A to B (and vice versa)? Both $A$ and $B$ have the same tensor rank (i.e. they both have rank-2 because they can be written as the sum of two rank-1 tensors), so is this transformation rank-preserving? And if, say, I had two canonical forms that were not rank-equivalent, is there a transformation that takes me from one canonical form to another? Thank you.
finite-fields tensor-products multilinear-algebra tensor-rank tensor-decomposition
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Suppose we have two canonical forms $A, B in mathbb{F}_2^{2 times 2 times 2}$ of a 3-dimensional tensor product space over the finite field with two elements, where
$A = e_1 otimes e_2 otimes e_1 + e_2 otimes e_1 otimes e_1$ and $B = e_1 otimes e_1 otimes e_1 + e_2 otimes e_2 otimes e_2$,
and $e_1, e_2$ are the usual standard basis of a 2-d vector space. Is it possible to "convert" from A to B? I am wondering if there are transformations that convert A to B (and vice versa)? Both $A$ and $B$ have the same tensor rank (i.e. they both have rank-2 because they can be written as the sum of two rank-1 tensors), so is this transformation rank-preserving? And if, say, I had two canonical forms that were not rank-equivalent, is there a transformation that takes me from one canonical form to another? Thank you.
finite-fields tensor-products multilinear-algebra tensor-rank tensor-decomposition
asked 2 days ago
SteveSteve
61
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Steve is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Suppose we have two canonical forms $A, B in mathbb{F}_2^{2 times 2 times 2}$ of a 3-dimensional tensor product space over the finite field with two elements, where
$A = e_1 otimes e_2 otimes e_1 + e_2 otimes e_1 otimes e_1$ and $B = e_1 otimes e_1 otimes e_1 + e_2 otimes e_2 otimes e_2$,
and $e_1, e_2$ are the usual standard basis of a 2-d vector space. Is it possible to "convert" from A to B? I am wondering if there are transformations that convert A to B (and vice versa)? Both $A$ and $B$ have the same tensor rank (i.e. they both have rank-2 because they can be written as the sum of two rank-1 tensors), so is this transformation rank-preserving? And if, say, I had two canonical forms that were not rank-equivalent, is there a transformation that takes me from one canonical form to another? Thank you.
finite-fields tensor-products multilinear-algebra tensor-rank tensor-decomposition
asked 2 days ago
SteveSteve
61
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Steve is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Suppose we have two canonical forms $A, B in mathbb{F}_2^{2 times 2 times 2}$ of a 3-dimensional tensor product space over the finite field with two elements, where
$A = e_1 otimes e_2 otimes e_1 + e_2 otimes e_1 otimes e_1$ and $B = e_1 otimes e_1 otimes e_1 + e_2 otimes e_2 otimes e_2$,
and $e_1, e_2$ are the usual standard basis of a 2-d vector space. Is it possible to "convert" from A to B? I am wondering if there are transformations that convert A to B (and vice versa)? Both $A$ and $B$ have the same tensor rank (i.e. they both have rank-2 because they can be written as the sum of two rank-1 tensors), so is this transformation rank-preserving? And if, say, I had two canonical forms that were not rank-equivalent, is there a transformation that takes me from one canonical form to another? Thank you.
finite-fields tensor-products multilinear-algebra tensor-rank tensor-decomposition
finite-fields tensor-products multilinear-algebra tensor-rank tensor-decomposition
asked 2 days ago
SteveSteve
61
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Steve is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 2 days ago
SteveSteve
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asked 2 days ago
SteveSteve
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asked 2 days ago
SteveSteve
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asked 2 days ago
SteveSteve
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61
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Steve is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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If you remove the origin, every vector space is homogenous, meaning for any two nonzero vectors $u, v$ there is an invertible linear map $A$ with $Au=v$. So the answer is yes.
If they're not rank equivalent, then unless you force them to they don't lie in the same vector space. However, without imposing restrictions on the maps the answer is still yes, because we can construct maps between vector spaces that send an arbitrary nonzero vector anywhere we want, as long as they are over the same field.
answered 2 days ago
Matt SamuelMatt Samuel
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1 Answer
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1 Answer
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If you remove the origin, every vector space is homogenous, meaning for any two nonzero vectors $u, v$ there is an invertible linear map $A$ with $Au=v$. So the answer is yes.
If they're not rank equivalent, then unless you force them to they don't lie in the same vector space. However, without imposing restrictions on the maps the answer is still yes, because we can construct maps between vector spaces that send an arbitrary nonzero vector anywhere we want, as long as they are over the same field.
answered 2 days ago
Matt SamuelMatt Samuel
37.5k63665
add a comment |
If you remove the origin, every vector space is homogenous, meaning for any two nonzero vectors $u, v$ there is an invertible linear map $A$ with $Au=v$. So the answer is yes.
If they're not rank equivalent, then unless you force them to they don't lie in the same vector space. However, without imposing restrictions on the maps the answer is still yes, because we can construct maps between vector spaces that send an arbitrary nonzero vector anywhere we want, as long as they are over the same field.
answered 2 days ago
Matt SamuelMatt Samuel
37.5k63665
add a comment |
If you remove the origin, every vector space is homogenous, meaning for any two nonzero vectors $u, v$ there is an invertible linear map $A$ with $Au=v$. So the answer is yes.
If they're not rank equivalent, then unless you force them to they don't lie in the same vector space. However, without imposing restrictions on the maps the answer is still yes, because we can construct maps between vector spaces that send an arbitrary nonzero vector anywhere we want, as long as they are over the same field.
answered 2 days ago
Matt SamuelMatt Samuel
37.5k63665
If you remove the origin, every vector space is homogenous, meaning for any two nonzero vectors $u, v$ there is an invertible linear map $A$ with $Au=v$. So the answer is yes.
If they're not rank equivalent, then unless you force them to they don't lie in the same vector space. However, without imposing restrictions on the maps the answer is still yes, because we can construct maps between vector spaces that send an arbitrary nonzero vector anywhere we want, as long as they are over the same field.
answered 2 days ago
Matt SamuelMatt Samuel
37.5k63665
answered 2 days ago
Matt SamuelMatt Samuel
37.5k63665
answered 2 days ago
Matt SamuelMatt Samuel
37.5k63665
answered 2 days ago
Matt SamuelMatt Samuel
37.5k63665
37.5k63665
add a comment |
add a comment |
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