Mixing
Convert from one tensor canonical form to another
$begingroup$
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 Jan 5 at 19:29
SteveSteve
61
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add a comment |
$begingroup$
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 Jan 5 at 19:29
SteveSteve
61
$endgroup$
add a comment |
$begingroup$
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 Jan 5 at 19:29
SteveSteve
61
$endgroup$
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 Jan 5 at 19:29
SteveSteve
61
asked Jan 5 at 19:29
SteveSteve
61
asked Jan 5 at 19:29
SteveSteve
61
asked Jan 5 at 19:29
SteveSteve
61
asked Jan 5 at 19:29
SteveSteve
61
61
add a comment |
add a comment |
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 Jan 5 at 19:33
Matt SamuelMatt Samuel
37.8k63665
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1 Answer
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1 Answer
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$begingroup$
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 Jan 5 at 19:33
Matt SamuelMatt Samuel
37.8k63665
$endgroup$
add a comment |
$begingroup$
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 Jan 5 at 19:33
Matt SamuelMatt Samuel
37.8k63665
$endgroup$
add a comment |
$begingroup$
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 Jan 5 at 19:33
Matt SamuelMatt Samuel
37.8k63665
$endgroup$
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 Jan 5 at 19:33
Matt SamuelMatt Samuel
37.8k63665
answered Jan 5 at 19:33
Matt SamuelMatt Samuel
37.8k63665
answered Jan 5 at 19:33
Matt SamuelMatt Samuel
37.8k63665
answered Jan 5 at 19:33
Matt SamuelMatt Samuel
37.8k63665
37.8k63665
add a comment |
add a comment |
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