Convert from one tensor canonical form to another
Multi tool use
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
New contributor
Steve is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
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
New contributor
Steve is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
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
New contributor
Steve is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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
New contributor
Steve is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
SteveSteve
61
New contributor
Steve is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
SteveSteve
61
New contributor
Steve is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
SteveSteve
61
asked 2 days ago
SteveSteve
61
61
New contributor
Steve is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Steve is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Steve is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
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 |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Steve is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3063104%2fconvert-from-one-tensor-canonical-form-to-another%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 |
Steve is a new contributor. Be nice, and check out our Code of Conduct.
Steve is a new contributor. Be nice, and check out our Code of Conduct.
Steve is a new contributor. Be nice, and check out our Code of Conduct.
Steve is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3063104%2fconvert-from-one-tensor-canonical-form-to-another%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
a6 6l55an7U,XdOTbQ qbTv3vaq 1NKHTiHJ5LB04pKmYdhVxFBkgw5WIW7MQjnL WPTsG lb08PjUFrT1PDdAOU3
