A linear combination of the set ${ {bf A},{bf A}^T}$
$begingroup$
Consider $mathbf{A}$, an $n times n$ matrix over $mathbb{F}_q$, the finite field with $q$ elements. The transpose of $mathbf{A}$ is denoted with $mathbf{A}^T$.
Let $mathbf{I}_n$ denote the identity matrix of order $n$.
Assume that the matrix $bf A$ has the following property over $mathbb{F}_q$,
$
mathbf{A}mathbf{A}^T=mathbf{I}_n.
$
We say $mathbf{B}$ is a linear combination of the set
${mathbf{A},mathbf{A}^T}$, if
$mathbf{B}=alpha_1mathbf{A}+alpha_2mathbf{A}^T$
where $alpha_i$ with $1leq i leq 2$ are elements in $mathbb{F}_q$.
My Question:
How to find an $ntimes n$ matrix $mathbf{B}$ such that
$mathbf{B}mathbf{B}^T=-mathbf{I}_n$ over $mathbb{F}_q$ whithout restriction on $q$?
Example: Consider $alpha in mathbb{F}_q$ such that $alpha^2=-1$, then
by considering $mathbf{B}=alphamathbf{A}$, we get $mathbf{B}mathbf{B}^T=-mathbf{I}_n$.
But due to the condition over $alpha$, we have $q equiv1pmod{4}$.
linear-algebra matrices finite-fields
edited Dec 1 '18 at 23:11
egreg
180k1485202
asked Dec 1 '18 at 22:34
Amin235Amin235
1,201621
$endgroup$
add a comment |
$begingroup$
Consider $mathbf{A}$, an $n times n$ matrix over $mathbb{F}_q$, the finite field with $q$ elements. The transpose of $mathbf{A}$ is denoted with $mathbf{A}^T$.
Let $mathbf{I}_n$ denote the identity matrix of order $n$.
Assume that the matrix $bf A$ has the following property over $mathbb{F}_q$,
$
mathbf{A}mathbf{A}^T=mathbf{I}_n.
$
We say $mathbf{B}$ is a linear combination of the set
${mathbf{A},mathbf{A}^T}$, if
$mathbf{B}=alpha_1mathbf{A}+alpha_2mathbf{A}^T$
where $alpha_i$ with $1leq i leq 2$ are elements in $mathbb{F}_q$.
My Question:
How to find an $ntimes n$ matrix $mathbf{B}$ such that
$mathbf{B}mathbf{B}^T=-mathbf{I}_n$ over $mathbb{F}_q$ whithout restriction on $q$?
Example: Consider $alpha in mathbb{F}_q$ such that $alpha^2=-1$, then
by considering $mathbf{B}=alphamathbf{A}$, we get $mathbf{B}mathbf{B}^T=-mathbf{I}_n$.
But due to the condition over $alpha$, we have $q equiv1pmod{4}$.
linear-algebra matrices finite-fields
edited Dec 1 '18 at 23:11
egreg
180k1485202
asked Dec 1 '18 at 22:34
Amin235Amin235
1,201621
$endgroup$
2
$begingroup$
If $A^2=I$, then $A=A^{-1}$, and so there is no need to include $alpha_3$. However, your question doesn't actually make reference to $A$. Do you need it to, i.e., do you require that $B$ is of the form $alpha_1 A + alpha_2 A^T$ for some $A$ satisfying $A^2=I$?
$endgroup$
– Aaron
Dec 1 '18 at 22:44
$begingroup$
@Aaron You right. Based on your comment i edit the question.
$endgroup$
– Amin235
Dec 1 '18 at 22:51
$begingroup$
@Aaron The condition $A=A^{-1}$ was not necessary and because of this I omitted it.
$endgroup$
– Amin235
Dec 1 '18 at 22:58
$begingroup$
@egreg I appreciate you for your nice edition.
$endgroup$
– Amin235
Dec 1 '18 at 23:21
add a comment |
$begingroup$
Consider $mathbf{A}$, an $n times n$ matrix over $mathbb{F}_q$, the finite field with $q$ elements. The transpose of $mathbf{A}$ is denoted with $mathbf{A}^T$.
Let $mathbf{I}_n$ denote the identity matrix of order $n$.
Assume that the matrix $bf A$ has the following property over $mathbb{F}_q$,
$
mathbf{A}mathbf{A}^T=mathbf{I}_n.
$
We say $mathbf{B}$ is a linear combination of the set
${mathbf{A},mathbf{A}^T}$, if
$mathbf{B}=alpha_1mathbf{A}+alpha_2mathbf{A}^T$
where $alpha_i$ with $1leq i leq 2$ are elements in $mathbb{F}_q$.
My Question:
How to find an $ntimes n$ matrix $mathbf{B}$ such that
$mathbf{B}mathbf{B}^T=-mathbf{I}_n$ over $mathbb{F}_q$ whithout restriction on $q$?
Example: Consider $alpha in mathbb{F}_q$ such that $alpha^2=-1$, then
by considering $mathbf{B}=alphamathbf{A}$, we get $mathbf{B}mathbf{B}^T=-mathbf{I}_n$.
But due to the condition over $alpha$, we have $q equiv1pmod{4}$.
linear-algebra matrices finite-fields
edited Dec 1 '18 at 23:11
egreg
180k1485202
asked Dec 1 '18 at 22:34
Amin235Amin235
1,201621
$endgroup$
Consider $mathbf{A}$, an $n times n$ matrix over $mathbb{F}_q$, the finite field with $q$ elements. The transpose of $mathbf{A}$ is denoted with $mathbf{A}^T$.
Let $mathbf{I}_n$ denote the identity matrix of order $n$.
Assume that the matrix $bf A$ has the following property over $mathbb{F}_q$,
$
mathbf{A}mathbf{A}^T=mathbf{I}_n.
$
We say $mathbf{B}$ is a linear combination of the set
${mathbf{A},mathbf{A}^T}$, if
$mathbf{B}=alpha_1mathbf{A}+alpha_2mathbf{A}^T$
where $alpha_i$ with $1leq i leq 2$ are elements in $mathbb{F}_q$.
My Question:
How to find an $ntimes n$ matrix $mathbf{B}$ such that
$mathbf{B}mathbf{B}^T=-mathbf{I}_n$ over $mathbb{F}_q$ whithout restriction on $q$?
Example: Consider $alpha in mathbb{F}_q$ such that $alpha^2=-1$, then
by considering $mathbf{B}=alphamathbf{A}$, we get $mathbf{B}mathbf{B}^T=-mathbf{I}_n$.
But due to the condition over $alpha$, we have $q equiv1pmod{4}$.
linear-algebra matrices finite-fields
linear-algebra matrices finite-fields
edited Dec 1 '18 at 23:11
egreg
180k1485202
asked Dec 1 '18 at 22:34
Amin235Amin235
1,201621
edited Dec 1 '18 at 23:11
egreg
180k1485202
asked Dec 1 '18 at 22:34
Amin235Amin235
1,201621
edited Dec 1 '18 at 23:11
egreg
180k1485202
edited Dec 1 '18 at 23:11
egreg
180k1485202
edited Dec 1 '18 at 23:11
egreg
180k1485202
180k1485202
asked Dec 1 '18 at 22:34
Amin235Amin235
1,201621
asked Dec 1 '18 at 22:34
Amin235Amin235
1,201621
asked Dec 1 '18 at 22:34
Amin235Amin235
1,201621
1,201621
2
$begingroup$
If $A^2=I$, then $A=A^{-1}$, and so there is no need to include $alpha_3$. However, your question doesn't actually make reference to $A$. Do you need it to, i.e., do you require that $B$ is of the form $alpha_1 A + alpha_2 A^T$ for some $A$ satisfying $A^2=I$?
$endgroup$
– Aaron
Dec 1 '18 at 22:44
$begingroup$
@Aaron You right. Based on your comment i edit the question.
$endgroup$
– Amin235
Dec 1 '18 at 22:51
$begingroup$
@Aaron The condition $A=A^{-1}$ was not necessary and because of this I omitted it.
$endgroup$
– Amin235
Dec 1 '18 at 22:58
$begingroup$
@egreg I appreciate you for your nice edition.
$endgroup$
– Amin235
Dec 1 '18 at 23:21
add a comment |
2
$begingroup$
If $A^2=I$, then $A=A^{-1}$, and so there is no need to include $alpha_3$. However, your question doesn't actually make reference to $A$. Do you need it to, i.e., do you require that $B$ is of the form $alpha_1 A + alpha_2 A^T$ for some $A$ satisfying $A^2=I$?
$endgroup$
– Aaron
Dec 1 '18 at 22:44
$begingroup$
@Aaron You right. Based on your comment i edit the question.
$endgroup$
– Amin235
Dec 1 '18 at 22:51
$begingroup$
@Aaron The condition $A=A^{-1}$ was not necessary and because of this I omitted it.
$endgroup$
– Amin235
Dec 1 '18 at 22:58
$begingroup$
@egreg I appreciate you for your nice edition.
$endgroup$
– Amin235
Dec 1 '18 at 23:21
2
2
$begingroup$
If $A^2=I$, then $A=A^{-1}$, and so there is no need to include $alpha_3$. However, your question doesn't actually make reference to $A$. Do you need it to, i.e., do you require that $B$ is of the form $alpha_1 A + alpha_2 A^T$ for some $A$ satisfying $A^2=I$?
$endgroup$
– Aaron
Dec 1 '18 at 22:44
$begingroup$
If $A^2=I$, then $A=A^{-1}$, and so there is no need to include $alpha_3$. However, your question doesn't actually make reference to $A$. Do you need it to, i.e., do you require that $B$ is of the form $alpha_1 A + alpha_2 A^T$ for some $A$ satisfying $A^2=I$?
$endgroup$
– Aaron
Dec 1 '18 at 22:44
$begingroup$
@Aaron You right. Based on your comment i edit the question.
$endgroup$
– Amin235
Dec 1 '18 at 22:51
$begingroup$
@Aaron You right. Based on your comment i edit the question.
$endgroup$
– Amin235
Dec 1 '18 at 22:51
$begingroup$
@Aaron The condition $A=A^{-1}$ was not necessary and because of this I omitted it.
$endgroup$
– Amin235
Dec 1 '18 at 22:58
$begingroup$
@Aaron The condition $A=A^{-1}$ was not necessary and because of this I omitted it.
$endgroup$
– Amin235
Dec 1 '18 at 22:58
$begingroup$
@egreg I appreciate you for your nice edition.
$endgroup$
– Amin235
Dec 1 '18 at 23:21
$begingroup$
@egreg I appreciate you for your nice edition.
$endgroup$
– Amin235
Dec 1 '18 at 23:21
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This isn't always possible. For example, if $A=I$, then the problem collapses to $(alpha A)(alpha A)^T=alpha^2 I=-I$, but $alpha^2=-1$ does not always have solutions.
However, if $B=alpha_1 A + alpha_2 A^T$, then $BB^T=(alpha_1^2+alpha_2^2)I+alpha_1 alpha_2 (A^2+(A^T)^2)$. If $alpha^2=-1$ has no solutions in our field, then the only way we can hope to solve is if $A^2+(A^T)^2$ is a scalar.
answered Dec 1 '18 at 23:48
AaronAaron
15.9k22754
$endgroup$
$begingroup$
"$A^2+(A^T)^2$ be a scalar" is an interesting note. Thanks.
$endgroup$
– Amin235
Dec 2 '18 at 8:24
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
});
}
});
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%2f3021930%2fa-linear-combination-of-the-set-bf-a-bf-at%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
$begingroup$
This isn't always possible. For example, if $A=I$, then the problem collapses to $(alpha A)(alpha A)^T=alpha^2 I=-I$, but $alpha^2=-1$ does not always have solutions.
However, if $B=alpha_1 A + alpha_2 A^T$, then $BB^T=(alpha_1^2+alpha_2^2)I+alpha_1 alpha_2 (A^2+(A^T)^2)$. If $alpha^2=-1$ has no solutions in our field, then the only way we can hope to solve is if $A^2+(A^T)^2$ is a scalar.
answered Dec 1 '18 at 23:48
AaronAaron
15.9k22754
$endgroup$
$begingroup$
"$A^2+(A^T)^2$ be a scalar" is an interesting note. Thanks.
$endgroup$
– Amin235
Dec 2 '18 at 8:24
add a comment |
$begingroup$
This isn't always possible. For example, if $A=I$, then the problem collapses to $(alpha A)(alpha A)^T=alpha^2 I=-I$, but $alpha^2=-1$ does not always have solutions.
However, if $B=alpha_1 A + alpha_2 A^T$, then $BB^T=(alpha_1^2+alpha_2^2)I+alpha_1 alpha_2 (A^2+(A^T)^2)$. If $alpha^2=-1$ has no solutions in our field, then the only way we can hope to solve is if $A^2+(A^T)^2$ is a scalar.
answered Dec 1 '18 at 23:48
AaronAaron
15.9k22754
$endgroup$
$begingroup$
"$A^2+(A^T)^2$ be a scalar" is an interesting note. Thanks.
$endgroup$
– Amin235
Dec 2 '18 at 8:24
add a comment |
$begingroup$
This isn't always possible. For example, if $A=I$, then the problem collapses to $(alpha A)(alpha A)^T=alpha^2 I=-I$, but $alpha^2=-1$ does not always have solutions.
However, if $B=alpha_1 A + alpha_2 A^T$, then $BB^T=(alpha_1^2+alpha_2^2)I+alpha_1 alpha_2 (A^2+(A^T)^2)$. If $alpha^2=-1$ has no solutions in our field, then the only way we can hope to solve is if $A^2+(A^T)^2$ is a scalar.
answered Dec 1 '18 at 23:48
AaronAaron
15.9k22754
$endgroup$
This isn't always possible. For example, if $A=I$, then the problem collapses to $(alpha A)(alpha A)^T=alpha^2 I=-I$, but $alpha^2=-1$ does not always have solutions.
However, if $B=alpha_1 A + alpha_2 A^T$, then $BB^T=(alpha_1^2+alpha_2^2)I+alpha_1 alpha_2 (A^2+(A^T)^2)$. If $alpha^2=-1$ has no solutions in our field, then the only way we can hope to solve is if $A^2+(A^T)^2$ is a scalar.
answered Dec 1 '18 at 23:48
AaronAaron
15.9k22754
answered Dec 1 '18 at 23:48
AaronAaron
15.9k22754
answered Dec 1 '18 at 23:48
AaronAaron
15.9k22754
answered Dec 1 '18 at 23:48
AaronAaron
15.9k22754
15.9k22754
$begingroup$
"$A^2+(A^T)^2$ be a scalar" is an interesting note. Thanks.
$endgroup$
– Amin235
Dec 2 '18 at 8:24
add a comment |
$begingroup$
"$A^2+(A^T)^2$ be a scalar" is an interesting note. Thanks.
$endgroup$
– Amin235
Dec 2 '18 at 8:24
$begingroup$
"$A^2+(A^T)^2$ be a scalar" is an interesting note. Thanks.
$endgroup$
– Amin235
Dec 2 '18 at 8:24
$begingroup$
"$A^2+(A^T)^2$ be a scalar" is an interesting note. Thanks.
$endgroup$
– Amin235
Dec 2 '18 at 8:24
add a comment |
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.
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%2f3021930%2fa-linear-combination-of-the-set-bf-a-bf-at%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
2
$begingroup$
If $A^2=I$, then $A=A^{-1}$, and so there is no need to include $alpha_3$. However, your question doesn't actually make reference to $A$. Do you need it to, i.e., do you require that $B$ is of the form $alpha_1 A + alpha_2 A^T$ for some $A$ satisfying $A^2=I$?
$endgroup$
– Aaron
Dec 1 '18 at 22:44
$begingroup$
@Aaron You right. Based on your comment i edit the question.
$endgroup$
– Amin235
Dec 1 '18 at 22:51
$begingroup$
@Aaron The condition $A=A^{-1}$ was not necessary and because of this I omitted it.
$endgroup$
– Amin235
Dec 1 '18 at 22:58
$begingroup$
@egreg I appreciate you for your nice edition.
$endgroup$
– Amin235
Dec 1 '18 at 23:21