System of homogen linear equations in a division ring
$begingroup$
Let $K$ be a division ring (one does not suppose that $K$ is commutative) and $m,n$ two positive integers such that $m<n$. Consider the system of homogen linear equations
$$left{begin{array}{rcl}
a_{11}x_1+a_{12}x_2+cdots+a_{1n}x_n&=&0\
a_{21}x_1+a_{22}x_2+cdots+a_{2n}x_n&=&0\
vdotsqquadqquadqquad&vdots&vdots\
a_{m1}x_1+a_{m2}x_2+cdots+a_{mn}x_n&=&0\
end{array}right.$$
where the $a_{ij}$'s are in $K$. Can we assert that this system admits an infinity of solutions. Obviously, it is true when $K$ is commutative. But when $K$ is not, I do not know if this results still holds.
Is it also true in the non commutative case?
Thanks in advance.
linear-algebra noncommutative-algebra
asked Dec 27 '18 at 2:37
joaopajoaopa
35618
$endgroup$
add a comment |
$begingroup$
Let $K$ be a division ring (one does not suppose that $K$ is commutative) and $m,n$ two positive integers such that $m<n$. Consider the system of homogen linear equations
$$left{begin{array}{rcl}
a_{11}x_1+a_{12}x_2+cdots+a_{1n}x_n&=&0\
a_{21}x_1+a_{22}x_2+cdots+a_{2n}x_n&=&0\
vdotsqquadqquadqquad&vdots&vdots\
a_{m1}x_1+a_{m2}x_2+cdots+a_{mn}x_n&=&0\
end{array}right.$$
where the $a_{ij}$'s are in $K$. Can we assert that this system admits an infinity of solutions. Obviously, it is true when $K$ is commutative. But when $K$ is not, I do not know if this results still holds.
Is it also true in the non commutative case?
Thanks in advance.
linear-algebra noncommutative-algebra
asked Dec 27 '18 at 2:37
joaopajoaopa
35618
$endgroup$
add a comment |
$begingroup$
Let $K$ be a division ring (one does not suppose that $K$ is commutative) and $m,n$ two positive integers such that $m<n$. Consider the system of homogen linear equations
$$left{begin{array}{rcl}
a_{11}x_1+a_{12}x_2+cdots+a_{1n}x_n&=&0\
a_{21}x_1+a_{22}x_2+cdots+a_{2n}x_n&=&0\
vdotsqquadqquadqquad&vdots&vdots\
a_{m1}x_1+a_{m2}x_2+cdots+a_{mn}x_n&=&0\
end{array}right.$$
where the $a_{ij}$'s are in $K$. Can we assert that this system admits an infinity of solutions. Obviously, it is true when $K$ is commutative. But when $K$ is not, I do not know if this results still holds.
Is it also true in the non commutative case?
Thanks in advance.
linear-algebra noncommutative-algebra
asked Dec 27 '18 at 2:37
joaopajoaopa
35618
$endgroup$
Let $K$ be a division ring (one does not suppose that $K$ is commutative) and $m,n$ two positive integers such that $m<n$. Consider the system of homogen linear equations
$$left{begin{array}{rcl}
a_{11}x_1+a_{12}x_2+cdots+a_{1n}x_n&=&0\
a_{21}x_1+a_{22}x_2+cdots+a_{2n}x_n&=&0\
vdotsqquadqquadqquad&vdots&vdots\
a_{m1}x_1+a_{m2}x_2+cdots+a_{mn}x_n&=&0\
end{array}right.$$
where the $a_{ij}$'s are in $K$. Can we assert that this system admits an infinity of solutions. Obviously, it is true when $K$ is commutative. But when $K$ is not, I do not know if this results still holds.
Is it also true in the non commutative case?
Thanks in advance.
linear-algebra noncommutative-algebra
linear-algebra noncommutative-algebra
asked Dec 27 '18 at 2:37
joaopajoaopa
35618
asked Dec 27 '18 at 2:37
joaopajoaopa
35618
asked Dec 27 '18 at 2:37
joaopajoaopa
35618
asked Dec 27 '18 at 2:37
joaopajoaopa
35618
asked Dec 27 '18 at 2:37
joaopajoaopa
35618
35618
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The system has infinitely many solutions, even when $K$ is non-commutative. The reason is that row reduction is still possible over a non-commutative division ring.
Side remark: if $K$ is commutative, the system has infinitely many solutions only if $K$ is infinite. If $K$ is not commutative, then it is automatically infinite by Wedderburn's little theorem.
answered Dec 27 '18 at 13:20
Pierre-Guy PlamondonPierre-Guy Plamondon
8,88511739
$endgroup$
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%2f3053539%2fsystem-of-homogen-linear-equations-in-a-division-ring%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
$begingroup$
The system has infinitely many solutions, even when $K$ is non-commutative. The reason is that row reduction is still possible over a non-commutative division ring.
Side remark: if $K$ is commutative, the system has infinitely many solutions only if $K$ is infinite. If $K$ is not commutative, then it is automatically infinite by Wedderburn's little theorem.
answered Dec 27 '18 at 13:20
Pierre-Guy PlamondonPierre-Guy Plamondon
8,88511739
$endgroup$
add a comment |
$begingroup$
The system has infinitely many solutions, even when $K$ is non-commutative. The reason is that row reduction is still possible over a non-commutative division ring.
Side remark: if $K$ is commutative, the system has infinitely many solutions only if $K$ is infinite. If $K$ is not commutative, then it is automatically infinite by Wedderburn's little theorem.
answered Dec 27 '18 at 13:20
Pierre-Guy PlamondonPierre-Guy Plamondon
8,88511739
$endgroup$
add a comment |
$begingroup$
The system has infinitely many solutions, even when $K$ is non-commutative. The reason is that row reduction is still possible over a non-commutative division ring.
Side remark: if $K$ is commutative, the system has infinitely many solutions only if $K$ is infinite. If $K$ is not commutative, then it is automatically infinite by Wedderburn's little theorem.
answered Dec 27 '18 at 13:20
Pierre-Guy PlamondonPierre-Guy Plamondon
8,88511739
$endgroup$
The system has infinitely many solutions, even when $K$ is non-commutative. The reason is that row reduction is still possible over a non-commutative division ring.
Side remark: if $K$ is commutative, the system has infinitely many solutions only if $K$ is infinite. If $K$ is not commutative, then it is automatically infinite by Wedderburn's little theorem.
answered Dec 27 '18 at 13:20
Pierre-Guy PlamondonPierre-Guy Plamondon
8,88511739
answered Dec 27 '18 at 13:20
Pierre-Guy PlamondonPierre-Guy Plamondon
8,88511739
answered Dec 27 '18 at 13:20
Pierre-Guy PlamondonPierre-Guy Plamondon
8,88511739
answered Dec 27 '18 at 13:20
Pierre-Guy PlamondonPierre-Guy Plamondon
8,88511739
8,88511739
add a comment |
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%2f3053539%2fsystem-of-homogen-linear-equations-in-a-division-ring%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
