How to prove surjectivity of determinant
$begingroup$
It may seem obvious that the determinant of a square matrix as a map is surjective (since there is always a choice of matrix entries that yields a real number). I can't prove this statement. Any clue please?
determinant
asked Jan 8 at 13:09
Zakariae BenslimaneZakariae Benslimane
11010
$endgroup$
add a comment |
$begingroup$
It may seem obvious that the determinant of a square matrix as a map is surjective (since there is always a choice of matrix entries that yields a real number). I can't prove this statement. Any clue please?
determinant
asked Jan 8 at 13:09
Zakariae BenslimaneZakariae Benslimane
11010
$endgroup$
$begingroup$
Hint: use a diagonal matrix.
$endgroup$
– Bernard
Jan 8 at 13:11
add a comment |
$begingroup$
It may seem obvious that the determinant of a square matrix as a map is surjective (since there is always a choice of matrix entries that yields a real number). I can't prove this statement. Any clue please?
determinant
asked Jan 8 at 13:09
Zakariae BenslimaneZakariae Benslimane
11010
$endgroup$
It may seem obvious that the determinant of a square matrix as a map is surjective (since there is always a choice of matrix entries that yields a real number). I can't prove this statement. Any clue please?
determinant
determinant
asked Jan 8 at 13:09
Zakariae BenslimaneZakariae Benslimane
11010
asked Jan 8 at 13:09
Zakariae BenslimaneZakariae Benslimane
11010
asked Jan 8 at 13:09
Zakariae BenslimaneZakariae Benslimane
11010
asked Jan 8 at 13:09
Zakariae BenslimaneZakariae Benslimane
11010
asked Jan 8 at 13:09
Zakariae BenslimaneZakariae Benslimane
11010
11010
$begingroup$
Hint: use a diagonal matrix.
$endgroup$
– Bernard
Jan 8 at 13:11
add a comment |
$begingroup$
Hint: use a diagonal matrix.
$endgroup$
– Bernard
Jan 8 at 13:11
$begingroup$
Hint: use a diagonal matrix.
$endgroup$
– Bernard
Jan 8 at 13:11
$begingroup$
Hint: use a diagonal matrix.
$endgroup$
– Bernard
Jan 8 at 13:11
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Take the identity matrix and change one of the values in the diagonal to be $x$. Then the determinant of the matrix will be $x$.
answered Jan 8 at 13:11
SamboSambo
2,3012532
$endgroup$
$begingroup$
obivous, but took me a while to find the same answer by myself. thank you for the hint
$endgroup$
– Zakariae Benslimane
Jan 8 at 13:43
add a comment |
$begingroup$
Hint:
Given $rin Bbb R $, take the diagonal matrix
$begin{bmatrix}r&0&cdots & 0\
0&1&cdots & 0\
vdots&vdots&ddots&vdots\0 & 0 & cdots &1end{bmatrix} $.
answered Jan 8 at 13:14
Thomas ShelbyThomas Shelby
4,5862727
$endgroup$
add a comment |
$begingroup$
Multiplying a row (or column) of a matrix by a constant results in multiplying the determinant by that constant. (The determinant is a $k$-multilinear function.)
So, given $rinBbb R$, just take a matrix with nonzero determinant, $d$, and multiply a row by the constant $frac rd$, to get a matrix with determinant $r$.
answered Jan 8 at 13:39
Chris CusterChris Custer
14.2k3827
$endgroup$
add a comment |
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%2f3066144%2fhow-to-prove-surjectivity-of-determinant%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Take the identity matrix and change one of the values in the diagonal to be $x$. Then the determinant of the matrix will be $x$.
answered Jan 8 at 13:11
SamboSambo
2,3012532
$endgroup$
$begingroup$
obivous, but took me a while to find the same answer by myself. thank you for the hint
$endgroup$
– Zakariae Benslimane
Jan 8 at 13:43
add a comment |
$begingroup$
Take the identity matrix and change one of the values in the diagonal to be $x$. Then the determinant of the matrix will be $x$.
answered Jan 8 at 13:11
SamboSambo
2,3012532
$endgroup$
$begingroup$
obivous, but took me a while to find the same answer by myself. thank you for the hint
$endgroup$
– Zakariae Benslimane
Jan 8 at 13:43
add a comment |
$begingroup$
Take the identity matrix and change one of the values in the diagonal to be $x$. Then the determinant of the matrix will be $x$.
answered Jan 8 at 13:11
SamboSambo
2,3012532
$endgroup$
Take the identity matrix and change one of the values in the diagonal to be $x$. Then the determinant of the matrix will be $x$.
answered Jan 8 at 13:11
SamboSambo
2,3012532
answered Jan 8 at 13:11
SamboSambo
2,3012532
answered Jan 8 at 13:11
SamboSambo
2,3012532
answered Jan 8 at 13:11
SamboSambo
2,3012532
2,3012532
$begingroup$
obivous, but took me a while to find the same answer by myself. thank you for the hint
$endgroup$
– Zakariae Benslimane
Jan 8 at 13:43
add a comment |
$begingroup$
obivous, but took me a while to find the same answer by myself. thank you for the hint
$endgroup$
– Zakariae Benslimane
Jan 8 at 13:43
$begingroup$
obivous, but took me a while to find the same answer by myself. thank you for the hint
$endgroup$
– Zakariae Benslimane
Jan 8 at 13:43
$begingroup$
obivous, but took me a while to find the same answer by myself. thank you for the hint
$endgroup$
– Zakariae Benslimane
Jan 8 at 13:43
add a comment |
$begingroup$
Hint:
Given $rin Bbb R $, take the diagonal matrix
$begin{bmatrix}r&0&cdots & 0\
0&1&cdots & 0\
vdots&vdots&ddots&vdots\0 & 0 & cdots &1end{bmatrix} $.
answered Jan 8 at 13:14
Thomas ShelbyThomas Shelby
4,5862727
$endgroup$
add a comment |
$begingroup$
Hint:
Given $rin Bbb R $, take the diagonal matrix
$begin{bmatrix}r&0&cdots & 0\
0&1&cdots & 0\
vdots&vdots&ddots&vdots\0 & 0 & cdots &1end{bmatrix} $.
answered Jan 8 at 13:14
Thomas ShelbyThomas Shelby
4,5862727
$endgroup$
add a comment |
$begingroup$
Hint:
Given $rin Bbb R $, take the diagonal matrix
$begin{bmatrix}r&0&cdots & 0\
0&1&cdots & 0\
vdots&vdots&ddots&vdots\0 & 0 & cdots &1end{bmatrix} $.
answered Jan 8 at 13:14
Thomas ShelbyThomas Shelby
4,5862727
$endgroup$
Hint:
Given $rin Bbb R $, take the diagonal matrix
$begin{bmatrix}r&0&cdots & 0\
0&1&cdots & 0\
vdots&vdots&ddots&vdots\0 & 0 & cdots &1end{bmatrix} $.
answered Jan 8 at 13:14
Thomas ShelbyThomas Shelby
4,5862727
edited Jan 8 at 13:19
answered Jan 8 at 13:14
Thomas ShelbyThomas Shelby
4,5862727
answered Jan 8 at 13:14
Thomas ShelbyThomas Shelby
4,5862727
answered Jan 8 at 13:14
Thomas ShelbyThomas Shelby
4,5862727
4,5862727
add a comment |
add a comment |
$begingroup$
Multiplying a row (or column) of a matrix by a constant results in multiplying the determinant by that constant. (The determinant is a $k$-multilinear function.)
So, given $rinBbb R$, just take a matrix with nonzero determinant, $d$, and multiply a row by the constant $frac rd$, to get a matrix with determinant $r$.
answered Jan 8 at 13:39
Chris CusterChris Custer
14.2k3827
$endgroup$
add a comment |
$begingroup$
Multiplying a row (or column) of a matrix by a constant results in multiplying the determinant by that constant. (The determinant is a $k$-multilinear function.)
So, given $rinBbb R$, just take a matrix with nonzero determinant, $d$, and multiply a row by the constant $frac rd$, to get a matrix with determinant $r$.
answered Jan 8 at 13:39
Chris CusterChris Custer
14.2k3827
$endgroup$
add a comment |
$begingroup$
Multiplying a row (or column) of a matrix by a constant results in multiplying the determinant by that constant. (The determinant is a $k$-multilinear function.)
So, given $rinBbb R$, just take a matrix with nonzero determinant, $d$, and multiply a row by the constant $frac rd$, to get a matrix with determinant $r$.
answered Jan 8 at 13:39
Chris CusterChris Custer
14.2k3827
$endgroup$
Multiplying a row (or column) of a matrix by a constant results in multiplying the determinant by that constant. (The determinant is a $k$-multilinear function.)
So, given $rinBbb R$, just take a matrix with nonzero determinant, $d$, and multiply a row by the constant $frac rd$, to get a matrix with determinant $r$.
answered Jan 8 at 13:39
Chris CusterChris Custer
14.2k3827
answered Jan 8 at 13:39
Chris CusterChris Custer
14.2k3827
answered Jan 8 at 13:39
Chris CusterChris Custer
14.2k3827
answered Jan 8 at 13:39
Chris CusterChris Custer
14.2k3827
14.2k3827
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%2f3066144%2fhow-to-prove-surjectivity-of-determinant%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
$begingroup$
Hint: use a diagonal matrix.
$endgroup$
– Bernard
Jan 8 at 13:11