Existence of orthogonal matrices with zero diagonal and non-zero off-diagonal values












0












$begingroup$


Does there exist an orthogonal matrix whose diagonal values are all zero but whose off-diagonal values are all non-zero for any $Bbb R^n$?



Furthermore, does this conclusion change if we are talking about unitary matrices and $Bbb C^n$?






linear-algebra orthogonality







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Yes. Consider $begin{pmatrix}0&1\1&0end{pmatrix}$. In $mathbb{R}^{ntimes n}$ you can create similar situation.
    $endgroup$
    – user9077
    Dec 15 '18 at 16:01










  • $begingroup$
    @user9077 How would you "create a similar situation" for $n = 3$?
    $endgroup$
    – Travis
    Dec 15 '18 at 16:47










  • $begingroup$
    @user9077 do you mean staking them diagonally to make the diagonal entries zero? But my main difficulty is how to ensure the off-diagonal entries are non-zero.
    $endgroup$
    – Vim
    Dec 15 '18 at 16:51










  • $begingroup$
    Let $e_1,e_2,ldots,e_n$ be the standard basis of $mathbb{R}^n$. Just create a matrix where the columns are $e_n,ldots,e_2,e_1$ in that order.
    $endgroup$
    – user9077
    Dec 15 '18 at 16:54










  • $begingroup$
    @user9077 For $n > 2$ the resulting matrix has some zero off-diagonal entries.
    $endgroup$
    – Travis
    Dec 15 '18 at 16:57
















0












$begingroup$


Does there exist an orthogonal matrix whose diagonal values are all zero but whose off-diagonal values are all non-zero for any $Bbb R^n$?



Furthermore, does this conclusion change if we are talking about unitary matrices and $Bbb C^n$?






linear-algebra orthogonality







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Yes. Consider $begin{pmatrix}0&1\1&0end{pmatrix}$. In $mathbb{R}^{ntimes n}$ you can create similar situation.
    $endgroup$
    – user9077
    Dec 15 '18 at 16:01










  • $begingroup$
    @user9077 How would you "create a similar situation" for $n = 3$?
    $endgroup$
    – Travis
    Dec 15 '18 at 16:47










  • $begingroup$
    @user9077 do you mean staking them diagonally to make the diagonal entries zero? But my main difficulty is how to ensure the off-diagonal entries are non-zero.
    $endgroup$
    – Vim
    Dec 15 '18 at 16:51










  • $begingroup$
    Let $e_1,e_2,ldots,e_n$ be the standard basis of $mathbb{R}^n$. Just create a matrix where the columns are $e_n,ldots,e_2,e_1$ in that order.
    $endgroup$
    – user9077
    Dec 15 '18 at 16:54










  • $begingroup$
    @user9077 For $n > 2$ the resulting matrix has some zero off-diagonal entries.
    $endgroup$
    – Travis
    Dec 15 '18 at 16:57














0












0








0





$begingroup$


Does there exist an orthogonal matrix whose diagonal values are all zero but whose off-diagonal values are all non-zero for any $Bbb R^n$?



Furthermore, does this conclusion change if we are talking about unitary matrices and $Bbb C^n$?






linear-algebra orthogonality







share|cite|improve this question









$endgroup$




Does there exist an orthogonal matrix whose diagonal values are all zero but whose off-diagonal values are all non-zero for any $Bbb R^n$?



Furthermore, does this conclusion change if we are talking about unitary matrices and $Bbb C^n$?






linear-algebra orthogonality




linear-algebra orthogonality






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 15 '18 at 15:51









VimVim

8,05631348




8,05631348








  • 1




    $begingroup$
    Yes. Consider $begin{pmatrix}0&1\1&0end{pmatrix}$. In $mathbb{R}^{ntimes n}$ you can create similar situation.
    $endgroup$
    – user9077
    Dec 15 '18 at 16:01










  • $begingroup$
    @user9077 How would you "create a similar situation" for $n = 3$?
    $endgroup$
    – Travis
    Dec 15 '18 at 16:47










  • $begingroup$
    @user9077 do you mean staking them diagonally to make the diagonal entries zero? But my main difficulty is how to ensure the off-diagonal entries are non-zero.
    $endgroup$
    – Vim
    Dec 15 '18 at 16:51










  • $begingroup$
    Let $e_1,e_2,ldots,e_n$ be the standard basis of $mathbb{R}^n$. Just create a matrix where the columns are $e_n,ldots,e_2,e_1$ in that order.
    $endgroup$
    – user9077
    Dec 15 '18 at 16:54










  • $begingroup$
    @user9077 For $n > 2$ the resulting matrix has some zero off-diagonal entries.
    $endgroup$
    – Travis
    Dec 15 '18 at 16:57














  • 1




    $begingroup$
    Yes. Consider $begin{pmatrix}0&1\1&0end{pmatrix}$. In $mathbb{R}^{ntimes n}$ you can create similar situation.
    $endgroup$
    – user9077
    Dec 15 '18 at 16:01










  • $begingroup$
    @user9077 How would you "create a similar situation" for $n = 3$?
    $endgroup$
    – Travis
    Dec 15 '18 at 16:47










  • $begingroup$
    @user9077 do you mean staking them diagonally to make the diagonal entries zero? But my main difficulty is how to ensure the off-diagonal entries are non-zero.
    $endgroup$
    – Vim
    Dec 15 '18 at 16:51










  • $begingroup$
    Let $e_1,e_2,ldots,e_n$ be the standard basis of $mathbb{R}^n$. Just create a matrix where the columns are $e_n,ldots,e_2,e_1$ in that order.
    $endgroup$
    – user9077
    Dec 15 '18 at 16:54










  • $begingroup$
    @user9077 For $n > 2$ the resulting matrix has some zero off-diagonal entries.
    $endgroup$
    – Travis
    Dec 15 '18 at 16:57








1




1




$begingroup$
Yes. Consider $begin{pmatrix}0&1\1&0end{pmatrix}$. In $mathbb{R}^{ntimes n}$ you can create similar situation.
$endgroup$
– user9077
Dec 15 '18 at 16:01




$begingroup$
Yes. Consider $begin{pmatrix}0&1\1&0end{pmatrix}$. In $mathbb{R}^{ntimes n}$ you can create similar situation.
$endgroup$
– user9077
Dec 15 '18 at 16:01












$begingroup$
@user9077 How would you "create a similar situation" for $n = 3$?
$endgroup$
– Travis
Dec 15 '18 at 16:47




$begingroup$
@user9077 How would you "create a similar situation" for $n = 3$?
$endgroup$
– Travis
Dec 15 '18 at 16:47












$begingroup$
@user9077 do you mean staking them diagonally to make the diagonal entries zero? But my main difficulty is how to ensure the off-diagonal entries are non-zero.
$endgroup$
– Vim
Dec 15 '18 at 16:51




$begingroup$
@user9077 do you mean staking them diagonally to make the diagonal entries zero? But my main difficulty is how to ensure the off-diagonal entries are non-zero.
$endgroup$
– Vim
Dec 15 '18 at 16:51












$begingroup$
Let $e_1,e_2,ldots,e_n$ be the standard basis of $mathbb{R}^n$. Just create a matrix where the columns are $e_n,ldots,e_2,e_1$ in that order.
$endgroup$
– user9077
Dec 15 '18 at 16:54




$begingroup$
Let $e_1,e_2,ldots,e_n$ be the standard basis of $mathbb{R}^n$. Just create a matrix where the columns are $e_n,ldots,e_2,e_1$ in that order.
$endgroup$
– user9077
Dec 15 '18 at 16:54












$begingroup$
@user9077 For $n > 2$ the resulting matrix has some zero off-diagonal entries.
$endgroup$
– Travis
Dec 15 '18 at 16:57




$begingroup$
@user9077 For $n > 2$ the resulting matrix has some zero off-diagonal entries.
$endgroup$
– Travis
Dec 15 '18 at 16:57










1 Answer
1






active

oldest

votes


















1












$begingroup$

Hint Recall that the columns of an orthogonal matrix are pairwise orthogonal. So, for any $3 times 3$ orthogonal matrix $A$ with zero diagonal entries, the dot product of, say, the first two columns is
$$0 = (0, A_{21}, A_{31}) cdot (A_{12}, 0, A_{32}) .$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks but how does this help with the existence, and how does it generalise to $n$?
    $endgroup$
    – Vim
    Dec 15 '18 at 16:49








  • 1




    $begingroup$
    When you expand the r.h.s. of the equation in the hint, what does it tell you about existence in the $n = 3$ case?
    $endgroup$
    – Travis
    Dec 15 '18 at 16:58












  • $begingroup$
    I got it. Its about non existence. Thanks.
    $endgroup$
    – Vim
    Dec 15 '18 at 17:02










  • $begingroup$
    You're welcome, I'm glad you found it helpful.
    $endgroup$
    – Travis
    Dec 15 '18 at 17:03











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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3041638%2fexistence-of-orthogonal-matrices-with-zero-diagonal-and-non-zero-off-diagonal-va%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









1












$begingroup$

Hint Recall that the columns of an orthogonal matrix are pairwise orthogonal. So, for any $3 times 3$ orthogonal matrix $A$ with zero diagonal entries, the dot product of, say, the first two columns is
$$0 = (0, A_{21}, A_{31}) cdot (A_{12}, 0, A_{32}) .$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks but how does this help with the existence, and how does it generalise to $n$?
    $endgroup$
    – Vim
    Dec 15 '18 at 16:49








  • 1




    $begingroup$
    When you expand the r.h.s. of the equation in the hint, what does it tell you about existence in the $n = 3$ case?
    $endgroup$
    – Travis
    Dec 15 '18 at 16:58












  • $begingroup$
    I got it. Its about non existence. Thanks.
    $endgroup$
    – Vim
    Dec 15 '18 at 17:02










  • $begingroup$
    You're welcome, I'm glad you found it helpful.
    $endgroup$
    – Travis
    Dec 15 '18 at 17:03
















1












$begingroup$

Hint Recall that the columns of an orthogonal matrix are pairwise orthogonal. So, for any $3 times 3$ orthogonal matrix $A$ with zero diagonal entries, the dot product of, say, the first two columns is
$$0 = (0, A_{21}, A_{31}) cdot (A_{12}, 0, A_{32}) .$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks but how does this help with the existence, and how does it generalise to $n$?
    $endgroup$
    – Vim
    Dec 15 '18 at 16:49








  • 1




    $begingroup$
    When you expand the r.h.s. of the equation in the hint, what does it tell you about existence in the $n = 3$ case?
    $endgroup$
    – Travis
    Dec 15 '18 at 16:58












  • $begingroup$
    I got it. Its about non existence. Thanks.
    $endgroup$
    – Vim
    Dec 15 '18 at 17:02










  • $begingroup$
    You're welcome, I'm glad you found it helpful.
    $endgroup$
    – Travis
    Dec 15 '18 at 17:03














1












1








1





$begingroup$

Hint Recall that the columns of an orthogonal matrix are pairwise orthogonal. So, for any $3 times 3$ orthogonal matrix $A$ with zero diagonal entries, the dot product of, say, the first two columns is
$$0 = (0, A_{21}, A_{31}) cdot (A_{12}, 0, A_{32}) .$$






share|cite|improve this answer









$endgroup$



Hint Recall that the columns of an orthogonal matrix are pairwise orthogonal. So, for any $3 times 3$ orthogonal matrix $A$ with zero diagonal entries, the dot product of, say, the first two columns is
$$0 = (0, A_{21}, A_{31}) cdot (A_{12}, 0, A_{32}) .$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 15 '18 at 15:58









TravisTravis

59.9k767146




59.9k767146












  • $begingroup$
    Thanks but how does this help with the existence, and how does it generalise to $n$?
    $endgroup$
    – Vim
    Dec 15 '18 at 16:49








  • 1




    $begingroup$
    When you expand the r.h.s. of the equation in the hint, what does it tell you about existence in the $n = 3$ case?
    $endgroup$
    – Travis
    Dec 15 '18 at 16:58












  • $begingroup$
    I got it. Its about non existence. Thanks.
    $endgroup$
    – Vim
    Dec 15 '18 at 17:02










  • $begingroup$
    You're welcome, I'm glad you found it helpful.
    $endgroup$
    – Travis
    Dec 15 '18 at 17:03


















  • $begingroup$
    Thanks but how does this help with the existence, and how does it generalise to $n$?
    $endgroup$
    – Vim
    Dec 15 '18 at 16:49








  • 1




    $begingroup$
    When you expand the r.h.s. of the equation in the hint, what does it tell you about existence in the $n = 3$ case?
    $endgroup$
    – Travis
    Dec 15 '18 at 16:58












  • $begingroup$
    I got it. Its about non existence. Thanks.
    $endgroup$
    – Vim
    Dec 15 '18 at 17:02










  • $begingroup$
    You're welcome, I'm glad you found it helpful.
    $endgroup$
    – Travis
    Dec 15 '18 at 17:03
















$begingroup$
Thanks but how does this help with the existence, and how does it generalise to $n$?
$endgroup$
– Vim
Dec 15 '18 at 16:49






$begingroup$
Thanks but how does this help with the existence, and how does it generalise to $n$?
$endgroup$
– Vim
Dec 15 '18 at 16:49






1




1




$begingroup$
When you expand the r.h.s. of the equation in the hint, what does it tell you about existence in the $n = 3$ case?
$endgroup$
– Travis
Dec 15 '18 at 16:58






$begingroup$
When you expand the r.h.s. of the equation in the hint, what does it tell you about existence in the $n = 3$ case?
$endgroup$
– Travis
Dec 15 '18 at 16:58














$begingroup$
I got it. Its about non existence. Thanks.
$endgroup$
– Vim
Dec 15 '18 at 17:02




$begingroup$
I got it. Its about non existence. Thanks.
$endgroup$
– Vim
Dec 15 '18 at 17:02












$begingroup$
You're welcome, I'm glad you found it helpful.
$endgroup$
– Travis
Dec 15 '18 at 17:03




$begingroup$
You're welcome, I'm glad you found it helpful.
$endgroup$
– Travis
Dec 15 '18 at 17:03


















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3041638%2fexistence-of-orthogonal-matrices-with-zero-diagonal-and-non-zero-off-diagonal-va%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

Bressuire

Image
.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid
Read more

Cabo Verde

Image
  Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º)
Read more

Gyllenstierna

Image
.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid
Read more