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











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1 Answer
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1 Answer
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active

oldest

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active

oldest

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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


















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