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
asked Dec 15 '18 at 15:51
VimVim
8,13931348
$endgroup$
|
show 1 more comment
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
asked Dec 15 '18 at 15:51
VimVim
8,13931348
$endgroup$
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
|
show 1 more comment
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
asked Dec 15 '18 at 15:51
VimVim
8,13931348
$endgroup$
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
asked Dec 15 '18 at 15:51
VimVim
8,13931348
asked Dec 15 '18 at 15:51
VimVim
8,13931348
asked Dec 15 '18 at 15:51
VimVim
8,13931348
asked Dec 15 '18 at 15:51
VimVim
8,13931348
asked Dec 15 '18 at 15:51
VimVim
8,13931348
8,13931348
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
|
show 1 more comment
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
|
show 1 more comment
1 Answer
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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
$$0 = (0, A_{21}, A_{31}) cdot (A_{12}, 0, A_{32}) .$$
answered Dec 15 '18 at 15:58
TravisTravis
60.5k767147
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
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1 Answer
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$begingroup$
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
$$0 = (0, A_{21}, A_{31}) cdot (A_{12}, 0, A_{32}) .$$
answered Dec 15 '18 at 15:58
TravisTravis
60.5k767147
$endgroup$
$begingroup$
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
add a comment |
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
$$0 = (0, A_{21}, A_{31}) cdot (A_{12}, 0, A_{32}) .$$
answered Dec 15 '18 at 15:58
TravisTravis
60.5k767147
$endgroup$
$begingroup$
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
add a comment |
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
$$0 = (0, A_{21}, A_{31}) cdot (A_{12}, 0, A_{32}) .$$
answered Dec 15 '18 at 15:58
TravisTravis
60.5k767147
$endgroup$
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
$$0 = (0, A_{21}, A_{31}) cdot (A_{12}, 0, A_{32}) .$$
answered Dec 15 '18 at 15:58
TravisTravis
60.5k767147
answered Dec 15 '18 at 15:58
TravisTravis
60.5k767147
answered Dec 15 '18 at 15:58
TravisTravis
60.5k767147
answered Dec 15 '18 at 15:58
TravisTravis
60.5k767147
60.5k767147
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
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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