Mixing
Number of possible zero entries in orthogonal matrices
$begingroup$
It's easy to check that in an orthogonal matrix $Q$ dimension $2 times 2$ if there is entry $0$ in the matrix then necessary one additional zero must be present and the total number of zeros is $2$.
In an orthogonal matrix dim. $3 times 3$ number of zeros can be (if they are present) , I suppose from observations, only $4$ or $6$ - once again we obtain an even number of possible zeros.
Examples:
$ begin{bmatrix}
0.6 & -0.8 & 0 \ 0.8 & 0.6 & 0 \ 0 & 0 & 1 \
end{bmatrix} $ , $ begin{bmatrix} 0 & 0 & 1 \
1 & 0 & 0 \ 0 & 1 & 0 \
end{bmatrix}$
Can this observation be extended for other orthogonal matrices of
greater dimensions? The number of zeros is always even? How to prove this?Maybe, it is known the explicit formula for the number of possible zeros in orthogonal matrices of any dimension?
linear-algebra matrices orthogonal-matrices
asked Jan 29 at 16:42
WidawensenWidawensen
4,74831446
$endgroup$
add a comment |
$begingroup$
It's easy to check that in an orthogonal matrix $Q$ dimension $2 times 2$ if there is entry $0$ in the matrix then necessary one additional zero must be present and the total number of zeros is $2$.
In an orthogonal matrix dim. $3 times 3$ number of zeros can be (if they are present) , I suppose from observations, only $4$ or $6$ - once again we obtain an even number of possible zeros.
Examples:
$ begin{bmatrix}
0.6 & -0.8 & 0 \ 0.8 & 0.6 & 0 \ 0 & 0 & 1 \
end{bmatrix} $ , $ begin{bmatrix} 0 & 0 & 1 \
1 & 0 & 0 \ 0 & 1 & 0 \
end{bmatrix}$
Can this observation be extended for other orthogonal matrices of
greater dimensions? The number of zeros is always even? How to prove this?Maybe, it is known the explicit formula for the number of possible zeros in orthogonal matrices of any dimension?
linear-algebra matrices orthogonal-matrices
asked Jan 29 at 16:42
WidawensenWidawensen
4,74831446
$endgroup$
add a comment |
$begingroup$
It's easy to check that in an orthogonal matrix $Q$ dimension $2 times 2$ if there is entry $0$ in the matrix then necessary one additional zero must be present and the total number of zeros is $2$.
In an orthogonal matrix dim. $3 times 3$ number of zeros can be (if they are present) , I suppose from observations, only $4$ or $6$ - once again we obtain an even number of possible zeros.
Examples:
$ begin{bmatrix}
0.6 & -0.8 & 0 \ 0.8 & 0.6 & 0 \ 0 & 0 & 1 \
end{bmatrix} $ , $ begin{bmatrix} 0 & 0 & 1 \
1 & 0 & 0 \ 0 & 1 & 0 \
end{bmatrix}$
Can this observation be extended for other orthogonal matrices of
greater dimensions? The number of zeros is always even? How to prove this?Maybe, it is known the explicit formula for the number of possible zeros in orthogonal matrices of any dimension?
linear-algebra matrices orthogonal-matrices
asked Jan 29 at 16:42
WidawensenWidawensen
4,74831446
$endgroup$
It's easy to check that in an orthogonal matrix $Q$ dimension $2 times 2$ if there is entry $0$ in the matrix then necessary one additional zero must be present and the total number of zeros is $2$.
In an orthogonal matrix dim. $3 times 3$ number of zeros can be (if they are present) , I suppose from observations, only $4$ or $6$ - once again we obtain an even number of possible zeros.
Examples:
$ begin{bmatrix}
0.6 & -0.8 & 0 \ 0.8 & 0.6 & 0 \ 0 & 0 & 1 \
end{bmatrix} $ , $ begin{bmatrix} 0 & 0 & 1 \
1 & 0 & 0 \ 0 & 1 & 0 \
end{bmatrix}$
Can this observation be extended for other orthogonal matrices of
greater dimensions? The number of zeros is always even? How to prove this?Maybe, it is known the explicit formula for the number of possible zeros in orthogonal matrices of any dimension?
linear-algebra matrices orthogonal-matrices
linear-algebra matrices orthogonal-matrices
asked Jan 29 at 16:42
WidawensenWidawensen
4,74831446
asked Jan 29 at 16:42
WidawensenWidawensen
4,74831446
edited Jan 29 at 18:00
Widawensen
asked Jan 29 at 16:42
WidawensenWidawensen
4,74831446
asked Jan 29 at 16:42
WidawensenWidawensen
4,74831446
asked Jan 29 at 16:42
WidawensenWidawensen
4,74831446
4,74831446
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
An interesting observation, but it doesn't pan out unfortunately! It already fails in dimension 3. Wikipedia has the following counterexample of a rotoinversion:
$$
begin{bmatrix}
0 & -0.8 & - 0.6 \
0.8 & -0.36 & 0.48 \
0.6&0.48&-0.64
end{bmatrix}
$$
(WolframAlpha agrees that this is indeed orthogonal, the example is from
here)
answered Jan 29 at 20:35
ffffforallffffforall
36028
$endgroup$
$begingroup$
Surprising example! Do you think that existence of such example is possible only for number of zeros $1$ or that also be possible to find, for example, for number $3$ or $5$? Can we show the all numbers for dimension $3$?
$endgroup$
– Widawensen
Jan 30 at 11:59
1
$begingroup$
I don't think 3 zeros are possible in the 3-dimensional case. The column vectors form an orthonormal basis. If one vector has two zeros (and hence is a standard unit vector) , then the other have either two or none, by the same argument as in the 2-dimensional case. Otherwise they would all have one zero each, so the matrix would have to act on each standard unit vector as a rotoinversion restricted to a plane, which I don't think works.
$endgroup$
– ffffforall
Jan 30 at 12:09
$begingroup$
What about $5$? Probably also not. It 's hard also for me to find an example with number of zeros $2$..
$endgroup$
– Widawensen
Jan 30 at 12:13
1
$begingroup$
5 zeros are impossible (for the 3-dimensional case). If the matrix has at least 5 zeros, then two of the column vectors must be standard unit vectors (up to sign). But then this forces the remaining column to also be a standard unit vector (up to sign), so that they form an orthonormal basis.
$endgroup$
– ffffforall
Jan 30 at 12:17
$begingroup$
I see that your counterexample give some insight for constructions of matrices with ($k_z=1$) : the first column is $ [0 a b ]^T$ , the second $ [-a -b^2 ab ]^T$, the third $ [-b ab -a^2 ]^T$ with additional condition $a^2+b^2=1$
$endgroup$
– Widawensen
Jan 30 at 12:27
|
show 3 more comments
$begingroup$
Rotations are orthogonal. Most examples of rotation matrices in any dimension will provide a counter example to this. In the $2 times 2$ case consider
$$begin{pmatrix}
Cos(x)&-Sin(x)\
Sin(x)&Cos(x)
end{pmatrix}$$
This is orthogonal for any $x$. There are many cases where there are no zeros.
answered Jan 29 at 21:04
WintermuteWintermute
2,35441832
$endgroup$
$begingroup$
Thank you Wintermute for the answer, but the question was for possible number of $0$ when they exist. In dimension of matrix $2$ it is only one possible number $2$.
$endgroup$
– Widawensen
Jan 30 at 12:03
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%2f3092403%2fnumber-of-possible-zero-entries-in-orthogonal-matrices%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
An interesting observation, but it doesn't pan out unfortunately! It already fails in dimension 3. Wikipedia has the following counterexample of a rotoinversion:
$$
begin{bmatrix}
0 & -0.8 & - 0.6 \
0.8 & -0.36 & 0.48 \
0.6&0.48&-0.64
end{bmatrix}
$$
(WolframAlpha agrees that this is indeed orthogonal, the example is from
here)
answered Jan 29 at 20:35
ffffforallffffforall
36028
$endgroup$
$begingroup$
Surprising example! Do you think that existence of such example is possible only for number of zeros $1$ or that also be possible to find, for example, for number $3$ or $5$? Can we show the all numbers for dimension $3$?
$endgroup$
– Widawensen
Jan 30 at 11:59
1
$begingroup$
I don't think 3 zeros are possible in the 3-dimensional case. The column vectors form an orthonormal basis. If one vector has two zeros (and hence is a standard unit vector) , then the other have either two or none, by the same argument as in the 2-dimensional case. Otherwise they would all have one zero each, so the matrix would have to act on each standard unit vector as a rotoinversion restricted to a plane, which I don't think works.
$endgroup$
– ffffforall
Jan 30 at 12:09
$begingroup$
What about $5$? Probably also not. It 's hard also for me to find an example with number of zeros $2$..
$endgroup$
– Widawensen
Jan 30 at 12:13
1
$begingroup$
5 zeros are impossible (for the 3-dimensional case). If the matrix has at least 5 zeros, then two of the column vectors must be standard unit vectors (up to sign). But then this forces the remaining column to also be a standard unit vector (up to sign), so that they form an orthonormal basis.
$endgroup$
– ffffforall
Jan 30 at 12:17
$begingroup$
I see that your counterexample give some insight for constructions of matrices with ($k_z=1$) : the first column is $ [0 a b ]^T$ , the second $ [-a -b^2 ab ]^T$, the third $ [-b ab -a^2 ]^T$ with additional condition $a^2+b^2=1$
$endgroup$
– Widawensen
Jan 30 at 12:27
|
show 3 more comments
$begingroup$
An interesting observation, but it doesn't pan out unfortunately! It already fails in dimension 3. Wikipedia has the following counterexample of a rotoinversion:
$$
begin{bmatrix}
0 & -0.8 & - 0.6 \
0.8 & -0.36 & 0.48 \
0.6&0.48&-0.64
end{bmatrix}
$$
(WolframAlpha agrees that this is indeed orthogonal, the example is from
here)
answered Jan 29 at 20:35
ffffforallffffforall
36028
$endgroup$
$begingroup$
Surprising example! Do you think that existence of such example is possible only for number of zeros $1$ or that also be possible to find, for example, for number $3$ or $5$? Can we show the all numbers for dimension $3$?
$endgroup$
– Widawensen
Jan 30 at 11:59
1
$begingroup$
I don't think 3 zeros are possible in the 3-dimensional case. The column vectors form an orthonormal basis. If one vector has two zeros (and hence is a standard unit vector) , then the other have either two or none, by the same argument as in the 2-dimensional case. Otherwise they would all have one zero each, so the matrix would have to act on each standard unit vector as a rotoinversion restricted to a plane, which I don't think works.
$endgroup$
– ffffforall
Jan 30 at 12:09
$begingroup$
What about $5$? Probably also not. It 's hard also for me to find an example with number of zeros $2$..
$endgroup$
– Widawensen
Jan 30 at 12:13
1
$begingroup$
5 zeros are impossible (for the 3-dimensional case). If the matrix has at least 5 zeros, then two of the column vectors must be standard unit vectors (up to sign). But then this forces the remaining column to also be a standard unit vector (up to sign), so that they form an orthonormal basis.
$endgroup$
– ffffforall
Jan 30 at 12:17
$begingroup$
I see that your counterexample give some insight for constructions of matrices with ($k_z=1$) : the first column is $ [0 a b ]^T$ , the second $ [-a -b^2 ab ]^T$, the third $ [-b ab -a^2 ]^T$ with additional condition $a^2+b^2=1$
$endgroup$
– Widawensen
Jan 30 at 12:27
|
show 3 more comments
$begingroup$
An interesting observation, but it doesn't pan out unfortunately! It already fails in dimension 3. Wikipedia has the following counterexample of a rotoinversion:
$$
begin{bmatrix}
0 & -0.8 & - 0.6 \
0.8 & -0.36 & 0.48 \
0.6&0.48&-0.64
end{bmatrix}
$$
(WolframAlpha agrees that this is indeed orthogonal, the example is from
here)
answered Jan 29 at 20:35
ffffforallffffforall
36028
$endgroup$
An interesting observation, but it doesn't pan out unfortunately! It already fails in dimension 3. Wikipedia has the following counterexample of a rotoinversion:
$$
begin{bmatrix}
0 & -0.8 & - 0.6 \
0.8 & -0.36 & 0.48 \
0.6&0.48&-0.64
end{bmatrix}
$$
(WolframAlpha agrees that this is indeed orthogonal, the example is from
here)
answered Jan 29 at 20:35
ffffforallffffforall
36028
edited Jan 29 at 20:40
answered Jan 29 at 20:35
ffffforallffffforall
36028
answered Jan 29 at 20:35
ffffforallffffforall
36028
answered Jan 29 at 20:35
ffffforallffffforall
36028
36028
$begingroup$
Surprising example! Do you think that existence of such example is possible only for number of zeros $1$ or that also be possible to find, for example, for number $3$ or $5$? Can we show the all numbers for dimension $3$?
$endgroup$
– Widawensen
Jan 30 at 11:59
1
$begingroup$
I don't think 3 zeros are possible in the 3-dimensional case. The column vectors form an orthonormal basis. If one vector has two zeros (and hence is a standard unit vector) , then the other have either two or none, by the same argument as in the 2-dimensional case. Otherwise they would all have one zero each, so the matrix would have to act on each standard unit vector as a rotoinversion restricted to a plane, which I don't think works.
$endgroup$
– ffffforall
Jan 30 at 12:09
$begingroup$
What about $5$? Probably also not. It 's hard also for me to find an example with number of zeros $2$..
$endgroup$
– Widawensen
Jan 30 at 12:13
1
$begingroup$
5 zeros are impossible (for the 3-dimensional case). If the matrix has at least 5 zeros, then two of the column vectors must be standard unit vectors (up to sign). But then this forces the remaining column to also be a standard unit vector (up to sign), so that they form an orthonormal basis.
$endgroup$
– ffffforall
Jan 30 at 12:17
$begingroup$
I see that your counterexample give some insight for constructions of matrices with ($k_z=1$) : the first column is $ [0 a b ]^T$ , the second $ [-a -b^2 ab ]^T$, the third $ [-b ab -a^2 ]^T$ with additional condition $a^2+b^2=1$
$endgroup$
– Widawensen
Jan 30 at 12:27
|
show 3 more comments
$begingroup$
Surprising example! Do you think that existence of such example is possible only for number of zeros $1$ or that also be possible to find, for example, for number $3$ or $5$? Can we show the all numbers for dimension $3$?
$endgroup$
– Widawensen
Jan 30 at 11:59
1
$begingroup$
I don't think 3 zeros are possible in the 3-dimensional case. The column vectors form an orthonormal basis. If one vector has two zeros (and hence is a standard unit vector) , then the other have either two or none, by the same argument as in the 2-dimensional case. Otherwise they would all have one zero each, so the matrix would have to act on each standard unit vector as a rotoinversion restricted to a plane, which I don't think works.
$endgroup$
– ffffforall
Jan 30 at 12:09
$begingroup$
What about $5$? Probably also not. It 's hard also for me to find an example with number of zeros $2$..
$endgroup$
– Widawensen
Jan 30 at 12:13
1
$begingroup$
5 zeros are impossible (for the 3-dimensional case). If the matrix has at least 5 zeros, then two of the column vectors must be standard unit vectors (up to sign). But then this forces the remaining column to also be a standard unit vector (up to sign), so that they form an orthonormal basis.
$endgroup$
– ffffforall
Jan 30 at 12:17
$begingroup$
I see that your counterexample give some insight for constructions of matrices with ($k_z=1$) : the first column is $ [0 a b ]^T$ , the second $ [-a -b^2 ab ]^T$, the third $ [-b ab -a^2 ]^T$ with additional condition $a^2+b^2=1$
$endgroup$
– Widawensen
Jan 30 at 12:27
$begingroup$
Surprising example! Do you think that existence of such example is possible only for number of zeros $1$ or that also be possible to find, for example, for number $3$ or $5$? Can we show the all numbers for dimension $3$?
$endgroup$
– Widawensen
Jan 30 at 11:59
$begingroup$
Surprising example! Do you think that existence of such example is possible only for number of zeros $1$ or that also be possible to find, for example, for number $3$ or $5$? Can we show the all numbers for dimension $3$?
$endgroup$
– Widawensen
Jan 30 at 11:59
1
1
$begingroup$
I don't think 3 zeros are possible in the 3-dimensional case. The column vectors form an orthonormal basis. If one vector has two zeros (and hence is a standard unit vector) , then the other have either two or none, by the same argument as in the 2-dimensional case. Otherwise they would all have one zero each, so the matrix would have to act on each standard unit vector as a rotoinversion restricted to a plane, which I don't think works.
$endgroup$
– ffffforall
Jan 30 at 12:09
$begingroup$
I don't think 3 zeros are possible in the 3-dimensional case. The column vectors form an orthonormal basis. If one vector has two zeros (and hence is a standard unit vector) , then the other have either two or none, by the same argument as in the 2-dimensional case. Otherwise they would all have one zero each, so the matrix would have to act on each standard unit vector as a rotoinversion restricted to a plane, which I don't think works.
$endgroup$
– ffffforall
Jan 30 at 12:09
$begingroup$
What about $5$? Probably also not. It 's hard also for me to find an example with number of zeros $2$..
$endgroup$
– Widawensen
Jan 30 at 12:13
$begingroup$
What about $5$? Probably also not. It 's hard also for me to find an example with number of zeros $2$..
$endgroup$
– Widawensen
Jan 30 at 12:13
1
1
$begingroup$
5 zeros are impossible (for the 3-dimensional case). If the matrix has at least 5 zeros, then two of the column vectors must be standard unit vectors (up to sign). But then this forces the remaining column to also be a standard unit vector (up to sign), so that they form an orthonormal basis.
$endgroup$
– ffffforall
Jan 30 at 12:17
$begingroup$
5 zeros are impossible (for the 3-dimensional case). If the matrix has at least 5 zeros, then two of the column vectors must be standard unit vectors (up to sign). But then this forces the remaining column to also be a standard unit vector (up to sign), so that they form an orthonormal basis.
$endgroup$
– ffffforall
Jan 30 at 12:17
$begingroup$
I see that your counterexample give some insight for constructions of matrices with ($k_z=1$) : the first column is $ [0 a b ]^T$ , the second $ [-a -b^2 ab ]^T$, the third $ [-b ab -a^2 ]^T$ with additional condition $a^2+b^2=1$
$endgroup$
– Widawensen
Jan 30 at 12:27
$begingroup$
I see that your counterexample give some insight for constructions of matrices with ($k_z=1$) : the first column is $ [0 a b ]^T$ , the second $ [-a -b^2 ab ]^T$, the third $ [-b ab -a^2 ]^T$ with additional condition $a^2+b^2=1$
$endgroup$
– Widawensen
Jan 30 at 12:27
|
show 3 more comments
$begingroup$
Rotations are orthogonal. Most examples of rotation matrices in any dimension will provide a counter example to this. In the $2 times 2$ case consider
$$begin{pmatrix}
Cos(x)&-Sin(x)\
Sin(x)&Cos(x)
end{pmatrix}$$
This is orthogonal for any $x$. There are many cases where there are no zeros.
answered Jan 29 at 21:04
WintermuteWintermute
2,35441832
$endgroup$
$begingroup$
Thank you Wintermute for the answer, but the question was for possible number of $0$ when they exist. In dimension of matrix $2$ it is only one possible number $2$.
$endgroup$
– Widawensen
Jan 30 at 12:03
add a comment |
$begingroup$
Rotations are orthogonal. Most examples of rotation matrices in any dimension will provide a counter example to this. In the $2 times 2$ case consider
$$begin{pmatrix}
Cos(x)&-Sin(x)\
Sin(x)&Cos(x)
end{pmatrix}$$
This is orthogonal for any $x$. There are many cases where there are no zeros.
answered Jan 29 at 21:04
WintermuteWintermute
2,35441832
$endgroup$
$begingroup$
Thank you Wintermute for the answer, but the question was for possible number of $0$ when they exist. In dimension of matrix $2$ it is only one possible number $2$.
$endgroup$
– Widawensen
Jan 30 at 12:03
add a comment |
$begingroup$
Rotations are orthogonal. Most examples of rotation matrices in any dimension will provide a counter example to this. In the $2 times 2$ case consider
$$begin{pmatrix}
Cos(x)&-Sin(x)\
Sin(x)&Cos(x)
end{pmatrix}$$
This is orthogonal for any $x$. There are many cases where there are no zeros.
answered Jan 29 at 21:04
WintermuteWintermute
2,35441832
$endgroup$
Rotations are orthogonal. Most examples of rotation matrices in any dimension will provide a counter example to this. In the $2 times 2$ case consider
$$begin{pmatrix}
Cos(x)&-Sin(x)\
Sin(x)&Cos(x)
end{pmatrix}$$
This is orthogonal for any $x$. There are many cases where there are no zeros.
answered Jan 29 at 21:04
WintermuteWintermute
2,35441832
answered Jan 29 at 21:04
WintermuteWintermute
2,35441832
answered Jan 29 at 21:04
WintermuteWintermute
2,35441832
answered Jan 29 at 21:04
WintermuteWintermute
2,35441832
2,35441832
$begingroup$
Thank you Wintermute for the answer, but the question was for possible number of $0$ when they exist. In dimension of matrix $2$ it is only one possible number $2$.
$endgroup$
– Widawensen
Jan 30 at 12:03
add a comment |
$begingroup$
Thank you Wintermute for the answer, but the question was for possible number of $0$ when they exist. In dimension of matrix $2$ it is only one possible number $2$.
$endgroup$
– Widawensen
Jan 30 at 12:03
$begingroup$
Thank you Wintermute for the answer, but the question was for possible number of $0$ when they exist. In dimension of matrix $2$ it is only one possible number $2$.
$endgroup$
– Widawensen
Jan 30 at 12:03
$begingroup$
Thank you Wintermute for the answer, but the question was for possible number of $0$ when they exist. In dimension of matrix $2$ it is only one possible number $2$.
$endgroup$
– Widawensen
Jan 30 at 12:03
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%2f3092403%2fnumber-of-possible-zero-entries-in-orthogonal-matrices%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
