Orthogonal matrices with eigenvalues equally spaced over unit circle?
$begingroup$
We know that the eigenvalues of orthogonal matrices have norm 1, and thus they are all on the unit circle. However, I wonder if there is a way to construct a orthogonal matrix (with real entries) whose eigenvalue is equally spaced on unit circle (or as uniformly distributed as possible)? "A way to construct" means given an $n$, we can find such a $n times n$ orthogonal matrix. Any help is appreciated!
linear-algebra matrices eigenvalues-eigenvectors orthonormal
asked Jan 10 at 20:13
dave2ddave2d
83
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add a comment |
$begingroup$
We know that the eigenvalues of orthogonal matrices have norm 1, and thus they are all on the unit circle. However, I wonder if there is a way to construct a orthogonal matrix (with real entries) whose eigenvalue is equally spaced on unit circle (or as uniformly distributed as possible)? "A way to construct" means given an $n$, we can find such a $n times n$ orthogonal matrix. Any help is appreciated!
linear-algebra matrices eigenvalues-eigenvectors orthonormal
asked Jan 10 at 20:13
dave2ddave2d
83
$endgroup$
add a comment |
$begingroup$
We know that the eigenvalues of orthogonal matrices have norm 1, and thus they are all on the unit circle. However, I wonder if there is a way to construct a orthogonal matrix (with real entries) whose eigenvalue is equally spaced on unit circle (or as uniformly distributed as possible)? "A way to construct" means given an $n$, we can find such a $n times n$ orthogonal matrix. Any help is appreciated!
linear-algebra matrices eigenvalues-eigenvectors orthonormal
asked Jan 10 at 20:13
dave2ddave2d
83
$endgroup$
We know that the eigenvalues of orthogonal matrices have norm 1, and thus they are all on the unit circle. However, I wonder if there is a way to construct a orthogonal matrix (with real entries) whose eigenvalue is equally spaced on unit circle (or as uniformly distributed as possible)? "A way to construct" means given an $n$, we can find such a $n times n$ orthogonal matrix. Any help is appreciated!
linear-algebra matrices eigenvalues-eigenvectors orthonormal
linear-algebra matrices eigenvalues-eigenvectors orthonormal
asked Jan 10 at 20:13
dave2ddave2d
83
asked Jan 10 at 20:13
dave2ddave2d
83
edited Jan 10 at 20:26
dave2d
asked Jan 10 at 20:13
dave2ddave2d
83
asked Jan 10 at 20:13
dave2ddave2d
83
asked Jan 10 at 20:13
dave2ddave2d
83
83
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
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The eigenvalues of the permutation matrix
$$
M = pmatrix{0&cdots & 0 & 0 & 1\
1&0\
&1&0\
&&ddots & ddots\
&&&1&0}
$$
will be all $n$th roots of unity, i.e. $e^{2 pi i k/n}$ for $i = 0,dots,n-1$. These are equally spaced over the unit circle.
answered Jan 10 at 21:18
OmnomnomnomOmnomnomnom
127k790178
$endgroup$
add a comment |
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You could make the eigenvalues be the $n$-th roots of unity, which are "evenly spaced" around the unit circle in $mathbb C$. For instance, the matrix $$begin{bmatrix}1&0&0\0&omega&0\0&0&omega^2end{bmatrix}$$ works for $n=3$ where $omega=e^{2pi i/3}$. I leave it to you to generalize this to arbitrary $n$.
answered Jan 10 at 20:24
DaveDave
8,76711033
$endgroup$
$begingroup$
Thanks, but sorry I forget to mention, I want the matrices to have real entries.
$endgroup$
– dave2d
Jan 10 at 20:27
$begingroup$
Ah, well then yes my answer won't work. I'll leave it up just in case people are interested.
$endgroup$
– Dave
Jan 10 at 20:27
add a comment |
$begingroup$
Here is the easiest way to do it with real entries (I think). Given a real number $theta$, the rotation matrix
$$
begin{bmatrix}cos theta&-sintheta\sintheta&costhetaend{bmatrix}
$$
has two complex eigenvalues, which have norm $1$ and argument $pmtheta$. Using this, we can build the matrix we're after.
Take the complex unit circle, and distribute $n$ points along it so that
- They are evenly spaced
- The configuration is symmetric with respect to mirroring across the $x$-axis (i.e. complex conjugation)
Pair up the points in complex conjugate pairs, and for each pair, construct the corresponding rotation matrix as shown above (you can choose the sign of $theta$ freely). Then take all those rotation matrices, and put them along the diagonal of an $ntimes n$ matrix (with zeroes in all other entries).
If $n$ is odd, one of the points is $1$ (or $-1$), and doesn't have a pair mate. Just put a $1$ (or $-1$) on the diagonal of the $ntimes n$ matrix instead of a rotation matrix when you get to that point.
answered Jan 10 at 20:24
ArthurArthur
112k7108192
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The eigenvalues of the permutation matrix
$$
M = pmatrix{0&cdots & 0 & 0 & 1\
1&0\
&1&0\
&&ddots & ddots\
&&&1&0}
$$
will be all $n$th roots of unity, i.e. $e^{2 pi i k/n}$ for $i = 0,dots,n-1$. These are equally spaced over the unit circle.
answered Jan 10 at 21:18
OmnomnomnomOmnomnomnom
127k790178
$endgroup$
add a comment |
$begingroup$
The eigenvalues of the permutation matrix
$$
M = pmatrix{0&cdots & 0 & 0 & 1\
1&0\
&1&0\
&&ddots & ddots\
&&&1&0}
$$
will be all $n$th roots of unity, i.e. $e^{2 pi i k/n}$ for $i = 0,dots,n-1$. These are equally spaced over the unit circle.
answered Jan 10 at 21:18
OmnomnomnomOmnomnomnom
127k790178
$endgroup$
add a comment |
$begingroup$
The eigenvalues of the permutation matrix
$$
M = pmatrix{0&cdots & 0 & 0 & 1\
1&0\
&1&0\
&&ddots & ddots\
&&&1&0}
$$
will be all $n$th roots of unity, i.e. $e^{2 pi i k/n}$ for $i = 0,dots,n-1$. These are equally spaced over the unit circle.
answered Jan 10 at 21:18
OmnomnomnomOmnomnomnom
127k790178
$endgroup$
The eigenvalues of the permutation matrix
$$
M = pmatrix{0&cdots & 0 & 0 & 1\
1&0\
&1&0\
&&ddots & ddots\
&&&1&0}
$$
will be all $n$th roots of unity, i.e. $e^{2 pi i k/n}$ for $i = 0,dots,n-1$. These are equally spaced over the unit circle.
answered Jan 10 at 21:18
OmnomnomnomOmnomnomnom
127k790178
answered Jan 10 at 21:18
OmnomnomnomOmnomnomnom
127k790178
answered Jan 10 at 21:18
OmnomnomnomOmnomnomnom
127k790178
answered Jan 10 at 21:18
OmnomnomnomOmnomnomnom
127k790178
127k790178
add a comment |
add a comment |
$begingroup$
You could make the eigenvalues be the $n$-th roots of unity, which are "evenly spaced" around the unit circle in $mathbb C$. For instance, the matrix $$begin{bmatrix}1&0&0\0&omega&0\0&0&omega^2end{bmatrix}$$ works for $n=3$ where $omega=e^{2pi i/3}$. I leave it to you to generalize this to arbitrary $n$.
answered Jan 10 at 20:24
DaveDave
8,76711033
$endgroup$
$begingroup$
Thanks, but sorry I forget to mention, I want the matrices to have real entries.
$endgroup$
– dave2d
Jan 10 at 20:27
$begingroup$
Ah, well then yes my answer won't work. I'll leave it up just in case people are interested.
$endgroup$
– Dave
Jan 10 at 20:27
add a comment |
$begingroup$
You could make the eigenvalues be the $n$-th roots of unity, which are "evenly spaced" around the unit circle in $mathbb C$. For instance, the matrix $$begin{bmatrix}1&0&0\0&omega&0\0&0&omega^2end{bmatrix}$$ works for $n=3$ where $omega=e^{2pi i/3}$. I leave it to you to generalize this to arbitrary $n$.
answered Jan 10 at 20:24
DaveDave
8,76711033
$endgroup$
$begingroup$
Thanks, but sorry I forget to mention, I want the matrices to have real entries.
$endgroup$
– dave2d
Jan 10 at 20:27
$begingroup$
Ah, well then yes my answer won't work. I'll leave it up just in case people are interested.
$endgroup$
– Dave
Jan 10 at 20:27
add a comment |
$begingroup$
You could make the eigenvalues be the $n$-th roots of unity, which are "evenly spaced" around the unit circle in $mathbb C$. For instance, the matrix $$begin{bmatrix}1&0&0\0&omega&0\0&0&omega^2end{bmatrix}$$ works for $n=3$ where $omega=e^{2pi i/3}$. I leave it to you to generalize this to arbitrary $n$.
answered Jan 10 at 20:24
DaveDave
8,76711033
$endgroup$
You could make the eigenvalues be the $n$-th roots of unity, which are "evenly spaced" around the unit circle in $mathbb C$. For instance, the matrix $$begin{bmatrix}1&0&0\0&omega&0\0&0&omega^2end{bmatrix}$$ works for $n=3$ where $omega=e^{2pi i/3}$. I leave it to you to generalize this to arbitrary $n$.
answered Jan 10 at 20:24
DaveDave
8,76711033
answered Jan 10 at 20:24
DaveDave
8,76711033
answered Jan 10 at 20:24
DaveDave
8,76711033
answered Jan 10 at 20:24
DaveDave
8,76711033
8,76711033
$begingroup$
Thanks, but sorry I forget to mention, I want the matrices to have real entries.
$endgroup$
– dave2d
Jan 10 at 20:27
$begingroup$
Ah, well then yes my answer won't work. I'll leave it up just in case people are interested.
$endgroup$
– Dave
Jan 10 at 20:27
add a comment |
$begingroup$
Thanks, but sorry I forget to mention, I want the matrices to have real entries.
$endgroup$
– dave2d
Jan 10 at 20:27
$begingroup$
Ah, well then yes my answer won't work. I'll leave it up just in case people are interested.
$endgroup$
– Dave
Jan 10 at 20:27
$begingroup$
Thanks, but sorry I forget to mention, I want the matrices to have real entries.
$endgroup$
– dave2d
Jan 10 at 20:27
$begingroup$
Thanks, but sorry I forget to mention, I want the matrices to have real entries.
$endgroup$
– dave2d
Jan 10 at 20:27
$begingroup$
Ah, well then yes my answer won't work. I'll leave it up just in case people are interested.
$endgroup$
– Dave
Jan 10 at 20:27
$begingroup$
Ah, well then yes my answer won't work. I'll leave it up just in case people are interested.
$endgroup$
– Dave
Jan 10 at 20:27
add a comment |
$begingroup$
Here is the easiest way to do it with real entries (I think). Given a real number $theta$, the rotation matrix
$$
begin{bmatrix}cos theta&-sintheta\sintheta&costhetaend{bmatrix}
$$
has two complex eigenvalues, which have norm $1$ and argument $pmtheta$. Using this, we can build the matrix we're after.
Take the complex unit circle, and distribute $n$ points along it so that
- They are evenly spaced
- The configuration is symmetric with respect to mirroring across the $x$-axis (i.e. complex conjugation)
Pair up the points in complex conjugate pairs, and for each pair, construct the corresponding rotation matrix as shown above (you can choose the sign of $theta$ freely). Then take all those rotation matrices, and put them along the diagonal of an $ntimes n$ matrix (with zeroes in all other entries).
If $n$ is odd, one of the points is $1$ (or $-1$), and doesn't have a pair mate. Just put a $1$ (or $-1$) on the diagonal of the $ntimes n$ matrix instead of a rotation matrix when you get to that point.
answered Jan 10 at 20:24
ArthurArthur
112k7108192
$endgroup$
add a comment |
$begingroup$
Here is the easiest way to do it with real entries (I think). Given a real number $theta$, the rotation matrix
$$
begin{bmatrix}cos theta&-sintheta\sintheta&costhetaend{bmatrix}
$$
has two complex eigenvalues, which have norm $1$ and argument $pmtheta$. Using this, we can build the matrix we're after.
Take the complex unit circle, and distribute $n$ points along it so that
- They are evenly spaced
- The configuration is symmetric with respect to mirroring across the $x$-axis (i.e. complex conjugation)
Pair up the points in complex conjugate pairs, and for each pair, construct the corresponding rotation matrix as shown above (you can choose the sign of $theta$ freely). Then take all those rotation matrices, and put them along the diagonal of an $ntimes n$ matrix (with zeroes in all other entries).
If $n$ is odd, one of the points is $1$ (or $-1$), and doesn't have a pair mate. Just put a $1$ (or $-1$) on the diagonal of the $ntimes n$ matrix instead of a rotation matrix when you get to that point.
answered Jan 10 at 20:24
ArthurArthur
112k7108192
$endgroup$
add a comment |
$begingroup$
Here is the easiest way to do it with real entries (I think). Given a real number $theta$, the rotation matrix
$$
begin{bmatrix}cos theta&-sintheta\sintheta&costhetaend{bmatrix}
$$
has two complex eigenvalues, which have norm $1$ and argument $pmtheta$. Using this, we can build the matrix we're after.
Take the complex unit circle, and distribute $n$ points along it so that
- They are evenly spaced
- The configuration is symmetric with respect to mirroring across the $x$-axis (i.e. complex conjugation)
Pair up the points in complex conjugate pairs, and for each pair, construct the corresponding rotation matrix as shown above (you can choose the sign of $theta$ freely). Then take all those rotation matrices, and put them along the diagonal of an $ntimes n$ matrix (with zeroes in all other entries).
If $n$ is odd, one of the points is $1$ (or $-1$), and doesn't have a pair mate. Just put a $1$ (or $-1$) on the diagonal of the $ntimes n$ matrix instead of a rotation matrix when you get to that point.
answered Jan 10 at 20:24
ArthurArthur
112k7108192
$endgroup$
Here is the easiest way to do it with real entries (I think). Given a real number $theta$, the rotation matrix
$$
begin{bmatrix}cos theta&-sintheta\sintheta&costhetaend{bmatrix}
$$
has two complex eigenvalues, which have norm $1$ and argument $pmtheta$. Using this, we can build the matrix we're after.
Take the complex unit circle, and distribute $n$ points along it so that
- They are evenly spaced
- The configuration is symmetric with respect to mirroring across the $x$-axis (i.e. complex conjugation)
Pair up the points in complex conjugate pairs, and for each pair, construct the corresponding rotation matrix as shown above (you can choose the sign of $theta$ freely). Then take all those rotation matrices, and put them along the diagonal of an $ntimes n$ matrix (with zeroes in all other entries).
If $n$ is odd, one of the points is $1$ (or $-1$), and doesn't have a pair mate. Just put a $1$ (or $-1$) on the diagonal of the $ntimes n$ matrix instead of a rotation matrix when you get to that point.
answered Jan 10 at 20:24
ArthurArthur
112k7108192
answered Jan 10 at 20:24
ArthurArthur
112k7108192
answered Jan 10 at 20:24
ArthurArthur
112k7108192
answered Jan 10 at 20:24
ArthurArthur
112k7108192
112k7108192
add a comment |
add a comment |
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