Orthogonal matrices with eigenvalues equally spaced over unit circle?












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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







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    1












    $begingroup$


    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







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    $endgroup$















      1












      1








      1





      $begingroup$


      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







      share|cite|improve this question











      $endgroup$




      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






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      share|cite|improve this question













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      share|cite|improve this question








      edited Jan 10 at 20:26







      dave2d

















      asked Jan 10 at 20:13









      dave2ddave2d

      83




      83






















          3 Answers
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          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            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 $n$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              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



















            0












            $begingroup$

            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.






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              3 Answers
              3






              active

              oldest

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              3 Answers
              3






              active

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              active

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              active

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              2












              $begingroup$

              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.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                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.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  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.






                  share|cite|improve this answer









                  $endgroup$



                  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.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 10 at 21:18









                  OmnomnomnomOmnomnomnom

                  127k790178




                  127k790178























                      0












                      $begingroup$

                      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 $n$.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        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
















                      0












                      $begingroup$

                      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 $n$.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        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














                      0












                      0








                      0





                      $begingroup$

                      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 $n$.






                      share|cite|improve this answer









                      $endgroup$



                      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 $n$.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      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


















                      • $begingroup$
                        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











                      0












                      $begingroup$

                      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.






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        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.






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          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.






                          share|cite|improve this answer









                          $endgroup$



                          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.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 10 at 20:24









                          ArthurArthur

                          112k7108192




                          112k7108192






























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