Are these square matrices always diagonalisable? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30 pm US/Eastern)Conditions for diagonalizability of $ntimes n$ anti-diagonal matricesFinding eigenvalues/vectors of a matrix and proving it is not diagonalisable.Finding if a linear transformation is diagonalisableThe diagonalisation of the two matricesIf $A in K^n times n$ is diagonalisable, the dimension of the subspace of its commuting matrices is $geq n$ nCalculating the eigenvalues of a diagonalisable linear operator $L$.How to find eigenvalues for mod 2 field?Find for which real parameter a matrix is diagonalisableSquare Roots of a Matrix: Diagonalisable Solutions.eigenvalues and eigenvectors of Diagonalisable matrices

Are these square matrices always diagonalisable?

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Are these square matrices always diagonalisable?



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30 pm US/Eastern)Conditions for diagonalizability of $ntimes n$ anti-diagonal matricesFinding eigenvalues/vectors of a matrix and proving it is not diagonalisable.Finding if a linear transformation is diagonalisableThe diagonalisation of the two matricesIf $A in K^n times n$ is diagonalisable, the dimension of the subspace of its commuting matrices is $geq n$ nCalculating the eigenvalues of a diagonalisable linear operator $L$.How to find eigenvalues for mod 2 field?Find for which real parameter a matrix is diagonalisableSquare Roots of a Matrix: Diagonalisable Solutions.eigenvalues and eigenvectors of Diagonalisable matrices










3












$begingroup$


When trying to solve a physics problem on decoupling a system of ODEs, I found myself needing to address the following problem:




Let $A_nin M_n(mathbb R)$ be the matrix with all $1$s above its main diagonal, all $-1$s below its diagonal, and $0$s everywhere else. Is $A_n$ always diagonalisable? If so, what is its diagonalisation (equivalently: what are its eigenvalues and corresponding eigenvectors)?




For example,
$$A_3=beginbmatrix0&1&0\-1&0&1\0&-1&0endbmatrix,quad A_5beginbmatrix0&1&0&0&0\-1&0&1&0&0\0&-1&0&1&0\0&0&-1&0&1\0&0&0&-1&0endbmatrix.$$




Assuming my code is correct, Mathematica has been able to verify that $A_n$ is always diagonalisable up to $n=1000$. If we use $chi_n(t)inmathbb Z[t]$ to denote the characteristic polynomial of $A_n$, a straightforward evaluation also shows that
$$chi_n(t)=-tchi_n-1(t)+chi_n-2(t)tag1$$
for all $ngeq4$. Furthermore, note that $A_n=-A_n^t$ so that, in the case where the dimension is even,
$$det(A_2n-lambda I)=det(A_2n^t-lambda I)=det(-A_2n-lambda I)=det(A_2n+lambda I).$$
This implies that whenever $lambda$ is an eigenvalue of $A_2n$, so is $-lambda$. In other words, $chi_2n(t)$ is always of the form $(t^2-lambda _1^2)(t^2-lambda_2^2)dotsm(t^2-lambda_n^2)$ for some $lambda_i$.



And this is where I am stuck. In order for $A_n$ to be diagonalisable, we must have that all the eigenvalues are distinct, but trying to use the recurrence $(1)$ and strong induction, or trying to use the formula for the even case have not helped at all. It seems like the most probable line of attack would be to somehow show that
$$chi_2n'(t)=2tsum_k=1^nfracchi_2n(t)t^2-lambda_k^2$$
never shares a common zero with $chi_2n$ (which would resolve the even case), though I don't see how to make this work.




Note: I do not have any clue how to actually find the eigenvalues/eigenvectors even in the case where the $A_n$ are diagonalisable. As such even if someone cannot answer the second part of the question, but can prove that the $A_n$ are diagonalisable, I would appreciate that as an answer as well. Above I tried to look at the special case where the dimension is even, though of course the proof for all odd and even $n$ is more valuable. Even if this is not possible, for my purposes I just need an unbounded subset $Ssubseteqmathbb Z$ for which the conclusion is proven for $nin S$, so any such approach is welcome too.



Thank you in advance!










share|cite|improve this question









$endgroup$











  • $begingroup$
    All eigenvalues distinct is a sufficient but not necessary condition for a matrix to be diagonalizable.
    $endgroup$
    – Henning Makholm
    1 hour ago















3












$begingroup$


When trying to solve a physics problem on decoupling a system of ODEs, I found myself needing to address the following problem:




Let $A_nin M_n(mathbb R)$ be the matrix with all $1$s above its main diagonal, all $-1$s below its diagonal, and $0$s everywhere else. Is $A_n$ always diagonalisable? If so, what is its diagonalisation (equivalently: what are its eigenvalues and corresponding eigenvectors)?




For example,
$$A_3=beginbmatrix0&1&0\-1&0&1\0&-1&0endbmatrix,quad A_5beginbmatrix0&1&0&0&0\-1&0&1&0&0\0&-1&0&1&0\0&0&-1&0&1\0&0&0&-1&0endbmatrix.$$




Assuming my code is correct, Mathematica has been able to verify that $A_n$ is always diagonalisable up to $n=1000$. If we use $chi_n(t)inmathbb Z[t]$ to denote the characteristic polynomial of $A_n$, a straightforward evaluation also shows that
$$chi_n(t)=-tchi_n-1(t)+chi_n-2(t)tag1$$
for all $ngeq4$. Furthermore, note that $A_n=-A_n^t$ so that, in the case where the dimension is even,
$$det(A_2n-lambda I)=det(A_2n^t-lambda I)=det(-A_2n-lambda I)=det(A_2n+lambda I).$$
This implies that whenever $lambda$ is an eigenvalue of $A_2n$, so is $-lambda$. In other words, $chi_2n(t)$ is always of the form $(t^2-lambda _1^2)(t^2-lambda_2^2)dotsm(t^2-lambda_n^2)$ for some $lambda_i$.



And this is where I am stuck. In order for $A_n$ to be diagonalisable, we must have that all the eigenvalues are distinct, but trying to use the recurrence $(1)$ and strong induction, or trying to use the formula for the even case have not helped at all. It seems like the most probable line of attack would be to somehow show that
$$chi_2n'(t)=2tsum_k=1^nfracchi_2n(t)t^2-lambda_k^2$$
never shares a common zero with $chi_2n$ (which would resolve the even case), though I don't see how to make this work.




Note: I do not have any clue how to actually find the eigenvalues/eigenvectors even in the case where the $A_n$ are diagonalisable. As such even if someone cannot answer the second part of the question, but can prove that the $A_n$ are diagonalisable, I would appreciate that as an answer as well. Above I tried to look at the special case where the dimension is even, though of course the proof for all odd and even $n$ is more valuable. Even if this is not possible, for my purposes I just need an unbounded subset $Ssubseteqmathbb Z$ for which the conclusion is proven for $nin S$, so any such approach is welcome too.



Thank you in advance!










share|cite|improve this question









$endgroup$











  • $begingroup$
    All eigenvalues distinct is a sufficient but not necessary condition for a matrix to be diagonalizable.
    $endgroup$
    – Henning Makholm
    1 hour ago













3












3








3





$begingroup$


When trying to solve a physics problem on decoupling a system of ODEs, I found myself needing to address the following problem:




Let $A_nin M_n(mathbb R)$ be the matrix with all $1$s above its main diagonal, all $-1$s below its diagonal, and $0$s everywhere else. Is $A_n$ always diagonalisable? If so, what is its diagonalisation (equivalently: what are its eigenvalues and corresponding eigenvectors)?




For example,
$$A_3=beginbmatrix0&1&0\-1&0&1\0&-1&0endbmatrix,quad A_5beginbmatrix0&1&0&0&0\-1&0&1&0&0\0&-1&0&1&0\0&0&-1&0&1\0&0&0&-1&0endbmatrix.$$




Assuming my code is correct, Mathematica has been able to verify that $A_n$ is always diagonalisable up to $n=1000$. If we use $chi_n(t)inmathbb Z[t]$ to denote the characteristic polynomial of $A_n$, a straightforward evaluation also shows that
$$chi_n(t)=-tchi_n-1(t)+chi_n-2(t)tag1$$
for all $ngeq4$. Furthermore, note that $A_n=-A_n^t$ so that, in the case where the dimension is even,
$$det(A_2n-lambda I)=det(A_2n^t-lambda I)=det(-A_2n-lambda I)=det(A_2n+lambda I).$$
This implies that whenever $lambda$ is an eigenvalue of $A_2n$, so is $-lambda$. In other words, $chi_2n(t)$ is always of the form $(t^2-lambda _1^2)(t^2-lambda_2^2)dotsm(t^2-lambda_n^2)$ for some $lambda_i$.



And this is where I am stuck. In order for $A_n$ to be diagonalisable, we must have that all the eigenvalues are distinct, but trying to use the recurrence $(1)$ and strong induction, or trying to use the formula for the even case have not helped at all. It seems like the most probable line of attack would be to somehow show that
$$chi_2n'(t)=2tsum_k=1^nfracchi_2n(t)t^2-lambda_k^2$$
never shares a common zero with $chi_2n$ (which would resolve the even case), though I don't see how to make this work.




Note: I do not have any clue how to actually find the eigenvalues/eigenvectors even in the case where the $A_n$ are diagonalisable. As such even if someone cannot answer the second part of the question, but can prove that the $A_n$ are diagonalisable, I would appreciate that as an answer as well. Above I tried to look at the special case where the dimension is even, though of course the proof for all odd and even $n$ is more valuable. Even if this is not possible, for my purposes I just need an unbounded subset $Ssubseteqmathbb Z$ for which the conclusion is proven for $nin S$, so any such approach is welcome too.



Thank you in advance!










share|cite|improve this question









$endgroup$




When trying to solve a physics problem on decoupling a system of ODEs, I found myself needing to address the following problem:




Let $A_nin M_n(mathbb R)$ be the matrix with all $1$s above its main diagonal, all $-1$s below its diagonal, and $0$s everywhere else. Is $A_n$ always diagonalisable? If so, what is its diagonalisation (equivalently: what are its eigenvalues and corresponding eigenvectors)?




For example,
$$A_3=beginbmatrix0&1&0\-1&0&1\0&-1&0endbmatrix,quad A_5beginbmatrix0&1&0&0&0\-1&0&1&0&0\0&-1&0&1&0\0&0&-1&0&1\0&0&0&-1&0endbmatrix.$$




Assuming my code is correct, Mathematica has been able to verify that $A_n$ is always diagonalisable up to $n=1000$. If we use $chi_n(t)inmathbb Z[t]$ to denote the characteristic polynomial of $A_n$, a straightforward evaluation also shows that
$$chi_n(t)=-tchi_n-1(t)+chi_n-2(t)tag1$$
for all $ngeq4$. Furthermore, note that $A_n=-A_n^t$ so that, in the case where the dimension is even,
$$det(A_2n-lambda I)=det(A_2n^t-lambda I)=det(-A_2n-lambda I)=det(A_2n+lambda I).$$
This implies that whenever $lambda$ is an eigenvalue of $A_2n$, so is $-lambda$. In other words, $chi_2n(t)$ is always of the form $(t^2-lambda _1^2)(t^2-lambda_2^2)dotsm(t^2-lambda_n^2)$ for some $lambda_i$.



And this is where I am stuck. In order for $A_n$ to be diagonalisable, we must have that all the eigenvalues are distinct, but trying to use the recurrence $(1)$ and strong induction, or trying to use the formula for the even case have not helped at all. It seems like the most probable line of attack would be to somehow show that
$$chi_2n'(t)=2tsum_k=1^nfracchi_2n(t)t^2-lambda_k^2$$
never shares a common zero with $chi_2n$ (which would resolve the even case), though I don't see how to make this work.




Note: I do not have any clue how to actually find the eigenvalues/eigenvectors even in the case where the $A_n$ are diagonalisable. As such even if someone cannot answer the second part of the question, but can prove that the $A_n$ are diagonalisable, I would appreciate that as an answer as well. Above I tried to look at the special case where the dimension is even, though of course the proof for all odd and even $n$ is more valuable. Even if this is not possible, for my purposes I just need an unbounded subset $Ssubseteqmathbb Z$ for which the conclusion is proven for $nin S$, so any such approach is welcome too.



Thank you in advance!







linear-algebra eigenvalues-eigenvectors determinant diagonalization






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 3 hours ago









YiFanYiFan

5,6252829




5,6252829











  • $begingroup$
    All eigenvalues distinct is a sufficient but not necessary condition for a matrix to be diagonalizable.
    $endgroup$
    – Henning Makholm
    1 hour ago
















  • $begingroup$
    All eigenvalues distinct is a sufficient but not necessary condition for a matrix to be diagonalizable.
    $endgroup$
    – Henning Makholm
    1 hour ago















$begingroup$
All eigenvalues distinct is a sufficient but not necessary condition for a matrix to be diagonalizable.
$endgroup$
– Henning Makholm
1 hour ago




$begingroup$
All eigenvalues distinct is a sufficient but not necessary condition for a matrix to be diagonalizable.
$endgroup$
– Henning Makholm
1 hour ago










3 Answers
3






active

oldest

votes


















5












$begingroup$

The matrix $A_n$ is a tridiagonal Toeplitz matrix with diagonal entries $delta = 0$ and off-diagonal entries $tau = 1$ and $sigma = -1$. Hence, we can use the formula in this paper to show that the eigenvalues are $$lambda_k = 2icosleft(dfrackpin+1right),$$ for $k = 1,ldots,n$, and the corresponding eigenvectors $v_1,ldots,v_n$ have entries $$v_k[m] = i^msinleft(dfracmkpin+1right).$$






share|cite|improve this answer











$endgroup$




















    5












    $begingroup$

    Using that your matrices are skew symmetric, you get that these matrices are diagonalizable. See the section spectral theory on this Wikipedia article.






    share|cite|improve this answer








    New contributor




    gcousin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






    $endgroup$












    • $begingroup$
      Thank you very much for the quick response! I hope you don't mind that I accepted JimmyK4542's answer, since it also gives explicitly the eigenvectors and eigenvalues.
      $endgroup$
      – YiFan
      2 hours ago


















    4












    $begingroup$

    All those matrices are anti-symmetric and therefore they are normal matrices. And every normal matrix is diagonalizable over $mathbb C$, by the spectral theorem.






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      Thank you very much for the quick response! I hope you don't mind that I accepted JimmyK4542's answer, since it also gives explicitly the eigenvectors and eigenvalues.
      $endgroup$
      – YiFan
      2 hours ago






    • 1




      $begingroup$
      I would have been surprised if you had not accepted that answer, since it provides more information than mine.
      $endgroup$
      – José Carlos Santos
      2 hours ago











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    3 Answers
    3






    active

    oldest

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    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5












    $begingroup$

    The matrix $A_n$ is a tridiagonal Toeplitz matrix with diagonal entries $delta = 0$ and off-diagonal entries $tau = 1$ and $sigma = -1$. Hence, we can use the formula in this paper to show that the eigenvalues are $$lambda_k = 2icosleft(dfrackpin+1right),$$ for $k = 1,ldots,n$, and the corresponding eigenvectors $v_1,ldots,v_n$ have entries $$v_k[m] = i^msinleft(dfracmkpin+1right).$$






    share|cite|improve this answer











    $endgroup$

















      5












      $begingroup$

      The matrix $A_n$ is a tridiagonal Toeplitz matrix with diagonal entries $delta = 0$ and off-diagonal entries $tau = 1$ and $sigma = -1$. Hence, we can use the formula in this paper to show that the eigenvalues are $$lambda_k = 2icosleft(dfrackpin+1right),$$ for $k = 1,ldots,n$, and the corresponding eigenvectors $v_1,ldots,v_n$ have entries $$v_k[m] = i^msinleft(dfracmkpin+1right).$$






      share|cite|improve this answer











      $endgroup$















        5












        5








        5





        $begingroup$

        The matrix $A_n$ is a tridiagonal Toeplitz matrix with diagonal entries $delta = 0$ and off-diagonal entries $tau = 1$ and $sigma = -1$. Hence, we can use the formula in this paper to show that the eigenvalues are $$lambda_k = 2icosleft(dfrackpin+1right),$$ for $k = 1,ldots,n$, and the corresponding eigenvectors $v_1,ldots,v_n$ have entries $$v_k[m] = i^msinleft(dfracmkpin+1right).$$






        share|cite|improve this answer











        $endgroup$



        The matrix $A_n$ is a tridiagonal Toeplitz matrix with diagonal entries $delta = 0$ and off-diagonal entries $tau = 1$ and $sigma = -1$. Hence, we can use the formula in this paper to show that the eigenvalues are $$lambda_k = 2icosleft(dfrackpin+1right),$$ for $k = 1,ldots,n$, and the corresponding eigenvectors $v_1,ldots,v_n$ have entries $$v_k[m] = i^msinleft(dfracmkpin+1right).$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 2 hours ago

























        answered 3 hours ago









        JimmyK4542JimmyK4542

        41.5k246108




        41.5k246108





















            5












            $begingroup$

            Using that your matrices are skew symmetric, you get that these matrices are diagonalizable. See the section spectral theory on this Wikipedia article.






            share|cite|improve this answer








            New contributor




            gcousin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.






            $endgroup$












            • $begingroup$
              Thank you very much for the quick response! I hope you don't mind that I accepted JimmyK4542's answer, since it also gives explicitly the eigenvectors and eigenvalues.
              $endgroup$
              – YiFan
              2 hours ago















            5












            $begingroup$

            Using that your matrices are skew symmetric, you get that these matrices are diagonalizable. See the section spectral theory on this Wikipedia article.






            share|cite|improve this answer








            New contributor




            gcousin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.






            $endgroup$












            • $begingroup$
              Thank you very much for the quick response! I hope you don't mind that I accepted JimmyK4542's answer, since it also gives explicitly the eigenvectors and eigenvalues.
              $endgroup$
              – YiFan
              2 hours ago













            5












            5








            5





            $begingroup$

            Using that your matrices are skew symmetric, you get that these matrices are diagonalizable. See the section spectral theory on this Wikipedia article.






            share|cite|improve this answer








            New contributor




            gcousin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.






            $endgroup$



            Using that your matrices are skew symmetric, you get that these matrices are diagonalizable. See the section spectral theory on this Wikipedia article.







            share|cite|improve this answer








            New contributor




            gcousin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.









            share|cite|improve this answer



            share|cite|improve this answer






            New contributor




            gcousin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.









            answered 2 hours ago









            gcousingcousin

            1312




            1312




            New contributor




            gcousin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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            New contributor





            gcousin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.






            gcousin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.











            • $begingroup$
              Thank you very much for the quick response! I hope you don't mind that I accepted JimmyK4542's answer, since it also gives explicitly the eigenvectors and eigenvalues.
              $endgroup$
              – YiFan
              2 hours ago
















            • $begingroup$
              Thank you very much for the quick response! I hope you don't mind that I accepted JimmyK4542's answer, since it also gives explicitly the eigenvectors and eigenvalues.
              $endgroup$
              – YiFan
              2 hours ago















            $begingroup$
            Thank you very much for the quick response! I hope you don't mind that I accepted JimmyK4542's answer, since it also gives explicitly the eigenvectors and eigenvalues.
            $endgroup$
            – YiFan
            2 hours ago




            $begingroup$
            Thank you very much for the quick response! I hope you don't mind that I accepted JimmyK4542's answer, since it also gives explicitly the eigenvectors and eigenvalues.
            $endgroup$
            – YiFan
            2 hours ago











            4












            $begingroup$

            All those matrices are anti-symmetric and therefore they are normal matrices. And every normal matrix is diagonalizable over $mathbb C$, by the spectral theorem.






            share|cite|improve this answer









            $endgroup$












            • $begingroup$
              Thank you very much for the quick response! I hope you don't mind that I accepted JimmyK4542's answer, since it also gives explicitly the eigenvectors and eigenvalues.
              $endgroup$
              – YiFan
              2 hours ago






            • 1




              $begingroup$
              I would have been surprised if you had not accepted that answer, since it provides more information than mine.
              $endgroup$
              – José Carlos Santos
              2 hours ago















            4












            $begingroup$

            All those matrices are anti-symmetric and therefore they are normal matrices. And every normal matrix is diagonalizable over $mathbb C$, by the spectral theorem.






            share|cite|improve this answer









            $endgroup$












            • $begingroup$
              Thank you very much for the quick response! I hope you don't mind that I accepted JimmyK4542's answer, since it also gives explicitly the eigenvectors and eigenvalues.
              $endgroup$
              – YiFan
              2 hours ago






            • 1




              $begingroup$
              I would have been surprised if you had not accepted that answer, since it provides more information than mine.
              $endgroup$
              – José Carlos Santos
              2 hours ago













            4












            4








            4





            $begingroup$

            All those matrices are anti-symmetric and therefore they are normal matrices. And every normal matrix is diagonalizable over $mathbb C$, by the spectral theorem.






            share|cite|improve this answer









            $endgroup$



            All those matrices are anti-symmetric and therefore they are normal matrices. And every normal matrix is diagonalizable over $mathbb C$, by the spectral theorem.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 3 hours ago









            José Carlos SantosJosé Carlos Santos

            177k24138250




            177k24138250











            • $begingroup$
              Thank you very much for the quick response! I hope you don't mind that I accepted JimmyK4542's answer, since it also gives explicitly the eigenvectors and eigenvalues.
              $endgroup$
              – YiFan
              2 hours ago






            • 1




              $begingroup$
              I would have been surprised if you had not accepted that answer, since it provides more information than mine.
              $endgroup$
              – José Carlos Santos
              2 hours ago
















            • $begingroup$
              Thank you very much for the quick response! I hope you don't mind that I accepted JimmyK4542's answer, since it also gives explicitly the eigenvectors and eigenvalues.
              $endgroup$
              – YiFan
              2 hours ago






            • 1




              $begingroup$
              I would have been surprised if you had not accepted that answer, since it provides more information than mine.
              $endgroup$
              – José Carlos Santos
              2 hours ago















            $begingroup$
            Thank you very much for the quick response! I hope you don't mind that I accepted JimmyK4542's answer, since it also gives explicitly the eigenvectors and eigenvalues.
            $endgroup$
            – YiFan
            2 hours ago




            $begingroup$
            Thank you very much for the quick response! I hope you don't mind that I accepted JimmyK4542's answer, since it also gives explicitly the eigenvectors and eigenvalues.
            $endgroup$
            – YiFan
            2 hours ago




            1




            1




            $begingroup$
            I would have been surprised if you had not accepted that answer, since it provides more information than mine.
            $endgroup$
            – José Carlos Santos
            2 hours ago




            $begingroup$
            I would have been surprised if you had not accepted that answer, since it provides more information than mine.
            $endgroup$
            – José Carlos Santos
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