Can a matrix be not a multiple of identity, have repeated eigen values and still be diagonalizable?












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The question: Diagonalisability of 2×2 matrices with repeated eigenvalues suggests that if a matrix has all its eigen values distinct, it must be diagonalizable. However, any multiple of the identity matrix will have all of its eigen values the same and yet be diagonalizable. I conjecture that if a general $n times n$ matrix has some non-zero off diagonal elements, and has any multiplicity of eigen values, it will not be diagonalizable. I haven't been able to find a counterexample to this. Can it be proven (or disproven with a counter example).



EDIT: Sorry, I was actually looking for a stochastic matrix (rows must sum to one) with these properties. I'll add another - matrix has to be full rank (so no zero eigen values). If no one answers with such an example in the next few hours, I'll accept the current one.






linear-algebra matrices eigenvalues-eigenvectors diagonalization







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  • $begingroup$
    You're right, edited the question.
    $endgroup$
    – Rohit Pandey
    Jan 17 at 2:44
















-1












$begingroup$


The question: Diagonalisability of 2×2 matrices with repeated eigenvalues suggests that if a matrix has all its eigen values distinct, it must be diagonalizable. However, any multiple of the identity matrix will have all of its eigen values the same and yet be diagonalizable. I conjecture that if a general $n times n$ matrix has some non-zero off diagonal elements, and has any multiplicity of eigen values, it will not be diagonalizable. I haven't been able to find a counterexample to this. Can it be proven (or disproven with a counter example).



EDIT: Sorry, I was actually looking for a stochastic matrix (rows must sum to one) with these properties. I'll add another - matrix has to be full rank (so no zero eigen values). If no one answers with such an example in the next few hours, I'll accept the current one.






linear-algebra matrices eigenvalues-eigenvectors diagonalization







share|cite|improve this question











$endgroup$












  • $begingroup$
    You're right, edited the question.
    $endgroup$
    – Rohit Pandey
    Jan 17 at 2:44














-1












-1








-1


1



$begingroup$


The question: Diagonalisability of 2×2 matrices with repeated eigenvalues suggests that if a matrix has all its eigen values distinct, it must be diagonalizable. However, any multiple of the identity matrix will have all of its eigen values the same and yet be diagonalizable. I conjecture that if a general $n times n$ matrix has some non-zero off diagonal elements, and has any multiplicity of eigen values, it will not be diagonalizable. I haven't been able to find a counterexample to this. Can it be proven (or disproven with a counter example).



EDIT: Sorry, I was actually looking for a stochastic matrix (rows must sum to one) with these properties. I'll add another - matrix has to be full rank (so no zero eigen values). If no one answers with such an example in the next few hours, I'll accept the current one.






linear-algebra matrices eigenvalues-eigenvectors diagonalization







share|cite|improve this question











$endgroup$




The question: Diagonalisability of 2×2 matrices with repeated eigenvalues suggests that if a matrix has all its eigen values distinct, it must be diagonalizable. However, any multiple of the identity matrix will have all of its eigen values the same and yet be diagonalizable. I conjecture that if a general $n times n$ matrix has some non-zero off diagonal elements, and has any multiplicity of eigen values, it will not be diagonalizable. I haven't been able to find a counterexample to this. Can it be proven (or disproven with a counter example).



EDIT: Sorry, I was actually looking for a stochastic matrix (rows must sum to one) with these properties. I'll add another - matrix has to be full rank (so no zero eigen values). If no one answers with such an example in the next few hours, I'll accept the current one.






linear-algebra matrices eigenvalues-eigenvectors diagonalization




linear-algebra matrices eigenvalues-eigenvectors diagonalization






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share|cite|improve this question













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edited Jan 17 at 2:44







Rohit Pandey

















asked Jan 16 at 0:18









Rohit PandeyRohit Pandey

1,6581024




1,6581024












  • $begingroup$
    You're right, edited the question.
    $endgroup$
    – Rohit Pandey
    Jan 17 at 2:44


















  • $begingroup$
    You're right, edited the question.
    $endgroup$
    – Rohit Pandey
    Jan 17 at 2:44
















$begingroup$
You're right, edited the question.
$endgroup$
– Rohit Pandey
Jan 17 at 2:44




$begingroup$
You're right, edited the question.
$endgroup$
– Rohit Pandey
Jan 17 at 2:44










3 Answers
3






active

oldest

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1












$begingroup$

Let $P$ be any full rank stochastic matrix (i.e. rows sum to one) which is not a multiple of the identity matrix. Then the following matrix will also not be a multiple of the identity matrix and have repeated eigenvalues $$
Q = begin{pmatrix}
P & 0 & 0\
0 & P & 0\
0 & 0 & P\
end{pmatrix}
$$
Here, $Q$ has the matrix "$P$" replicated on its diagonal and has zeros outside the $P$-blocks.



Notice that each $P$-block describes a distinct subcomponent of the larger markov chain described by $Q$. And since the eigenvalues of each $P$-block are the same, $Q$ has distinct eigenvectors with the same eigenvalue. $Q$ is also stochastic.






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$endgroup$













  • $begingroup$
    Sorry for the edit, but I'm looking for a full rank, stochastic matrix with multiplicity in some of its eigen values and not a multiple of the identity that is diagonalizable.
    $endgroup$
    – Rohit Pandey
    Jan 16 at 0:55



















1












$begingroup$

Try for example, $$ pmatrix{1 & 1 & 1cr 1 & 1 & 1cr 1 & 1 & 1cr}$$



In fact any real $n times n$ symmetric matrix is diagonalizable, but these can have repeated eigenvalues.



EDIT: A full rank stochastic example is
$$ pmatrix{1/2 & 1/4 & 1/4cr
1/4 & 1/2 & 1/4cr
1/4 & 1/4 & 1/2cr} $$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Sorry, is there an example of a full rank, stochastic matrix like this you can think of (played around a little with symmetric matrices but couldn't construct one), if not, I'll accept this answer in a while.
    $endgroup$
    – Rohit Pandey
    Jan 16 at 0:49



















1












$begingroup$

A simple counterexample: $$begin{bmatrix}1&0&0\0&0&1\0&1&0end{bmatrix}.$$






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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

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    1












    $begingroup$

    Let $P$ be any full rank stochastic matrix (i.e. rows sum to one) which is not a multiple of the identity matrix. Then the following matrix will also not be a multiple of the identity matrix and have repeated eigenvalues $$
    Q = begin{pmatrix}
    P & 0 & 0\
    0 & P & 0\
    0 & 0 & P\
    end{pmatrix}
    $$
    Here, $Q$ has the matrix "$P$" replicated on its diagonal and has zeros outside the $P$-blocks.



    Notice that each $P$-block describes a distinct subcomponent of the larger markov chain described by $Q$. And since the eigenvalues of each $P$-block are the same, $Q$ has distinct eigenvectors with the same eigenvalue. $Q$ is also stochastic.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Sorry for the edit, but I'm looking for a full rank, stochastic matrix with multiplicity in some of its eigen values and not a multiple of the identity that is diagonalizable.
      $endgroup$
      – Rohit Pandey
      Jan 16 at 0:55
















    1












    $begingroup$

    Let $P$ be any full rank stochastic matrix (i.e. rows sum to one) which is not a multiple of the identity matrix. Then the following matrix will also not be a multiple of the identity matrix and have repeated eigenvalues $$
    Q = begin{pmatrix}
    P & 0 & 0\
    0 & P & 0\
    0 & 0 & P\
    end{pmatrix}
    $$
    Here, $Q$ has the matrix "$P$" replicated on its diagonal and has zeros outside the $P$-blocks.



    Notice that each $P$-block describes a distinct subcomponent of the larger markov chain described by $Q$. And since the eigenvalues of each $P$-block are the same, $Q$ has distinct eigenvectors with the same eigenvalue. $Q$ is also stochastic.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Sorry for the edit, but I'm looking for a full rank, stochastic matrix with multiplicity in some of its eigen values and not a multiple of the identity that is diagonalizable.
      $endgroup$
      – Rohit Pandey
      Jan 16 at 0:55














    1












    1








    1





    $begingroup$

    Let $P$ be any full rank stochastic matrix (i.e. rows sum to one) which is not a multiple of the identity matrix. Then the following matrix will also not be a multiple of the identity matrix and have repeated eigenvalues $$
    Q = begin{pmatrix}
    P & 0 & 0\
    0 & P & 0\
    0 & 0 & P\
    end{pmatrix}
    $$
    Here, $Q$ has the matrix "$P$" replicated on its diagonal and has zeros outside the $P$-blocks.



    Notice that each $P$-block describes a distinct subcomponent of the larger markov chain described by $Q$. And since the eigenvalues of each $P$-block are the same, $Q$ has distinct eigenvectors with the same eigenvalue. $Q$ is also stochastic.






    share|cite|improve this answer











    $endgroup$



    Let $P$ be any full rank stochastic matrix (i.e. rows sum to one) which is not a multiple of the identity matrix. Then the following matrix will also not be a multiple of the identity matrix and have repeated eigenvalues $$
    Q = begin{pmatrix}
    P & 0 & 0\
    0 & P & 0\
    0 & 0 & P\
    end{pmatrix}
    $$
    Here, $Q$ has the matrix "$P$" replicated on its diagonal and has zeros outside the $P$-blocks.



    Notice that each $P$-block describes a distinct subcomponent of the larger markov chain described by $Q$. And since the eigenvalues of each $P$-block are the same, $Q$ has distinct eigenvectors with the same eigenvalue. $Q$ is also stochastic.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 16 at 1:03

























    answered Jan 16 at 0:54









    tedted

    579312




    579312












    • $begingroup$
      Sorry for the edit, but I'm looking for a full rank, stochastic matrix with multiplicity in some of its eigen values and not a multiple of the identity that is diagonalizable.
      $endgroup$
      – Rohit Pandey
      Jan 16 at 0:55


















    • $begingroup$
      Sorry for the edit, but I'm looking for a full rank, stochastic matrix with multiplicity in some of its eigen values and not a multiple of the identity that is diagonalizable.
      $endgroup$
      – Rohit Pandey
      Jan 16 at 0:55
















    $begingroup$
    Sorry for the edit, but I'm looking for a full rank, stochastic matrix with multiplicity in some of its eigen values and not a multiple of the identity that is diagonalizable.
    $endgroup$
    – Rohit Pandey
    Jan 16 at 0:55




    $begingroup$
    Sorry for the edit, but I'm looking for a full rank, stochastic matrix with multiplicity in some of its eigen values and not a multiple of the identity that is diagonalizable.
    $endgroup$
    – Rohit Pandey
    Jan 16 at 0:55











    1












    $begingroup$

    Try for example, $$ pmatrix{1 & 1 & 1cr 1 & 1 & 1cr 1 & 1 & 1cr}$$



    In fact any real $n times n$ symmetric matrix is diagonalizable, but these can have repeated eigenvalues.



    EDIT: A full rank stochastic example is
    $$ pmatrix{1/2 & 1/4 & 1/4cr
    1/4 & 1/2 & 1/4cr
    1/4 & 1/4 & 1/2cr} $$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Sorry, is there an example of a full rank, stochastic matrix like this you can think of (played around a little with symmetric matrices but couldn't construct one), if not, I'll accept this answer in a while.
      $endgroup$
      – Rohit Pandey
      Jan 16 at 0:49
















    1












    $begingroup$

    Try for example, $$ pmatrix{1 & 1 & 1cr 1 & 1 & 1cr 1 & 1 & 1cr}$$



    In fact any real $n times n$ symmetric matrix is diagonalizable, but these can have repeated eigenvalues.



    EDIT: A full rank stochastic example is
    $$ pmatrix{1/2 & 1/4 & 1/4cr
    1/4 & 1/2 & 1/4cr
    1/4 & 1/4 & 1/2cr} $$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Sorry, is there an example of a full rank, stochastic matrix like this you can think of (played around a little with symmetric matrices but couldn't construct one), if not, I'll accept this answer in a while.
      $endgroup$
      – Rohit Pandey
      Jan 16 at 0:49














    1












    1








    1





    $begingroup$

    Try for example, $$ pmatrix{1 & 1 & 1cr 1 & 1 & 1cr 1 & 1 & 1cr}$$



    In fact any real $n times n$ symmetric matrix is diagonalizable, but these can have repeated eigenvalues.



    EDIT: A full rank stochastic example is
    $$ pmatrix{1/2 & 1/4 & 1/4cr
    1/4 & 1/2 & 1/4cr
    1/4 & 1/4 & 1/2cr} $$






    share|cite|improve this answer











    $endgroup$



    Try for example, $$ pmatrix{1 & 1 & 1cr 1 & 1 & 1cr 1 & 1 & 1cr}$$



    In fact any real $n times n$ symmetric matrix is diagonalizable, but these can have repeated eigenvalues.



    EDIT: A full rank stochastic example is
    $$ pmatrix{1/2 & 1/4 & 1/4cr
    1/4 & 1/2 & 1/4cr
    1/4 & 1/4 & 1/2cr} $$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 16 at 1:14

























    answered Jan 16 at 0:24









    Robert IsraelRobert Israel

    330k23219473




    330k23219473












    • $begingroup$
      Sorry, is there an example of a full rank, stochastic matrix like this you can think of (played around a little with symmetric matrices but couldn't construct one), if not, I'll accept this answer in a while.
      $endgroup$
      – Rohit Pandey
      Jan 16 at 0:49


















    • $begingroup$
      Sorry, is there an example of a full rank, stochastic matrix like this you can think of (played around a little with symmetric matrices but couldn't construct one), if not, I'll accept this answer in a while.
      $endgroup$
      – Rohit Pandey
      Jan 16 at 0:49
















    $begingroup$
    Sorry, is there an example of a full rank, stochastic matrix like this you can think of (played around a little with symmetric matrices but couldn't construct one), if not, I'll accept this answer in a while.
    $endgroup$
    – Rohit Pandey
    Jan 16 at 0:49




    $begingroup$
    Sorry, is there an example of a full rank, stochastic matrix like this you can think of (played around a little with symmetric matrices but couldn't construct one), if not, I'll accept this answer in a while.
    $endgroup$
    – Rohit Pandey
    Jan 16 at 0:49











    1












    $begingroup$

    A simple counterexample: $$begin{bmatrix}1&0&0\0&0&1\0&1&0end{bmatrix}.$$






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      A simple counterexample: $$begin{bmatrix}1&0&0\0&0&1\0&1&0end{bmatrix}.$$






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        A simple counterexample: $$begin{bmatrix}1&0&0\0&0&1\0&1&0end{bmatrix}.$$






        share|cite|improve this answer











        $endgroup$



        A simple counterexample: $$begin{bmatrix}1&0&0\0&0&1\0&1&0end{bmatrix}.$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 17 at 1:22

























        answered Jan 16 at 0:43









        amdamd

        31.4k21052




        31.4k21052






























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