Does every (complex) $ntimes n$ matrix have n linearly independent eigenvectors?












1












$begingroup$


Every $ntimes n$ matrix will produce an $n$-th order characteristic equation. This polynomial will have $n$ distinct roots or less than $n$ distinct roots (with some repeated).



In the former case, each eigenvalue corresponds to a distinct eigenvector, so you can form an eigenbasis of $n$ linearly independent eigenvectors.



In the latter case, I can see two possibilities:




  1. There are still $n$ linearly independent eigenvectors, but some share the same eigenvalue, so a $k$-fold root of the characteristic equation (an eigenvalue repeated $k$ times) will correspond to $k$ linearly independent eigenvectors.


  2. There are less than $n$ linearly independent eigenvectors, so a $k$-fold root may correspond to less than $k$ linearly independent eigenvectors.



Is one of these two possibilities always true, for all complex $ntimes n$ matrices, or could either possibility be true depending on the matrix in question?



Is there a proof of this, or perhaps a counter example?






linear-algebra matrices eigenvalues-eigenvectors







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$endgroup$












  • $begingroup$
    $begin{bmatrix}1&1\0&1end{bmatrix}$
    $endgroup$
    – David C. Ullrich
    Dec 24 '18 at 17:23


















1












$begingroup$


Every $ntimes n$ matrix will produce an $n$-th order characteristic equation. This polynomial will have $n$ distinct roots or less than $n$ distinct roots (with some repeated).



In the former case, each eigenvalue corresponds to a distinct eigenvector, so you can form an eigenbasis of $n$ linearly independent eigenvectors.



In the latter case, I can see two possibilities:




  1. There are still $n$ linearly independent eigenvectors, but some share the same eigenvalue, so a $k$-fold root of the characteristic equation (an eigenvalue repeated $k$ times) will correspond to $k$ linearly independent eigenvectors.


  2. There are less than $n$ linearly independent eigenvectors, so a $k$-fold root may correspond to less than $k$ linearly independent eigenvectors.



Is one of these two possibilities always true, for all complex $ntimes n$ matrices, or could either possibility be true depending on the matrix in question?



Is there a proof of this, or perhaps a counter example?






linear-algebra matrices eigenvalues-eigenvectors







share|cite|improve this question











$endgroup$












  • $begingroup$
    $begin{bmatrix}1&1\0&1end{bmatrix}$
    $endgroup$
    – David C. Ullrich
    Dec 24 '18 at 17:23
















1












1








1





$begingroup$


Every $ntimes n$ matrix will produce an $n$-th order characteristic equation. This polynomial will have $n$ distinct roots or less than $n$ distinct roots (with some repeated).



In the former case, each eigenvalue corresponds to a distinct eigenvector, so you can form an eigenbasis of $n$ linearly independent eigenvectors.



In the latter case, I can see two possibilities:




  1. There are still $n$ linearly independent eigenvectors, but some share the same eigenvalue, so a $k$-fold root of the characteristic equation (an eigenvalue repeated $k$ times) will correspond to $k$ linearly independent eigenvectors.


  2. There are less than $n$ linearly independent eigenvectors, so a $k$-fold root may correspond to less than $k$ linearly independent eigenvectors.



Is one of these two possibilities always true, for all complex $ntimes n$ matrices, or could either possibility be true depending on the matrix in question?



Is there a proof of this, or perhaps a counter example?






linear-algebra matrices eigenvalues-eigenvectors







share|cite|improve this question











$endgroup$




Every $ntimes n$ matrix will produce an $n$-th order characteristic equation. This polynomial will have $n$ distinct roots or less than $n$ distinct roots (with some repeated).



In the former case, each eigenvalue corresponds to a distinct eigenvector, so you can form an eigenbasis of $n$ linearly independent eigenvectors.



In the latter case, I can see two possibilities:




  1. There are still $n$ linearly independent eigenvectors, but some share the same eigenvalue, so a $k$-fold root of the characteristic equation (an eigenvalue repeated $k$ times) will correspond to $k$ linearly independent eigenvectors.


  2. There are less than $n$ linearly independent eigenvectors, so a $k$-fold root may correspond to less than $k$ linearly independent eigenvectors.



Is one of these two possibilities always true, for all complex $ntimes n$ matrices, or could either possibility be true depending on the matrix in question?



Is there a proof of this, or perhaps a counter example?






linear-algebra matrices eigenvalues-eigenvectors




linear-algebra matrices eigenvalues-eigenvectors






share|cite|improve this question















share|cite|improve this question













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edited Dec 24 '18 at 17:26









Namaste

1




1










asked Dec 24 '18 at 17:19









Pancake_SenpaiPancake_Senpai

25116




25116












  • $begingroup$
    $begin{bmatrix}1&1\0&1end{bmatrix}$
    $endgroup$
    – David C. Ullrich
    Dec 24 '18 at 17:23




















  • $begingroup$
    $begin{bmatrix}1&1\0&1end{bmatrix}$
    $endgroup$
    – David C. Ullrich
    Dec 24 '18 at 17:23


















$begingroup$
$begin{bmatrix}1&1\0&1end{bmatrix}$
$endgroup$
– David C. Ullrich
Dec 24 '18 at 17:23






$begingroup$
$begin{bmatrix}1&1\0&1end{bmatrix}$
$endgroup$
– David C. Ullrich
Dec 24 '18 at 17:23












2 Answers
2






active

oldest

votes


















1












$begingroup$

Both possibilities apply. Case (1) means the matrix is diagonalisable over $mathbb{C}$, of which there clearly are examples. However matrices like $$begin{bmatrix}1 & 1 \ 0 & 1end{bmatrix}$$ are ones that are not diagonalisable over $mathbb{C}$; this one would fit your case (2).



The reason why we would consider eigenvalues over $mathbb{C}$ rather than over $mathbb{R}$ is precisely because the characteristic polynomial will split as a products of its roots, which does not always happen over $mathbb{R}$. Hence we always find (counting multiplicities) $n$ eigenvalues, but it turns out this is not sufficient for diagonalisability.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks. I've also actually just managed to find $begin{bmatrix} 0 & 1 & 1 \ 1 & 0 & 1 \ 1 & 1 & 0 \ end{bmatrix}$ as an example of a matrix with one distinct eigenvalue and one degenerate eigenvalue, but three linearly independent eigenvectors.
    $endgroup$
    – Pancake_Senpai
    Dec 24 '18 at 17:38








  • 1




    $begingroup$
    @Pancake_Senpai If you want to look into this stuff in more detail, I suggest learning something about the Jordan canonical form of a matrix. The Jordan canonical form looks to the eye to almost be diagonalisation of it. Every matrix can be brought into a Jordan canonical form, even though not all matrices are diagonalisable. Moreover, you can read off from the Jordan canonical form whether a matrix is diagonalisable or not.
    $endgroup$
    – SvanN
    Dec 24 '18 at 18:49










  • $begingroup$
    Thanks for the info!
    $endgroup$
    – Pancake_Senpai
    Dec 25 '18 at 0:30



















0












$begingroup$

Standard example: $$ pmatrix{0 & 1cr 0 & 0cr}$$
with only one linearly independent eigenvector.






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

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    1












    $begingroup$

    Both possibilities apply. Case (1) means the matrix is diagonalisable over $mathbb{C}$, of which there clearly are examples. However matrices like $$begin{bmatrix}1 & 1 \ 0 & 1end{bmatrix}$$ are ones that are not diagonalisable over $mathbb{C}$; this one would fit your case (2).



    The reason why we would consider eigenvalues over $mathbb{C}$ rather than over $mathbb{R}$ is precisely because the characteristic polynomial will split as a products of its roots, which does not always happen over $mathbb{R}$. Hence we always find (counting multiplicities) $n$ eigenvalues, but it turns out this is not sufficient for diagonalisability.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thanks. I've also actually just managed to find $begin{bmatrix} 0 & 1 & 1 \ 1 & 0 & 1 \ 1 & 1 & 0 \ end{bmatrix}$ as an example of a matrix with one distinct eigenvalue and one degenerate eigenvalue, but three linearly independent eigenvectors.
      $endgroup$
      – Pancake_Senpai
      Dec 24 '18 at 17:38








    • 1




      $begingroup$
      @Pancake_Senpai If you want to look into this stuff in more detail, I suggest learning something about the Jordan canonical form of a matrix. The Jordan canonical form looks to the eye to almost be diagonalisation of it. Every matrix can be brought into a Jordan canonical form, even though not all matrices are diagonalisable. Moreover, you can read off from the Jordan canonical form whether a matrix is diagonalisable or not.
      $endgroup$
      – SvanN
      Dec 24 '18 at 18:49










    • $begingroup$
      Thanks for the info!
      $endgroup$
      – Pancake_Senpai
      Dec 25 '18 at 0:30
















    1












    $begingroup$

    Both possibilities apply. Case (1) means the matrix is diagonalisable over $mathbb{C}$, of which there clearly are examples. However matrices like $$begin{bmatrix}1 & 1 \ 0 & 1end{bmatrix}$$ are ones that are not diagonalisable over $mathbb{C}$; this one would fit your case (2).



    The reason why we would consider eigenvalues over $mathbb{C}$ rather than over $mathbb{R}$ is precisely because the characteristic polynomial will split as a products of its roots, which does not always happen over $mathbb{R}$. Hence we always find (counting multiplicities) $n$ eigenvalues, but it turns out this is not sufficient for diagonalisability.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thanks. I've also actually just managed to find $begin{bmatrix} 0 & 1 & 1 \ 1 & 0 & 1 \ 1 & 1 & 0 \ end{bmatrix}$ as an example of a matrix with one distinct eigenvalue and one degenerate eigenvalue, but three linearly independent eigenvectors.
      $endgroup$
      – Pancake_Senpai
      Dec 24 '18 at 17:38








    • 1




      $begingroup$
      @Pancake_Senpai If you want to look into this stuff in more detail, I suggest learning something about the Jordan canonical form of a matrix. The Jordan canonical form looks to the eye to almost be diagonalisation of it. Every matrix can be brought into a Jordan canonical form, even though not all matrices are diagonalisable. Moreover, you can read off from the Jordan canonical form whether a matrix is diagonalisable or not.
      $endgroup$
      – SvanN
      Dec 24 '18 at 18:49










    • $begingroup$
      Thanks for the info!
      $endgroup$
      – Pancake_Senpai
      Dec 25 '18 at 0:30














    1












    1








    1





    $begingroup$

    Both possibilities apply. Case (1) means the matrix is diagonalisable over $mathbb{C}$, of which there clearly are examples. However matrices like $$begin{bmatrix}1 & 1 \ 0 & 1end{bmatrix}$$ are ones that are not diagonalisable over $mathbb{C}$; this one would fit your case (2).



    The reason why we would consider eigenvalues over $mathbb{C}$ rather than over $mathbb{R}$ is precisely because the characteristic polynomial will split as a products of its roots, which does not always happen over $mathbb{R}$. Hence we always find (counting multiplicities) $n$ eigenvalues, but it turns out this is not sufficient for diagonalisability.






    share|cite|improve this answer









    $endgroup$



    Both possibilities apply. Case (1) means the matrix is diagonalisable over $mathbb{C}$, of which there clearly are examples. However matrices like $$begin{bmatrix}1 & 1 \ 0 & 1end{bmatrix}$$ are ones that are not diagonalisable over $mathbb{C}$; this one would fit your case (2).



    The reason why we would consider eigenvalues over $mathbb{C}$ rather than over $mathbb{R}$ is precisely because the characteristic polynomial will split as a products of its roots, which does not always happen over $mathbb{R}$. Hence we always find (counting multiplicities) $n$ eigenvalues, but it turns out this is not sufficient for diagonalisability.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 24 '18 at 17:24









    SvanNSvanN

    2,0661422




    2,0661422












    • $begingroup$
      Thanks. I've also actually just managed to find $begin{bmatrix} 0 & 1 & 1 \ 1 & 0 & 1 \ 1 & 1 & 0 \ end{bmatrix}$ as an example of a matrix with one distinct eigenvalue and one degenerate eigenvalue, but three linearly independent eigenvectors.
      $endgroup$
      – Pancake_Senpai
      Dec 24 '18 at 17:38








    • 1




      $begingroup$
      @Pancake_Senpai If you want to look into this stuff in more detail, I suggest learning something about the Jordan canonical form of a matrix. The Jordan canonical form looks to the eye to almost be diagonalisation of it. Every matrix can be brought into a Jordan canonical form, even though not all matrices are diagonalisable. Moreover, you can read off from the Jordan canonical form whether a matrix is diagonalisable or not.
      $endgroup$
      – SvanN
      Dec 24 '18 at 18:49










    • $begingroup$
      Thanks for the info!
      $endgroup$
      – Pancake_Senpai
      Dec 25 '18 at 0:30


















    • $begingroup$
      Thanks. I've also actually just managed to find $begin{bmatrix} 0 & 1 & 1 \ 1 & 0 & 1 \ 1 & 1 & 0 \ end{bmatrix}$ as an example of a matrix with one distinct eigenvalue and one degenerate eigenvalue, but three linearly independent eigenvectors.
      $endgroup$
      – Pancake_Senpai
      Dec 24 '18 at 17:38








    • 1




      $begingroup$
      @Pancake_Senpai If you want to look into this stuff in more detail, I suggest learning something about the Jordan canonical form of a matrix. The Jordan canonical form looks to the eye to almost be diagonalisation of it. Every matrix can be brought into a Jordan canonical form, even though not all matrices are diagonalisable. Moreover, you can read off from the Jordan canonical form whether a matrix is diagonalisable or not.
      $endgroup$
      – SvanN
      Dec 24 '18 at 18:49










    • $begingroup$
      Thanks for the info!
      $endgroup$
      – Pancake_Senpai
      Dec 25 '18 at 0:30
















    $begingroup$
    Thanks. I've also actually just managed to find $begin{bmatrix} 0 & 1 & 1 \ 1 & 0 & 1 \ 1 & 1 & 0 \ end{bmatrix}$ as an example of a matrix with one distinct eigenvalue and one degenerate eigenvalue, but three linearly independent eigenvectors.
    $endgroup$
    – Pancake_Senpai
    Dec 24 '18 at 17:38






    $begingroup$
    Thanks. I've also actually just managed to find $begin{bmatrix} 0 & 1 & 1 \ 1 & 0 & 1 \ 1 & 1 & 0 \ end{bmatrix}$ as an example of a matrix with one distinct eigenvalue and one degenerate eigenvalue, but three linearly independent eigenvectors.
    $endgroup$
    – Pancake_Senpai
    Dec 24 '18 at 17:38






    1




    1




    $begingroup$
    @Pancake_Senpai If you want to look into this stuff in more detail, I suggest learning something about the Jordan canonical form of a matrix. The Jordan canonical form looks to the eye to almost be diagonalisation of it. Every matrix can be brought into a Jordan canonical form, even though not all matrices are diagonalisable. Moreover, you can read off from the Jordan canonical form whether a matrix is diagonalisable or not.
    $endgroup$
    – SvanN
    Dec 24 '18 at 18:49




    $begingroup$
    @Pancake_Senpai If you want to look into this stuff in more detail, I suggest learning something about the Jordan canonical form of a matrix. The Jordan canonical form looks to the eye to almost be diagonalisation of it. Every matrix can be brought into a Jordan canonical form, even though not all matrices are diagonalisable. Moreover, you can read off from the Jordan canonical form whether a matrix is diagonalisable or not.
    $endgroup$
    – SvanN
    Dec 24 '18 at 18:49












    $begingroup$
    Thanks for the info!
    $endgroup$
    – Pancake_Senpai
    Dec 25 '18 at 0:30




    $begingroup$
    Thanks for the info!
    $endgroup$
    – Pancake_Senpai
    Dec 25 '18 at 0:30











    0












    $begingroup$

    Standard example: $$ pmatrix{0 & 1cr 0 & 0cr}$$
    with only one linearly independent eigenvector.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Standard example: $$ pmatrix{0 & 1cr 0 & 0cr}$$
      with only one linearly independent eigenvector.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Standard example: $$ pmatrix{0 & 1cr 0 & 0cr}$$
        with only one linearly independent eigenvector.






        share|cite|improve this answer









        $endgroup$



        Standard example: $$ pmatrix{0 & 1cr 0 & 0cr}$$
        with only one linearly independent eigenvector.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 24 '18 at 17:23









        Robert IsraelRobert Israel

        327k23216470




        327k23216470






























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