find the Jordan basis of a matrix












2












$begingroup$


I'm trying to find the Jordan basis of the matrix $$A =begin{bmatrix} 8 & 1 & 2 \ -3 & 4 & -2\ -3 & -1 & 3end{bmatrix}$$ I've got the characteristic equation to be $CA(x) = (5-x)^3$ and hence the eigenvalue to be $5$. I started by finding $v_1$ such that $(A-5I)v_1=0$ and chose $v_1 = begin{bmatrix} 1 \ -3 \ 0end{bmatrix}$. I then tried to find a $v_2$ such that $(A-5I)^2v_2 = 0$ but $(A-5I)^2$ is the zero matrix so can $v_2$ be any vector in $mathbb{R}^3$?



More generally, when trying to find a Jordan basis for a matrix $A$, what do you do when $(A-lambda I)^i=0$ for some $i$? Also what do you do when there is no solution to $(A-lambda I)^iv_i = 0$ for a particular $i$?






linear-algebra jordan-normal-form







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  • $begingroup$
    Please use MathJax to format your posts.
    $endgroup$
    – saulspatz
    Dec 31 '18 at 23:27










  • $begingroup$
    Ok, what is this? but for now, any ideas/help?
    $endgroup$
    – Sam.S
    Dec 31 '18 at 23:29
















2












$begingroup$


I'm trying to find the Jordan basis of the matrix $$A =begin{bmatrix} 8 & 1 & 2 \ -3 & 4 & -2\ -3 & -1 & 3end{bmatrix}$$ I've got the characteristic equation to be $CA(x) = (5-x)^3$ and hence the eigenvalue to be $5$. I started by finding $v_1$ such that $(A-5I)v_1=0$ and chose $v_1 = begin{bmatrix} 1 \ -3 \ 0end{bmatrix}$. I then tried to find a $v_2$ such that $(A-5I)^2v_2 = 0$ but $(A-5I)^2$ is the zero matrix so can $v_2$ be any vector in $mathbb{R}^3$?



More generally, when trying to find a Jordan basis for a matrix $A$, what do you do when $(A-lambda I)^i=0$ for some $i$? Also what do you do when there is no solution to $(A-lambda I)^iv_i = 0$ for a particular $i$?






linear-algebra jordan-normal-form







share|cite|improve this question











$endgroup$












  • $begingroup$
    Please use MathJax to format your posts.
    $endgroup$
    – saulspatz
    Dec 31 '18 at 23:27










  • $begingroup$
    Ok, what is this? but for now, any ideas/help?
    $endgroup$
    – Sam.S
    Dec 31 '18 at 23:29














2












2








2


0



$begingroup$


I'm trying to find the Jordan basis of the matrix $$A =begin{bmatrix} 8 & 1 & 2 \ -3 & 4 & -2\ -3 & -1 & 3end{bmatrix}$$ I've got the characteristic equation to be $CA(x) = (5-x)^3$ and hence the eigenvalue to be $5$. I started by finding $v_1$ such that $(A-5I)v_1=0$ and chose $v_1 = begin{bmatrix} 1 \ -3 \ 0end{bmatrix}$. I then tried to find a $v_2$ such that $(A-5I)^2v_2 = 0$ but $(A-5I)^2$ is the zero matrix so can $v_2$ be any vector in $mathbb{R}^3$?



More generally, when trying to find a Jordan basis for a matrix $A$, what do you do when $(A-lambda I)^i=0$ for some $i$? Also what do you do when there is no solution to $(A-lambda I)^iv_i = 0$ for a particular $i$?






linear-algebra jordan-normal-form







share|cite|improve this question











$endgroup$




I'm trying to find the Jordan basis of the matrix $$A =begin{bmatrix} 8 & 1 & 2 \ -3 & 4 & -2\ -3 & -1 & 3end{bmatrix}$$ I've got the characteristic equation to be $CA(x) = (5-x)^3$ and hence the eigenvalue to be $5$. I started by finding $v_1$ such that $(A-5I)v_1=0$ and chose $v_1 = begin{bmatrix} 1 \ -3 \ 0end{bmatrix}$. I then tried to find a $v_2$ such that $(A-5I)^2v_2 = 0$ but $(A-5I)^2$ is the zero matrix so can $v_2$ be any vector in $mathbb{R}^3$?



More generally, when trying to find a Jordan basis for a matrix $A$, what do you do when $(A-lambda I)^i=0$ for some $i$? Also what do you do when there is no solution to $(A-lambda I)^iv_i = 0$ for a particular $i$?






linear-algebra jordan-normal-form




linear-algebra jordan-normal-form






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 1 at 0:02







Sam.S

















asked Dec 31 '18 at 23:19









Sam.SSam.S

649




649












  • $begingroup$
    Please use MathJax to format your posts.
    $endgroup$
    – saulspatz
    Dec 31 '18 at 23:27










  • $begingroup$
    Ok, what is this? but for now, any ideas/help?
    $endgroup$
    – Sam.S
    Dec 31 '18 at 23:29


















  • $begingroup$
    Please use MathJax to format your posts.
    $endgroup$
    – saulspatz
    Dec 31 '18 at 23:27










  • $begingroup$
    Ok, what is this? but for now, any ideas/help?
    $endgroup$
    – Sam.S
    Dec 31 '18 at 23:29
















$begingroup$
Please use MathJax to format your posts.
$endgroup$
– saulspatz
Dec 31 '18 at 23:27




$begingroup$
Please use MathJax to format your posts.
$endgroup$
– saulspatz
Dec 31 '18 at 23:27












$begingroup$
Ok, what is this? but for now, any ideas/help?
$endgroup$
– Sam.S
Dec 31 '18 at 23:29




$begingroup$
Ok, what is this? but for now, any ideas/help?
$endgroup$
– Sam.S
Dec 31 '18 at 23:29










3 Answers
3






active

oldest

votes


















0












$begingroup$

The eigenspace $E_5$ has dimension $2$ (so the Jordan form of the matrix will have $2$ Jordan blocks) since
$$A-5I=begin{bmatrix}
3&1&2\-3&-1&-2\-3&-1&-2end{bmatrix}
$$
and it is defined by the single equation $;3x+y+2z=0$.



You should attack the problem backwards: begin with choosing a vector $v_3=(x,y,z)$,
in $ker(A-5I)^2smallsetminus E_5=mathbf R^3smallsetminus E_5$, i.e. such that
$$3x+y+2zne 0,enspace text{say }enspace v_3=(1,0,-1),$$



and set $;(A-5I)v_3=v_2=(1,-1,-1)$ ; $v_2$ belongs to the eigenspace $E_5$.



Last you have to complete $v_2$ with another vector $v_1$ in the eigenspace which is linearly independent from $v_2$. The vector you've found – $v_1=(1,-3,0)$ is fine. In the basis $mathcal B=(v_1,v_2,v_3)$ the matrix of $A$ becomes by construction:
$$J=begin{bmatrix}
5&0&0\0&5&1\0&0&5end{bmatrix}. $$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Why in the world did someone downvote this?
    $endgroup$
    – Moo
    Jan 1 at 1:54










  • $begingroup$
    I understand why v3 must be in Ker(A-5I)^2, but why mustn't v3 be in E5?
    $endgroup$
    – Sam.S
    Jan 1 at 15:42








  • 1




    $begingroup$
    This is because the algorithm starts with the generalised eigenvectors and ends with the real eigenvectorrs, obtained as images of the above level generalised eigenvectors – unless the matrix is diagonalisable.
    $endgroup$
    – Bernard
    Jan 1 at 16:08





















2












$begingroup$

If the minimal polynomial is $(lambda -5)^2,$ you may, indeed, pick any column vector $w$ you like that is not already an eigenvector, make it the right hand column of $P. ; $ The rule is that the middle column must be $v = (A-5I) w.$ The left column of $P$ is then $u,$ a different eigenvector from $v.$ Sometimes it takes a little ingenuity to take the pair of eigenvectors you first calculated and revise to get $u.$



Then, with matrix $P,$ you get $J = P^{-1}AP.$



Try it.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Much appreciated. apologies for the lack of Mathjax. I will give this a go.
    $endgroup$
    – Sam.S
    Dec 31 '18 at 23:47



















2












$begingroup$

First take some $v in mathbb{C}^3$ such that $(A-5I)v ne 0$, for example take $v = e_2$. Then calculate $$(A-5I)v = begin{bmatrix} 1 \ -1 \ -1end{bmatrix}$$ and find some $w in ker (A-5I)$ which is linearly independent with $(A-5I)v$. For example we can take the vector you already found: $$w = begin{bmatrix} 1 \ -3 \ 0end{bmatrix}$$



Then the Jordan basis is ${(A-5I)v, v, w} = left{begin{bmatrix} 1 \ -1 \ -1end{bmatrix}, begin{bmatrix} 0 \ 1 \ 0end{bmatrix}, begin{bmatrix} 1 \ -3 \ 0end{bmatrix}right}$.






share|cite|improve this answer









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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    The eigenspace $E_5$ has dimension $2$ (so the Jordan form of the matrix will have $2$ Jordan blocks) since
    $$A-5I=begin{bmatrix}
    3&1&2\-3&-1&-2\-3&-1&-2end{bmatrix}
    $$
    and it is defined by the single equation $;3x+y+2z=0$.



    You should attack the problem backwards: begin with choosing a vector $v_3=(x,y,z)$,
    in $ker(A-5I)^2smallsetminus E_5=mathbf R^3smallsetminus E_5$, i.e. such that
    $$3x+y+2zne 0,enspace text{say }enspace v_3=(1,0,-1),$$



    and set $;(A-5I)v_3=v_2=(1,-1,-1)$ ; $v_2$ belongs to the eigenspace $E_5$.



    Last you have to complete $v_2$ with another vector $v_1$ in the eigenspace which is linearly independent from $v_2$. The vector you've found – $v_1=(1,-3,0)$ is fine. In the basis $mathcal B=(v_1,v_2,v_3)$ the matrix of $A$ becomes by construction:
    $$J=begin{bmatrix}
    5&0&0\0&5&1\0&0&5end{bmatrix}. $$






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Why in the world did someone downvote this?
      $endgroup$
      – Moo
      Jan 1 at 1:54










    • $begingroup$
      I understand why v3 must be in Ker(A-5I)^2, but why mustn't v3 be in E5?
      $endgroup$
      – Sam.S
      Jan 1 at 15:42








    • 1




      $begingroup$
      This is because the algorithm starts with the generalised eigenvectors and ends with the real eigenvectorrs, obtained as images of the above level generalised eigenvectors – unless the matrix is diagonalisable.
      $endgroup$
      – Bernard
      Jan 1 at 16:08


















    0












    $begingroup$

    The eigenspace $E_5$ has dimension $2$ (so the Jordan form of the matrix will have $2$ Jordan blocks) since
    $$A-5I=begin{bmatrix}
    3&1&2\-3&-1&-2\-3&-1&-2end{bmatrix}
    $$
    and it is defined by the single equation $;3x+y+2z=0$.



    You should attack the problem backwards: begin with choosing a vector $v_3=(x,y,z)$,
    in $ker(A-5I)^2smallsetminus E_5=mathbf R^3smallsetminus E_5$, i.e. such that
    $$3x+y+2zne 0,enspace text{say }enspace v_3=(1,0,-1),$$



    and set $;(A-5I)v_3=v_2=(1,-1,-1)$ ; $v_2$ belongs to the eigenspace $E_5$.



    Last you have to complete $v_2$ with another vector $v_1$ in the eigenspace which is linearly independent from $v_2$. The vector you've found – $v_1=(1,-3,0)$ is fine. In the basis $mathcal B=(v_1,v_2,v_3)$ the matrix of $A$ becomes by construction:
    $$J=begin{bmatrix}
    5&0&0\0&5&1\0&0&5end{bmatrix}. $$






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Why in the world did someone downvote this?
      $endgroup$
      – Moo
      Jan 1 at 1:54










    • $begingroup$
      I understand why v3 must be in Ker(A-5I)^2, but why mustn't v3 be in E5?
      $endgroup$
      – Sam.S
      Jan 1 at 15:42








    • 1




      $begingroup$
      This is because the algorithm starts with the generalised eigenvectors and ends with the real eigenvectorrs, obtained as images of the above level generalised eigenvectors – unless the matrix is diagonalisable.
      $endgroup$
      – Bernard
      Jan 1 at 16:08
















    0












    0








    0





    $begingroup$

    The eigenspace $E_5$ has dimension $2$ (so the Jordan form of the matrix will have $2$ Jordan blocks) since
    $$A-5I=begin{bmatrix}
    3&1&2\-3&-1&-2\-3&-1&-2end{bmatrix}
    $$
    and it is defined by the single equation $;3x+y+2z=0$.



    You should attack the problem backwards: begin with choosing a vector $v_3=(x,y,z)$,
    in $ker(A-5I)^2smallsetminus E_5=mathbf R^3smallsetminus E_5$, i.e. such that
    $$3x+y+2zne 0,enspace text{say }enspace v_3=(1,0,-1),$$



    and set $;(A-5I)v_3=v_2=(1,-1,-1)$ ; $v_2$ belongs to the eigenspace $E_5$.



    Last you have to complete $v_2$ with another vector $v_1$ in the eigenspace which is linearly independent from $v_2$. The vector you've found – $v_1=(1,-3,0)$ is fine. In the basis $mathcal B=(v_1,v_2,v_3)$ the matrix of $A$ becomes by construction:
    $$J=begin{bmatrix}
    5&0&0\0&5&1\0&0&5end{bmatrix}. $$






    share|cite|improve this answer











    $endgroup$



    The eigenspace $E_5$ has dimension $2$ (so the Jordan form of the matrix will have $2$ Jordan blocks) since
    $$A-5I=begin{bmatrix}
    3&1&2\-3&-1&-2\-3&-1&-2end{bmatrix}
    $$
    and it is defined by the single equation $;3x+y+2z=0$.



    You should attack the problem backwards: begin with choosing a vector $v_3=(x,y,z)$,
    in $ker(A-5I)^2smallsetminus E_5=mathbf R^3smallsetminus E_5$, i.e. such that
    $$3x+y+2zne 0,enspace text{say }enspace v_3=(1,0,-1),$$



    and set $;(A-5I)v_3=v_2=(1,-1,-1)$ ; $v_2$ belongs to the eigenspace $E_5$.



    Last you have to complete $v_2$ with another vector $v_1$ in the eigenspace which is linearly independent from $v_2$. The vector you've found – $v_1=(1,-3,0)$ is fine. In the basis $mathcal B=(v_1,v_2,v_3)$ the matrix of $A$ becomes by construction:
    $$J=begin{bmatrix}
    5&0&0\0&5&1\0&0&5end{bmatrix}. $$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 1 at 9:36

























    answered Jan 1 at 0:01









    BernardBernard

    123k741117




    123k741117








    • 1




      $begingroup$
      Why in the world did someone downvote this?
      $endgroup$
      – Moo
      Jan 1 at 1:54










    • $begingroup$
      I understand why v3 must be in Ker(A-5I)^2, but why mustn't v3 be in E5?
      $endgroup$
      – Sam.S
      Jan 1 at 15:42








    • 1




      $begingroup$
      This is because the algorithm starts with the generalised eigenvectors and ends with the real eigenvectorrs, obtained as images of the above level generalised eigenvectors – unless the matrix is diagonalisable.
      $endgroup$
      – Bernard
      Jan 1 at 16:08
















    • 1




      $begingroup$
      Why in the world did someone downvote this?
      $endgroup$
      – Moo
      Jan 1 at 1:54










    • $begingroup$
      I understand why v3 must be in Ker(A-5I)^2, but why mustn't v3 be in E5?
      $endgroup$
      – Sam.S
      Jan 1 at 15:42








    • 1




      $begingroup$
      This is because the algorithm starts with the generalised eigenvectors and ends with the real eigenvectorrs, obtained as images of the above level generalised eigenvectors – unless the matrix is diagonalisable.
      $endgroup$
      – Bernard
      Jan 1 at 16:08










    1




    1




    $begingroup$
    Why in the world did someone downvote this?
    $endgroup$
    – Moo
    Jan 1 at 1:54




    $begingroup$
    Why in the world did someone downvote this?
    $endgroup$
    – Moo
    Jan 1 at 1:54












    $begingroup$
    I understand why v3 must be in Ker(A-5I)^2, but why mustn't v3 be in E5?
    $endgroup$
    – Sam.S
    Jan 1 at 15:42






    $begingroup$
    I understand why v3 must be in Ker(A-5I)^2, but why mustn't v3 be in E5?
    $endgroup$
    – Sam.S
    Jan 1 at 15:42






    1




    1




    $begingroup$
    This is because the algorithm starts with the generalised eigenvectors and ends with the real eigenvectorrs, obtained as images of the above level generalised eigenvectors – unless the matrix is diagonalisable.
    $endgroup$
    – Bernard
    Jan 1 at 16:08






    $begingroup$
    This is because the algorithm starts with the generalised eigenvectors and ends with the real eigenvectorrs, obtained as images of the above level generalised eigenvectors – unless the matrix is diagonalisable.
    $endgroup$
    – Bernard
    Jan 1 at 16:08













    2












    $begingroup$

    If the minimal polynomial is $(lambda -5)^2,$ you may, indeed, pick any column vector $w$ you like that is not already an eigenvector, make it the right hand column of $P. ; $ The rule is that the middle column must be $v = (A-5I) w.$ The left column of $P$ is then $u,$ a different eigenvector from $v.$ Sometimes it takes a little ingenuity to take the pair of eigenvectors you first calculated and revise to get $u.$



    Then, with matrix $P,$ you get $J = P^{-1}AP.$



    Try it.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Much appreciated. apologies for the lack of Mathjax. I will give this a go.
      $endgroup$
      – Sam.S
      Dec 31 '18 at 23:47
















    2












    $begingroup$

    If the minimal polynomial is $(lambda -5)^2,$ you may, indeed, pick any column vector $w$ you like that is not already an eigenvector, make it the right hand column of $P. ; $ The rule is that the middle column must be $v = (A-5I) w.$ The left column of $P$ is then $u,$ a different eigenvector from $v.$ Sometimes it takes a little ingenuity to take the pair of eigenvectors you first calculated and revise to get $u.$



    Then, with matrix $P,$ you get $J = P^{-1}AP.$



    Try it.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Much appreciated. apologies for the lack of Mathjax. I will give this a go.
      $endgroup$
      – Sam.S
      Dec 31 '18 at 23:47














    2












    2








    2





    $begingroup$

    If the minimal polynomial is $(lambda -5)^2,$ you may, indeed, pick any column vector $w$ you like that is not already an eigenvector, make it the right hand column of $P. ; $ The rule is that the middle column must be $v = (A-5I) w.$ The left column of $P$ is then $u,$ a different eigenvector from $v.$ Sometimes it takes a little ingenuity to take the pair of eigenvectors you first calculated and revise to get $u.$



    Then, with matrix $P,$ you get $J = P^{-1}AP.$



    Try it.






    share|cite|improve this answer









    $endgroup$



    If the minimal polynomial is $(lambda -5)^2,$ you may, indeed, pick any column vector $w$ you like that is not already an eigenvector, make it the right hand column of $P. ; $ The rule is that the middle column must be $v = (A-5I) w.$ The left column of $P$ is then $u,$ a different eigenvector from $v.$ Sometimes it takes a little ingenuity to take the pair of eigenvectors you first calculated and revise to get $u.$



    Then, with matrix $P,$ you get $J = P^{-1}AP.$



    Try it.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 31 '18 at 23:39









    Will JagyWill Jagy

    104k5102201




    104k5102201












    • $begingroup$
      Much appreciated. apologies for the lack of Mathjax. I will give this a go.
      $endgroup$
      – Sam.S
      Dec 31 '18 at 23:47


















    • $begingroup$
      Much appreciated. apologies for the lack of Mathjax. I will give this a go.
      $endgroup$
      – Sam.S
      Dec 31 '18 at 23:47
















    $begingroup$
    Much appreciated. apologies for the lack of Mathjax. I will give this a go.
    $endgroup$
    – Sam.S
    Dec 31 '18 at 23:47




    $begingroup$
    Much appreciated. apologies for the lack of Mathjax. I will give this a go.
    $endgroup$
    – Sam.S
    Dec 31 '18 at 23:47











    2












    $begingroup$

    First take some $v in mathbb{C}^3$ such that $(A-5I)v ne 0$, for example take $v = e_2$. Then calculate $$(A-5I)v = begin{bmatrix} 1 \ -1 \ -1end{bmatrix}$$ and find some $w in ker (A-5I)$ which is linearly independent with $(A-5I)v$. For example we can take the vector you already found: $$w = begin{bmatrix} 1 \ -3 \ 0end{bmatrix}$$



    Then the Jordan basis is ${(A-5I)v, v, w} = left{begin{bmatrix} 1 \ -1 \ -1end{bmatrix}, begin{bmatrix} 0 \ 1 \ 0end{bmatrix}, begin{bmatrix} 1 \ -3 \ 0end{bmatrix}right}$.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      First take some $v in mathbb{C}^3$ such that $(A-5I)v ne 0$, for example take $v = e_2$. Then calculate $$(A-5I)v = begin{bmatrix} 1 \ -1 \ -1end{bmatrix}$$ and find some $w in ker (A-5I)$ which is linearly independent with $(A-5I)v$. For example we can take the vector you already found: $$w = begin{bmatrix} 1 \ -3 \ 0end{bmatrix}$$



      Then the Jordan basis is ${(A-5I)v, v, w} = left{begin{bmatrix} 1 \ -1 \ -1end{bmatrix}, begin{bmatrix} 0 \ 1 \ 0end{bmatrix}, begin{bmatrix} 1 \ -3 \ 0end{bmatrix}right}$.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        First take some $v in mathbb{C}^3$ such that $(A-5I)v ne 0$, for example take $v = e_2$. Then calculate $$(A-5I)v = begin{bmatrix} 1 \ -1 \ -1end{bmatrix}$$ and find some $w in ker (A-5I)$ which is linearly independent with $(A-5I)v$. For example we can take the vector you already found: $$w = begin{bmatrix} 1 \ -3 \ 0end{bmatrix}$$



        Then the Jordan basis is ${(A-5I)v, v, w} = left{begin{bmatrix} 1 \ -1 \ -1end{bmatrix}, begin{bmatrix} 0 \ 1 \ 0end{bmatrix}, begin{bmatrix} 1 \ -3 \ 0end{bmatrix}right}$.






        share|cite|improve this answer









        $endgroup$



        First take some $v in mathbb{C}^3$ such that $(A-5I)v ne 0$, for example take $v = e_2$. Then calculate $$(A-5I)v = begin{bmatrix} 1 \ -1 \ -1end{bmatrix}$$ and find some $w in ker (A-5I)$ which is linearly independent with $(A-5I)v$. For example we can take the vector you already found: $$w = begin{bmatrix} 1 \ -3 \ 0end{bmatrix}$$



        Then the Jordan basis is ${(A-5I)v, v, w} = left{begin{bmatrix} 1 \ -1 \ -1end{bmatrix}, begin{bmatrix} 0 \ 1 \ 0end{bmatrix}, begin{bmatrix} 1 \ -3 \ 0end{bmatrix}right}$.







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        answered Dec 31 '18 at 23:57









        mechanodroidmechanodroid

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