A linear map $T: mathbb{R^3 to mathbb{R^3}}$ has a two dimensional invariant subspace.












4












$begingroup$



Let $T: mathbb{R^3 to mathbb{R^3}}$ be an $mathbb{R}$-linear map. Then I want to show that $T$ has a $2$ dimensional invariant subspace of $mathbb{R^3}.$




I considered all possible minimal polynomial of $T$ and applying canonical forms I found some obvious $2$ dimensional invariant subspaces.




I stuck when the minimal polynomial is of the form $(X-a)^3$ for some real number $a.$




In this situation since the minimal polynomial and the characteristic polynomial coincides $T$ has a cyclic vector. But I can't complete it further. I need some help. Thanks.






linear-algebra abstract-algebra matrices linear-transformations jordan-normal-form







share|cite|improve this question









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    4












    $begingroup$



    Let $T: mathbb{R^3 to mathbb{R^3}}$ be an $mathbb{R}$-linear map. Then I want to show that $T$ has a $2$ dimensional invariant subspace of $mathbb{R^3}.$




    I considered all possible minimal polynomial of $T$ and applying canonical forms I found some obvious $2$ dimensional invariant subspaces.




    I stuck when the minimal polynomial is of the form $(X-a)^3$ for some real number $a.$




    In this situation since the minimal polynomial and the characteristic polynomial coincides $T$ has a cyclic vector. But I can't complete it further. I need some help. Thanks.






    linear-algebra abstract-algebra matrices linear-transformations jordan-normal-form







    share|cite|improve this question









    $endgroup$















      4












      4








      4


      2



      $begingroup$



      Let $T: mathbb{R^3 to mathbb{R^3}}$ be an $mathbb{R}$-linear map. Then I want to show that $T$ has a $2$ dimensional invariant subspace of $mathbb{R^3}.$




      I considered all possible minimal polynomial of $T$ and applying canonical forms I found some obvious $2$ dimensional invariant subspaces.




      I stuck when the minimal polynomial is of the form $(X-a)^3$ for some real number $a.$




      In this situation since the minimal polynomial and the characteristic polynomial coincides $T$ has a cyclic vector. But I can't complete it further. I need some help. Thanks.






      linear-algebra abstract-algebra matrices linear-transformations jordan-normal-form







      share|cite|improve this question









      $endgroup$





      Let $T: mathbb{R^3 to mathbb{R^3}}$ be an $mathbb{R}$-linear map. Then I want to show that $T$ has a $2$ dimensional invariant subspace of $mathbb{R^3}.$




      I considered all possible minimal polynomial of $T$ and applying canonical forms I found some obvious $2$ dimensional invariant subspaces.




      I stuck when the minimal polynomial is of the form $(X-a)^3$ for some real number $a.$




      In this situation since the minimal polynomial and the characteristic polynomial coincides $T$ has a cyclic vector. But I can't complete it further. I need some help. Thanks.






      linear-algebra abstract-algebra matrices linear-transformations jordan-normal-form




      linear-algebra abstract-algebra matrices linear-transformations jordan-normal-form






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 10 at 11:17









      user371231user371231

      826511




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          1 Answer
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          $begingroup$

          If the minimal polynomial is of the form $(X-a)^3$, then there is some basis $B={e_1,e_2,e_3}$ of $mathbb{R}^3$ such that the matrix of $T$ with respect to $B$ is$$begin{bmatrix}a&1&0\0&a&1\0&0&aend{bmatrix}.$$So, consider the space spanned by $e_1$ and $e_2$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            How can it be a diagonal matrix ? also second matrix how do you get ?
            $endgroup$
            – user371231
            Jan 10 at 11:41










          • $begingroup$
            I wrote “minimal polynomial” but I was thinking about characteristic polynomials. I hope that everything is correct now.
            $endgroup$
            – José Carlos Santos
            Jan 10 at 11:43













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          1 Answer
          1






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          active

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          $begingroup$

          If the minimal polynomial is of the form $(X-a)^3$, then there is some basis $B={e_1,e_2,e_3}$ of $mathbb{R}^3$ such that the matrix of $T$ with respect to $B$ is$$begin{bmatrix}a&1&0\0&a&1\0&0&aend{bmatrix}.$$So, consider the space spanned by $e_1$ and $e_2$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            How can it be a diagonal matrix ? also second matrix how do you get ?
            $endgroup$
            – user371231
            Jan 10 at 11:41










          • $begingroup$
            I wrote “minimal polynomial” but I was thinking about characteristic polynomials. I hope that everything is correct now.
            $endgroup$
            – José Carlos Santos
            Jan 10 at 11:43


















          2












          $begingroup$

          If the minimal polynomial is of the form $(X-a)^3$, then there is some basis $B={e_1,e_2,e_3}$ of $mathbb{R}^3$ such that the matrix of $T$ with respect to $B$ is$$begin{bmatrix}a&1&0\0&a&1\0&0&aend{bmatrix}.$$So, consider the space spanned by $e_1$ and $e_2$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            How can it be a diagonal matrix ? also second matrix how do you get ?
            $endgroup$
            – user371231
            Jan 10 at 11:41










          • $begingroup$
            I wrote “minimal polynomial” but I was thinking about characteristic polynomials. I hope that everything is correct now.
            $endgroup$
            – José Carlos Santos
            Jan 10 at 11:43
















          2












          2








          2





          $begingroup$

          If the minimal polynomial is of the form $(X-a)^3$, then there is some basis $B={e_1,e_2,e_3}$ of $mathbb{R}^3$ such that the matrix of $T$ with respect to $B$ is$$begin{bmatrix}a&1&0\0&a&1\0&0&aend{bmatrix}.$$So, consider the space spanned by $e_1$ and $e_2$.






          share|cite|improve this answer











          $endgroup$



          If the minimal polynomial is of the form $(X-a)^3$, then there is some basis $B={e_1,e_2,e_3}$ of $mathbb{R}^3$ such that the matrix of $T$ with respect to $B$ is$$begin{bmatrix}a&1&0\0&a&1\0&0&aend{bmatrix}.$$So, consider the space spanned by $e_1$ and $e_2$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 10 at 11:42

























          answered Jan 10 at 11:36









          José Carlos SantosJosé Carlos Santos

          159k22126231




          159k22126231












          • $begingroup$
            How can it be a diagonal matrix ? also second matrix how do you get ?
            $endgroup$
            – user371231
            Jan 10 at 11:41










          • $begingroup$
            I wrote “minimal polynomial” but I was thinking about characteristic polynomials. I hope that everything is correct now.
            $endgroup$
            – José Carlos Santos
            Jan 10 at 11:43




















          • $begingroup$
            How can it be a diagonal matrix ? also second matrix how do you get ?
            $endgroup$
            – user371231
            Jan 10 at 11:41










          • $begingroup$
            I wrote “minimal polynomial” but I was thinking about characteristic polynomials. I hope that everything is correct now.
            $endgroup$
            – José Carlos Santos
            Jan 10 at 11:43


















          $begingroup$
          How can it be a diagonal matrix ? also second matrix how do you get ?
          $endgroup$
          – user371231
          Jan 10 at 11:41




          $begingroup$
          How can it be a diagonal matrix ? also second matrix how do you get ?
          $endgroup$
          – user371231
          Jan 10 at 11:41












          $begingroup$
          I wrote “minimal polynomial” but I was thinking about characteristic polynomials. I hope that everything is correct now.
          $endgroup$
          – José Carlos Santos
          Jan 10 at 11:43






          $begingroup$
          I wrote “minimal polynomial” but I was thinking about characteristic polynomials. I hope that everything is correct now.
          $endgroup$
          – José Carlos Santos
          Jan 10 at 11:43




















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