About a some basis for linear mapping on a plane












1












$begingroup$


Let $f: Vrightarrow V$ be a linear mapping on a real 2-dimensional vector space $V$ with characteristic polynomial
$p(x)=x^2+bx+c$ and with negative discriminant $Delta=b^2-4ac<0$.
Let $t_1=alpha+ibeta$, $t_2=alpha-ibeta$ be its roots.



Then I know that there exists a basis $e_1, e_2$ in $V$ in which thematrix of $f$ is of the form



$
left [ begin {array}{lr}
alpha & -beta\
beta & alpha
end{array} right ].
$



Is there a basis in which matrix of $f$ is of the form



$
left [ begin {array}{lr}
0 & -c \
1& -b
end{array} right ]?
$



Thanks.






linear-algebra







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Let $f: Vrightarrow V$ be a linear mapping on a real 2-dimensional vector space $V$ with characteristic polynomial
    $p(x)=x^2+bx+c$ and with negative discriminant $Delta=b^2-4ac<0$.
    Let $t_1=alpha+ibeta$, $t_2=alpha-ibeta$ be its roots.



    Then I know that there exists a basis $e_1, e_2$ in $V$ in which thematrix of $f$ is of the form



    $
    left [ begin {array}{lr}
    alpha & -beta\
    beta & alpha
    end{array} right ].
    $



    Is there a basis in which matrix of $f$ is of the form



    $
    left [ begin {array}{lr}
    0 & -c \
    1& -b
    end{array} right ]?
    $



    Thanks.






    linear-algebra







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Let $f: Vrightarrow V$ be a linear mapping on a real 2-dimensional vector space $V$ with characteristic polynomial
      $p(x)=x^2+bx+c$ and with negative discriminant $Delta=b^2-4ac<0$.
      Let $t_1=alpha+ibeta$, $t_2=alpha-ibeta$ be its roots.



      Then I know that there exists a basis $e_1, e_2$ in $V$ in which thematrix of $f$ is of the form



      $
      left [ begin {array}{lr}
      alpha & -beta\
      beta & alpha
      end{array} right ].
      $



      Is there a basis in which matrix of $f$ is of the form



      $
      left [ begin {array}{lr}
      0 & -c \
      1& -b
      end{array} right ]?
      $



      Thanks.






      linear-algebra







      share|cite|improve this question









      $endgroup$




      Let $f: Vrightarrow V$ be a linear mapping on a real 2-dimensional vector space $V$ with characteristic polynomial
      $p(x)=x^2+bx+c$ and with negative discriminant $Delta=b^2-4ac<0$.
      Let $t_1=alpha+ibeta$, $t_2=alpha-ibeta$ be its roots.



      Then I know that there exists a basis $e_1, e_2$ in $V$ in which thematrix of $f$ is of the form



      $
      left [ begin {array}{lr}
      alpha & -beta\
      beta & alpha
      end{array} right ].
      $



      Is there a basis in which matrix of $f$ is of the form



      $
      left [ begin {array}{lr}
      0 & -c \
      1& -b
      end{array} right ]?
      $



      Thanks.






      linear-algebra




      linear-algebra






      share|cite|improve this question













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      share|cite|improve this question










      asked Jan 31 at 20:06









      AlexAlex

      699519




      699519






















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          $begingroup$

          If for all $x in V$, $x$ and $f(x)$ are collinear, $f$ would be an homothetic transformation. As this is not the case, it exists $a in V$ such that ${a, f(a)}$ is linearly independent.



          What is the matrix of $f$ in the basis $(a,f(a))$?



          We have $f(a)= 0.a + 1.f(a)$ and $f(f(a))= f^2(a) = (-c).a +(-b).f(a)$ using the equation of the characteristic polynomial. This is exactly the matrix we were looking for!



          The answer to your question is positive.






          share|cite|improve this answer









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            1 Answer
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            $begingroup$

            If for all $x in V$, $x$ and $f(x)$ are collinear, $f$ would be an homothetic transformation. As this is not the case, it exists $a in V$ such that ${a, f(a)}$ is linearly independent.



            What is the matrix of $f$ in the basis $(a,f(a))$?



            We have $f(a)= 0.a + 1.f(a)$ and $f(f(a))= f^2(a) = (-c).a +(-b).f(a)$ using the equation of the characteristic polynomial. This is exactly the matrix we were looking for!



            The answer to your question is positive.






            share|cite|improve this answer









            $endgroup$


















              3












              $begingroup$

              If for all $x in V$, $x$ and $f(x)$ are collinear, $f$ would be an homothetic transformation. As this is not the case, it exists $a in V$ such that ${a, f(a)}$ is linearly independent.



              What is the matrix of $f$ in the basis $(a,f(a))$?



              We have $f(a)= 0.a + 1.f(a)$ and $f(f(a))= f^2(a) = (-c).a +(-b).f(a)$ using the equation of the characteristic polynomial. This is exactly the matrix we were looking for!



              The answer to your question is positive.






              share|cite|improve this answer









              $endgroup$
















                3












                3








                3





                $begingroup$

                If for all $x in V$, $x$ and $f(x)$ are collinear, $f$ would be an homothetic transformation. As this is not the case, it exists $a in V$ such that ${a, f(a)}$ is linearly independent.



                What is the matrix of $f$ in the basis $(a,f(a))$?



                We have $f(a)= 0.a + 1.f(a)$ and $f(f(a))= f^2(a) = (-c).a +(-b).f(a)$ using the equation of the characteristic polynomial. This is exactly the matrix we were looking for!



                The answer to your question is positive.






                share|cite|improve this answer









                $endgroup$



                If for all $x in V$, $x$ and $f(x)$ are collinear, $f$ would be an homothetic transformation. As this is not the case, it exists $a in V$ such that ${a, f(a)}$ is linearly independent.



                What is the matrix of $f$ in the basis $(a,f(a))$?



                We have $f(a)= 0.a + 1.f(a)$ and $f(f(a))= f^2(a) = (-c).a +(-b).f(a)$ using the equation of the characteristic polynomial. This is exactly the matrix we were looking for!



                The answer to your question is positive.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 31 at 20:39









                mathcounterexamples.netmathcounterexamples.net

                26.9k22158




                26.9k22158






























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