Diagonalize $f(A)= begin{pmatrix} 1 & 0 \ -1 & 3 end{pmatrix} A $












4












$begingroup$


I have to diagonalize the endomorphism $fin mathrm{End(M_2}(mathbb{R}))$ defined by
$f(A)=
begin{pmatrix}
1 & 0 \
-1 & 3
end{pmatrix}
A
$



I know I can rewrite it, if $A=
begin{pmatrix}
a & b \
c & d
end{pmatrix}$, as $f(A)=
begin{pmatrix}
a & b \
3c-a & 3d-b
end{pmatrix}
$,



but I don't know how to continue. Could you help me? Thanks in advance!






linear-algebra diagonalization







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Pick up a basis of $mathrm M_2(Bbb R)$, find out the action of $f$ on these, and write down the matrix of $f$, then do the diagonalization.
    $endgroup$
    – xbh
    Dec 3 '18 at 18:50
















4












$begingroup$


I have to diagonalize the endomorphism $fin mathrm{End(M_2}(mathbb{R}))$ defined by
$f(A)=
begin{pmatrix}
1 & 0 \
-1 & 3
end{pmatrix}
A
$



I know I can rewrite it, if $A=
begin{pmatrix}
a & b \
c & d
end{pmatrix}$, as $f(A)=
begin{pmatrix}
a & b \
3c-a & 3d-b
end{pmatrix}
$,



but I don't know how to continue. Could you help me? Thanks in advance!






linear-algebra diagonalization







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Pick up a basis of $mathrm M_2(Bbb R)$, find out the action of $f$ on these, and write down the matrix of $f$, then do the diagonalization.
    $endgroup$
    – xbh
    Dec 3 '18 at 18:50














4












4








4


2



$begingroup$


I have to diagonalize the endomorphism $fin mathrm{End(M_2}(mathbb{R}))$ defined by
$f(A)=
begin{pmatrix}
1 & 0 \
-1 & 3
end{pmatrix}
A
$



I know I can rewrite it, if $A=
begin{pmatrix}
a & b \
c & d
end{pmatrix}$, as $f(A)=
begin{pmatrix}
a & b \
3c-a & 3d-b
end{pmatrix}
$,



but I don't know how to continue. Could you help me? Thanks in advance!






linear-algebra diagonalization







share|cite|improve this question









$endgroup$




I have to diagonalize the endomorphism $fin mathrm{End(M_2}(mathbb{R}))$ defined by
$f(A)=
begin{pmatrix}
1 & 0 \
-1 & 3
end{pmatrix}
A
$



I know I can rewrite it, if $A=
begin{pmatrix}
a & b \
c & d
end{pmatrix}$, as $f(A)=
begin{pmatrix}
a & b \
3c-a & 3d-b
end{pmatrix}
$,



but I don't know how to continue. Could you help me? Thanks in advance!






linear-algebra diagonalization




linear-algebra diagonalization






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 3 '18 at 18:44









GibbsGibbs

13110




13110








  • 1




    $begingroup$
    Pick up a basis of $mathrm M_2(Bbb R)$, find out the action of $f$ on these, and write down the matrix of $f$, then do the diagonalization.
    $endgroup$
    – xbh
    Dec 3 '18 at 18:50














  • 1




    $begingroup$
    Pick up a basis of $mathrm M_2(Bbb R)$, find out the action of $f$ on these, and write down the matrix of $f$, then do the diagonalization.
    $endgroup$
    – xbh
    Dec 3 '18 at 18:50








1




1




$begingroup$
Pick up a basis of $mathrm M_2(Bbb R)$, find out the action of $f$ on these, and write down the matrix of $f$, then do the diagonalization.
$endgroup$
– xbh
Dec 3 '18 at 18:50




$begingroup$
Pick up a basis of $mathrm M_2(Bbb R)$, find out the action of $f$ on these, and write down the matrix of $f$, then do the diagonalization.
$endgroup$
– xbh
Dec 3 '18 at 18:50










2 Answers
2






active

oldest

votes


















3












$begingroup$

Hint: Choose a basis for $mathrm M_2(mathbb R)$, say ${E_{11},E_{12},E_{21},E_{22}}$. Calculate $f(E_{ij})$ and write it in the basis. This will give you matrix for $f$ (it's $4times 4$). Proceed to diagonalize it as usual.



It might help if you think of $mathrm M_2(mathbb R)cong mathbb R^4$. Then $f$ becomes $$f(a,b,c,d) = (a,b,3c-a,3d-b).$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So then, I just have to diagonalize the following matrix, right? $ begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ -1 & 0 & 3 & 0 \ 0 & -1 & 0 & 3 end{pmatrix} $
    $endgroup$
    – Gibbs
    Dec 3 '18 at 18:58










  • $begingroup$
    @Gibbs, correct.
    $endgroup$
    – Ennar
    Dec 3 '18 at 18:58










  • $begingroup$
    This was easier than I thought! Thanks!!
    $endgroup$
    – Gibbs
    Dec 3 '18 at 19:00










  • $begingroup$
    @Gibbs, you are welcome. Don't let algebra structure of $mathrm M_2$ confuse you - it's just $4$-dimensional vector space with canonical basis.
    $endgroup$
    – Ennar
    Dec 3 '18 at 19:00





















1












$begingroup$

Hint Let $lambda$ an eigenvalue and $Ane 0$ an associated eigenvector so



$$f(A)=lambda Aiff begin{pmatrix}1-lambda &0\ -1 & 3-lambdaend{pmatrix}A=0$$



Since $Ane0$ so $begin{pmatrix}1-lambda &0\ -1 & 3-lambdaend{pmatrix}$ should not be invertible hence $lambdain{1,3}$. Now for every value of $lambda$ solve the equation $f(A)=lambda A$ for $A=begin{pmatrix}a &b\ c & dend{pmatrix}$ to find the associated eigenspace.






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

    votes









    3












    $begingroup$

    Hint: Choose a basis for $mathrm M_2(mathbb R)$, say ${E_{11},E_{12},E_{21},E_{22}}$. Calculate $f(E_{ij})$ and write it in the basis. This will give you matrix for $f$ (it's $4times 4$). Proceed to diagonalize it as usual.



    It might help if you think of $mathrm M_2(mathbb R)cong mathbb R^4$. Then $f$ becomes $$f(a,b,c,d) = (a,b,3c-a,3d-b).$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      So then, I just have to diagonalize the following matrix, right? $ begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ -1 & 0 & 3 & 0 \ 0 & -1 & 0 & 3 end{pmatrix} $
      $endgroup$
      – Gibbs
      Dec 3 '18 at 18:58










    • $begingroup$
      @Gibbs, correct.
      $endgroup$
      – Ennar
      Dec 3 '18 at 18:58










    • $begingroup$
      This was easier than I thought! Thanks!!
      $endgroup$
      – Gibbs
      Dec 3 '18 at 19:00










    • $begingroup$
      @Gibbs, you are welcome. Don't let algebra structure of $mathrm M_2$ confuse you - it's just $4$-dimensional vector space with canonical basis.
      $endgroup$
      – Ennar
      Dec 3 '18 at 19:00


















    3












    $begingroup$

    Hint: Choose a basis for $mathrm M_2(mathbb R)$, say ${E_{11},E_{12},E_{21},E_{22}}$. Calculate $f(E_{ij})$ and write it in the basis. This will give you matrix for $f$ (it's $4times 4$). Proceed to diagonalize it as usual.



    It might help if you think of $mathrm M_2(mathbb R)cong mathbb R^4$. Then $f$ becomes $$f(a,b,c,d) = (a,b,3c-a,3d-b).$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      So then, I just have to diagonalize the following matrix, right? $ begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ -1 & 0 & 3 & 0 \ 0 & -1 & 0 & 3 end{pmatrix} $
      $endgroup$
      – Gibbs
      Dec 3 '18 at 18:58










    • $begingroup$
      @Gibbs, correct.
      $endgroup$
      – Ennar
      Dec 3 '18 at 18:58










    • $begingroup$
      This was easier than I thought! Thanks!!
      $endgroup$
      – Gibbs
      Dec 3 '18 at 19:00










    • $begingroup$
      @Gibbs, you are welcome. Don't let algebra structure of $mathrm M_2$ confuse you - it's just $4$-dimensional vector space with canonical basis.
      $endgroup$
      – Ennar
      Dec 3 '18 at 19:00
















    3












    3








    3





    $begingroup$

    Hint: Choose a basis for $mathrm M_2(mathbb R)$, say ${E_{11},E_{12},E_{21},E_{22}}$. Calculate $f(E_{ij})$ and write it in the basis. This will give you matrix for $f$ (it's $4times 4$). Proceed to diagonalize it as usual.



    It might help if you think of $mathrm M_2(mathbb R)cong mathbb R^4$. Then $f$ becomes $$f(a,b,c,d) = (a,b,3c-a,3d-b).$$






    share|cite|improve this answer











    $endgroup$



    Hint: Choose a basis for $mathrm M_2(mathbb R)$, say ${E_{11},E_{12},E_{21},E_{22}}$. Calculate $f(E_{ij})$ and write it in the basis. This will give you matrix for $f$ (it's $4times 4$). Proceed to diagonalize it as usual.



    It might help if you think of $mathrm M_2(mathbb R)cong mathbb R^4$. Then $f$ becomes $$f(a,b,c,d) = (a,b,3c-a,3d-b).$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 3 '18 at 18:57

























    answered Dec 3 '18 at 18:52









    EnnarEnnar

    14.4k32443




    14.4k32443












    • $begingroup$
      So then, I just have to diagonalize the following matrix, right? $ begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ -1 & 0 & 3 & 0 \ 0 & -1 & 0 & 3 end{pmatrix} $
      $endgroup$
      – Gibbs
      Dec 3 '18 at 18:58










    • $begingroup$
      @Gibbs, correct.
      $endgroup$
      – Ennar
      Dec 3 '18 at 18:58










    • $begingroup$
      This was easier than I thought! Thanks!!
      $endgroup$
      – Gibbs
      Dec 3 '18 at 19:00










    • $begingroup$
      @Gibbs, you are welcome. Don't let algebra structure of $mathrm M_2$ confuse you - it's just $4$-dimensional vector space with canonical basis.
      $endgroup$
      – Ennar
      Dec 3 '18 at 19:00




















    • $begingroup$
      So then, I just have to diagonalize the following matrix, right? $ begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ -1 & 0 & 3 & 0 \ 0 & -1 & 0 & 3 end{pmatrix} $
      $endgroup$
      – Gibbs
      Dec 3 '18 at 18:58










    • $begingroup$
      @Gibbs, correct.
      $endgroup$
      – Ennar
      Dec 3 '18 at 18:58










    • $begingroup$
      This was easier than I thought! Thanks!!
      $endgroup$
      – Gibbs
      Dec 3 '18 at 19:00










    • $begingroup$
      @Gibbs, you are welcome. Don't let algebra structure of $mathrm M_2$ confuse you - it's just $4$-dimensional vector space with canonical basis.
      $endgroup$
      – Ennar
      Dec 3 '18 at 19:00


















    $begingroup$
    So then, I just have to diagonalize the following matrix, right? $ begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ -1 & 0 & 3 & 0 \ 0 & -1 & 0 & 3 end{pmatrix} $
    $endgroup$
    – Gibbs
    Dec 3 '18 at 18:58




    $begingroup$
    So then, I just have to diagonalize the following matrix, right? $ begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ -1 & 0 & 3 & 0 \ 0 & -1 & 0 & 3 end{pmatrix} $
    $endgroup$
    – Gibbs
    Dec 3 '18 at 18:58












    $begingroup$
    @Gibbs, correct.
    $endgroup$
    – Ennar
    Dec 3 '18 at 18:58




    $begingroup$
    @Gibbs, correct.
    $endgroup$
    – Ennar
    Dec 3 '18 at 18:58












    $begingroup$
    This was easier than I thought! Thanks!!
    $endgroup$
    – Gibbs
    Dec 3 '18 at 19:00




    $begingroup$
    This was easier than I thought! Thanks!!
    $endgroup$
    – Gibbs
    Dec 3 '18 at 19:00












    $begingroup$
    @Gibbs, you are welcome. Don't let algebra structure of $mathrm M_2$ confuse you - it's just $4$-dimensional vector space with canonical basis.
    $endgroup$
    – Ennar
    Dec 3 '18 at 19:00






    $begingroup$
    @Gibbs, you are welcome. Don't let algebra structure of $mathrm M_2$ confuse you - it's just $4$-dimensional vector space with canonical basis.
    $endgroup$
    – Ennar
    Dec 3 '18 at 19:00













    1












    $begingroup$

    Hint Let $lambda$ an eigenvalue and $Ane 0$ an associated eigenvector so



    $$f(A)=lambda Aiff begin{pmatrix}1-lambda &0\ -1 & 3-lambdaend{pmatrix}A=0$$



    Since $Ane0$ so $begin{pmatrix}1-lambda &0\ -1 & 3-lambdaend{pmatrix}$ should not be invertible hence $lambdain{1,3}$. Now for every value of $lambda$ solve the equation $f(A)=lambda A$ for $A=begin{pmatrix}a &b\ c & dend{pmatrix}$ to find the associated eigenspace.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Hint Let $lambda$ an eigenvalue and $Ane 0$ an associated eigenvector so



      $$f(A)=lambda Aiff begin{pmatrix}1-lambda &0\ -1 & 3-lambdaend{pmatrix}A=0$$



      Since $Ane0$ so $begin{pmatrix}1-lambda &0\ -1 & 3-lambdaend{pmatrix}$ should not be invertible hence $lambdain{1,3}$. Now for every value of $lambda$ solve the equation $f(A)=lambda A$ for $A=begin{pmatrix}a &b\ c & dend{pmatrix}$ to find the associated eigenspace.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Hint Let $lambda$ an eigenvalue and $Ane 0$ an associated eigenvector so



        $$f(A)=lambda Aiff begin{pmatrix}1-lambda &0\ -1 & 3-lambdaend{pmatrix}A=0$$



        Since $Ane0$ so $begin{pmatrix}1-lambda &0\ -1 & 3-lambdaend{pmatrix}$ should not be invertible hence $lambdain{1,3}$. Now for every value of $lambda$ solve the equation $f(A)=lambda A$ for $A=begin{pmatrix}a &b\ c & dend{pmatrix}$ to find the associated eigenspace.






        share|cite|improve this answer











        $endgroup$



        Hint Let $lambda$ an eigenvalue and $Ane 0$ an associated eigenvector so



        $$f(A)=lambda Aiff begin{pmatrix}1-lambda &0\ -1 & 3-lambdaend{pmatrix}A=0$$



        Since $Ane0$ so $begin{pmatrix}1-lambda &0\ -1 & 3-lambdaend{pmatrix}$ should not be invertible hence $lambdain{1,3}$. Now for every value of $lambda$ solve the equation $f(A)=lambda A$ for $A=begin{pmatrix}a &b\ c & dend{pmatrix}$ to find the associated eigenspace.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 3 '18 at 19:00

























        answered Dec 3 '18 at 18:55









        As soon as possibleAs soon as possible

        38218




        38218






























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