Find a matrix $B$ such that $B^3 = A$












2












$begingroup$



$$A=begin{pmatrix} 1 & -1 \ -2 & 1 end{pmatrix}$$



Find a matrix $B$ such that $B^3$ = A




My attempt:



I found $lambda_1= 1+{sqrt 2}$ and $lambda_2= 1-{sqrt 2}$



I also found their corresponding eigenvectors $vec v_1 =begin{pmatrix} frac{-sqrt 2}{2} \ 1 end{pmatrix}$ and $vec v_2 = begin{pmatrix} frac{sqrt 2}{2} \ 1 end{pmatrix}$



I know the Power function of a matrix formula $A=PDP^{-1}$



Because it's the cubed root I'm looking for I don't know how to get the cubed root of the eigenvaules and keep the maths neat. Is there another way to solve this problem or an I going the wrong way about doing it ?










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$endgroup$








  • 1




    $begingroup$
    What's the problem with cube roots? Just write $sqrt[3]{1pm sqrt{2}}$. These numbers are unique if you want them to be real.
    $endgroup$
    – Peter Franek
    Jun 21 '16 at 18:15












  • $begingroup$
    What numbers are the components of $B$ allowed from?
    $endgroup$
    – mvw
    Jun 21 '16 at 18:38
















2












$begingroup$



$$A=begin{pmatrix} 1 & -1 \ -2 & 1 end{pmatrix}$$



Find a matrix $B$ such that $B^3$ = A




My attempt:



I found $lambda_1= 1+{sqrt 2}$ and $lambda_2= 1-{sqrt 2}$



I also found their corresponding eigenvectors $vec v_1 =begin{pmatrix} frac{-sqrt 2}{2} \ 1 end{pmatrix}$ and $vec v_2 = begin{pmatrix} frac{sqrt 2}{2} \ 1 end{pmatrix}$



I know the Power function of a matrix formula $A=PDP^{-1}$



Because it's the cubed root I'm looking for I don't know how to get the cubed root of the eigenvaules and keep the maths neat. Is there another way to solve this problem or an I going the wrong way about doing it ?










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What's the problem with cube roots? Just write $sqrt[3]{1pm sqrt{2}}$. These numbers are unique if you want them to be real.
    $endgroup$
    – Peter Franek
    Jun 21 '16 at 18:15












  • $begingroup$
    What numbers are the components of $B$ allowed from?
    $endgroup$
    – mvw
    Jun 21 '16 at 18:38














2












2








2


1



$begingroup$



$$A=begin{pmatrix} 1 & -1 \ -2 & 1 end{pmatrix}$$



Find a matrix $B$ such that $B^3$ = A




My attempt:



I found $lambda_1= 1+{sqrt 2}$ and $lambda_2= 1-{sqrt 2}$



I also found their corresponding eigenvectors $vec v_1 =begin{pmatrix} frac{-sqrt 2}{2} \ 1 end{pmatrix}$ and $vec v_2 = begin{pmatrix} frac{sqrt 2}{2} \ 1 end{pmatrix}$



I know the Power function of a matrix formula $A=PDP^{-1}$



Because it's the cubed root I'm looking for I don't know how to get the cubed root of the eigenvaules and keep the maths neat. Is there another way to solve this problem or an I going the wrong way about doing it ?










share|cite|improve this question











$endgroup$





$$A=begin{pmatrix} 1 & -1 \ -2 & 1 end{pmatrix}$$



Find a matrix $B$ such that $B^3$ = A




My attempt:



I found $lambda_1= 1+{sqrt 2}$ and $lambda_2= 1-{sqrt 2}$



I also found their corresponding eigenvectors $vec v_1 =begin{pmatrix} frac{-sqrt 2}{2} \ 1 end{pmatrix}$ and $vec v_2 = begin{pmatrix} frac{sqrt 2}{2} \ 1 end{pmatrix}$



I know the Power function of a matrix formula $A=PDP^{-1}$



Because it's the cubed root I'm looking for I don't know how to get the cubed root of the eigenvaules and keep the maths neat. Is there another way to solve this problem or an I going the wrong way about doing it ?







linear-algebra matrices






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edited Jun 21 '16 at 20:01









Jennifer

8,41721737




8,41721737










asked Jun 21 '16 at 18:12









Patrick MoloneyPatrick Moloney

312114




312114








  • 1




    $begingroup$
    What's the problem with cube roots? Just write $sqrt[3]{1pm sqrt{2}}$. These numbers are unique if you want them to be real.
    $endgroup$
    – Peter Franek
    Jun 21 '16 at 18:15












  • $begingroup$
    What numbers are the components of $B$ allowed from?
    $endgroup$
    – mvw
    Jun 21 '16 at 18:38














  • 1




    $begingroup$
    What's the problem with cube roots? Just write $sqrt[3]{1pm sqrt{2}}$. These numbers are unique if you want them to be real.
    $endgroup$
    – Peter Franek
    Jun 21 '16 at 18:15












  • $begingroup$
    What numbers are the components of $B$ allowed from?
    $endgroup$
    – mvw
    Jun 21 '16 at 18:38








1




1




$begingroup$
What's the problem with cube roots? Just write $sqrt[3]{1pm sqrt{2}}$. These numbers are unique if you want them to be real.
$endgroup$
– Peter Franek
Jun 21 '16 at 18:15






$begingroup$
What's the problem with cube roots? Just write $sqrt[3]{1pm sqrt{2}}$. These numbers are unique if you want them to be real.
$endgroup$
– Peter Franek
Jun 21 '16 at 18:15














$begingroup$
What numbers are the components of $B$ allowed from?
$endgroup$
– mvw
Jun 21 '16 at 18:38




$begingroup$
What numbers are the components of $B$ allowed from?
$endgroup$
– mvw
Jun 21 '16 at 18:38










4 Answers
4






active

oldest

votes


















6












$begingroup$

you find a matrix $P=left(
begin{array}{cc}
frac{-sqrt{2}}2 & frac{sqrt{2}}2 \
1 & 1
end{array}
right) $ diagonalizes $ A $ that is
$P^{-1}AP=left(begin{array}{cc}
1+sqrt{2} & 0 \
0 & 1-sqrt{2}
end{array}right) $,
then we look
for a simple matrix $ M =PBP^{-1} $ such that it is diagonal and $
M^3 = D$ an simple solution is $M=left(
begin{array}{cc}
sqrt[3]{left( 1+sqrt{2}right) } & 0 \
0 & -sqrt[3]{left( -1+sqrt{2}right) }
end{array}
right) $ and so $B=PDP^{-1}$ is a solution. Precisely $$B= left(
begin{array}{cc}
frac 12sqrt[3]{left( 1+sqrt{2}right) }-frac 12sqrt[3]{left( -1+sqrt{%
2}right) } & -frac 14sqrt{2}sqrt[3]{left( 1+sqrt{2}right) }-frac 14%
sqrt{2}sqrt[3]{left( -1+sqrt{2}right) } \
-frac 12sqrt{2}sqrt[3]{left( 1+sqrt{2}right) }-frac 12sqrt{2}sqrt[3%
]{left( -1+sqrt{2}right) } & frac 12sqrt[3]{left( 1+sqrt{2}right) }%
-frac 12sqrt[3]{left( -1+sqrt{2}right) }
end{array}
right) $$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Suppose that there is a matrix $B=begin{pmatrix} b_1 & b_3 \ b_2 & b_4end{pmatrix}$ such that $B^3=A$. Then we obtain two linear equations by applying the Buchberger algorithm to the four equations, namely
    $$
    b_2=2b_3,; b_4=b_1.
    $$
    This yields $b_1= - 4b_3^2 - 2b_3 + 1$, and all equations are satisfied if and only if $b_3$ satisfies
    $$
    8b_3^3 - 3b_3 + 1=0.
    $$
    We obtain exactly one real solution for $b_3$, and hence for $B$ with $B^3=A$.






    share|cite|improve this answer











    $endgroup$





















      1












      $begingroup$

      You're almost done.



      Take $B = PD^{1/3}P^{-1}$ where $D^{1/3}=operatorname{diag}(lambda_1^{1/3},lambda_2^{1/3})$ and try to compute $B^3$.






      share|cite|improve this answer









      $endgroup$





















        0












        $begingroup$

        By the Cayley-Hamilton theorem, any analytic function $f$ of an $ntimes n$ matrix can be expressed as a polynomial $p(A)$ of degree at most $n-1$. So, a cube root $B$ of $A$ can be expressed in the form $aI+bA$ for some to-be-determined coefficients $a$ and $b$. Now, if $lambda$ is an eigenvalue of $A$, then we also have that $f(lambda)=p(lambda)$. For this problem, this second property generates a system of two linear equations in the unknown coefficients $a$ and $b$. Solve this system.






        share|cite|improve this answer









        $endgroup$













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          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          6












          $begingroup$

          you find a matrix $P=left(
          begin{array}{cc}
          frac{-sqrt{2}}2 & frac{sqrt{2}}2 \
          1 & 1
          end{array}
          right) $ diagonalizes $ A $ that is
          $P^{-1}AP=left(begin{array}{cc}
          1+sqrt{2} & 0 \
          0 & 1-sqrt{2}
          end{array}right) $,
          then we look
          for a simple matrix $ M =PBP^{-1} $ such that it is diagonal and $
          M^3 = D$ an simple solution is $M=left(
          begin{array}{cc}
          sqrt[3]{left( 1+sqrt{2}right) } & 0 \
          0 & -sqrt[3]{left( -1+sqrt{2}right) }
          end{array}
          right) $ and so $B=PDP^{-1}$ is a solution. Precisely $$B= left(
          begin{array}{cc}
          frac 12sqrt[3]{left( 1+sqrt{2}right) }-frac 12sqrt[3]{left( -1+sqrt{%
          2}right) } & -frac 14sqrt{2}sqrt[3]{left( 1+sqrt{2}right) }-frac 14%
          sqrt{2}sqrt[3]{left( -1+sqrt{2}right) } \
          -frac 12sqrt{2}sqrt[3]{left( 1+sqrt{2}right) }-frac 12sqrt{2}sqrt[3%
          ]{left( -1+sqrt{2}right) } & frac 12sqrt[3]{left( 1+sqrt{2}right) }%
          -frac 12sqrt[3]{left( -1+sqrt{2}right) }
          end{array}
          right) $$






          share|cite|improve this answer









          $endgroup$


















            6












            $begingroup$

            you find a matrix $P=left(
            begin{array}{cc}
            frac{-sqrt{2}}2 & frac{sqrt{2}}2 \
            1 & 1
            end{array}
            right) $ diagonalizes $ A $ that is
            $P^{-1}AP=left(begin{array}{cc}
            1+sqrt{2} & 0 \
            0 & 1-sqrt{2}
            end{array}right) $,
            then we look
            for a simple matrix $ M =PBP^{-1} $ such that it is diagonal and $
            M^3 = D$ an simple solution is $M=left(
            begin{array}{cc}
            sqrt[3]{left( 1+sqrt{2}right) } & 0 \
            0 & -sqrt[3]{left( -1+sqrt{2}right) }
            end{array}
            right) $ and so $B=PDP^{-1}$ is a solution. Precisely $$B= left(
            begin{array}{cc}
            frac 12sqrt[3]{left( 1+sqrt{2}right) }-frac 12sqrt[3]{left( -1+sqrt{%
            2}right) } & -frac 14sqrt{2}sqrt[3]{left( 1+sqrt{2}right) }-frac 14%
            sqrt{2}sqrt[3]{left( -1+sqrt{2}right) } \
            -frac 12sqrt{2}sqrt[3]{left( 1+sqrt{2}right) }-frac 12sqrt{2}sqrt[3%
            ]{left( -1+sqrt{2}right) } & frac 12sqrt[3]{left( 1+sqrt{2}right) }%
            -frac 12sqrt[3]{left( -1+sqrt{2}right) }
            end{array}
            right) $$






            share|cite|improve this answer









            $endgroup$
















              6












              6








              6





              $begingroup$

              you find a matrix $P=left(
              begin{array}{cc}
              frac{-sqrt{2}}2 & frac{sqrt{2}}2 \
              1 & 1
              end{array}
              right) $ diagonalizes $ A $ that is
              $P^{-1}AP=left(begin{array}{cc}
              1+sqrt{2} & 0 \
              0 & 1-sqrt{2}
              end{array}right) $,
              then we look
              for a simple matrix $ M =PBP^{-1} $ such that it is diagonal and $
              M^3 = D$ an simple solution is $M=left(
              begin{array}{cc}
              sqrt[3]{left( 1+sqrt{2}right) } & 0 \
              0 & -sqrt[3]{left( -1+sqrt{2}right) }
              end{array}
              right) $ and so $B=PDP^{-1}$ is a solution. Precisely $$B= left(
              begin{array}{cc}
              frac 12sqrt[3]{left( 1+sqrt{2}right) }-frac 12sqrt[3]{left( -1+sqrt{%
              2}right) } & -frac 14sqrt{2}sqrt[3]{left( 1+sqrt{2}right) }-frac 14%
              sqrt{2}sqrt[3]{left( -1+sqrt{2}right) } \
              -frac 12sqrt{2}sqrt[3]{left( 1+sqrt{2}right) }-frac 12sqrt{2}sqrt[3%
              ]{left( -1+sqrt{2}right) } & frac 12sqrt[3]{left( 1+sqrt{2}right) }%
              -frac 12sqrt[3]{left( -1+sqrt{2}right) }
              end{array}
              right) $$






              share|cite|improve this answer









              $endgroup$



              you find a matrix $P=left(
              begin{array}{cc}
              frac{-sqrt{2}}2 & frac{sqrt{2}}2 \
              1 & 1
              end{array}
              right) $ diagonalizes $ A $ that is
              $P^{-1}AP=left(begin{array}{cc}
              1+sqrt{2} & 0 \
              0 & 1-sqrt{2}
              end{array}right) $,
              then we look
              for a simple matrix $ M =PBP^{-1} $ such that it is diagonal and $
              M^3 = D$ an simple solution is $M=left(
              begin{array}{cc}
              sqrt[3]{left( 1+sqrt{2}right) } & 0 \
              0 & -sqrt[3]{left( -1+sqrt{2}right) }
              end{array}
              right) $ and so $B=PDP^{-1}$ is a solution. Precisely $$B= left(
              begin{array}{cc}
              frac 12sqrt[3]{left( 1+sqrt{2}right) }-frac 12sqrt[3]{left( -1+sqrt{%
              2}right) } & -frac 14sqrt{2}sqrt[3]{left( 1+sqrt{2}right) }-frac 14%
              sqrt{2}sqrt[3]{left( -1+sqrt{2}right) } \
              -frac 12sqrt{2}sqrt[3]{left( 1+sqrt{2}right) }-frac 12sqrt{2}sqrt[3%
              ]{left( -1+sqrt{2}right) } & frac 12sqrt[3]{left( 1+sqrt{2}right) }%
              -frac 12sqrt[3]{left( -1+sqrt{2}right) }
              end{array}
              right) $$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Jun 21 '16 at 18:54









              m.idayam.idaya

              1,371410




              1,371410























                  1












                  $begingroup$

                  Suppose that there is a matrix $B=begin{pmatrix} b_1 & b_3 \ b_2 & b_4end{pmatrix}$ such that $B^3=A$. Then we obtain two linear equations by applying the Buchberger algorithm to the four equations, namely
                  $$
                  b_2=2b_3,; b_4=b_1.
                  $$
                  This yields $b_1= - 4b_3^2 - 2b_3 + 1$, and all equations are satisfied if and only if $b_3$ satisfies
                  $$
                  8b_3^3 - 3b_3 + 1=0.
                  $$
                  We obtain exactly one real solution for $b_3$, and hence for $B$ with $B^3=A$.






                  share|cite|improve this answer











                  $endgroup$


















                    1












                    $begingroup$

                    Suppose that there is a matrix $B=begin{pmatrix} b_1 & b_3 \ b_2 & b_4end{pmatrix}$ such that $B^3=A$. Then we obtain two linear equations by applying the Buchberger algorithm to the four equations, namely
                    $$
                    b_2=2b_3,; b_4=b_1.
                    $$
                    This yields $b_1= - 4b_3^2 - 2b_3 + 1$, and all equations are satisfied if and only if $b_3$ satisfies
                    $$
                    8b_3^3 - 3b_3 + 1=0.
                    $$
                    We obtain exactly one real solution for $b_3$, and hence for $B$ with $B^3=A$.






                    share|cite|improve this answer











                    $endgroup$
















                      1












                      1








                      1





                      $begingroup$

                      Suppose that there is a matrix $B=begin{pmatrix} b_1 & b_3 \ b_2 & b_4end{pmatrix}$ such that $B^3=A$. Then we obtain two linear equations by applying the Buchberger algorithm to the four equations, namely
                      $$
                      b_2=2b_3,; b_4=b_1.
                      $$
                      This yields $b_1= - 4b_3^2 - 2b_3 + 1$, and all equations are satisfied if and only if $b_3$ satisfies
                      $$
                      8b_3^3 - 3b_3 + 1=0.
                      $$
                      We obtain exactly one real solution for $b_3$, and hence for $B$ with $B^3=A$.






                      share|cite|improve this answer











                      $endgroup$



                      Suppose that there is a matrix $B=begin{pmatrix} b_1 & b_3 \ b_2 & b_4end{pmatrix}$ such that $B^3=A$. Then we obtain two linear equations by applying the Buchberger algorithm to the four equations, namely
                      $$
                      b_2=2b_3,; b_4=b_1.
                      $$
                      This yields $b_1= - 4b_3^2 - 2b_3 + 1$, and all equations are satisfied if and only if $b_3$ satisfies
                      $$
                      8b_3^3 - 3b_3 + 1=0.
                      $$
                      We obtain exactly one real solution for $b_3$, and hence for $B$ with $B^3=A$.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Jun 21 '16 at 18:40

























                      answered Jun 21 '16 at 18:34









                      Dietrich BurdeDietrich Burde

                      78.6k64386




                      78.6k64386























                          1












                          $begingroup$

                          You're almost done.



                          Take $B = PD^{1/3}P^{-1}$ where $D^{1/3}=operatorname{diag}(lambda_1^{1/3},lambda_2^{1/3})$ and try to compute $B^3$.






                          share|cite|improve this answer









                          $endgroup$


















                            1












                            $begingroup$

                            You're almost done.



                            Take $B = PD^{1/3}P^{-1}$ where $D^{1/3}=operatorname{diag}(lambda_1^{1/3},lambda_2^{1/3})$ and try to compute $B^3$.






                            share|cite|improve this answer









                            $endgroup$
















                              1












                              1








                              1





                              $begingroup$

                              You're almost done.



                              Take $B = PD^{1/3}P^{-1}$ where $D^{1/3}=operatorname{diag}(lambda_1^{1/3},lambda_2^{1/3})$ and try to compute $B^3$.






                              share|cite|improve this answer









                              $endgroup$



                              You're almost done.



                              Take $B = PD^{1/3}P^{-1}$ where $D^{1/3}=operatorname{diag}(lambda_1^{1/3},lambda_2^{1/3})$ and try to compute $B^3$.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Jun 21 '16 at 18:44









                              SurbSurb

                              37.5k94375




                              37.5k94375























                                  0












                                  $begingroup$

                                  By the Cayley-Hamilton theorem, any analytic function $f$ of an $ntimes n$ matrix can be expressed as a polynomial $p(A)$ of degree at most $n-1$. So, a cube root $B$ of $A$ can be expressed in the form $aI+bA$ for some to-be-determined coefficients $a$ and $b$. Now, if $lambda$ is an eigenvalue of $A$, then we also have that $f(lambda)=p(lambda)$. For this problem, this second property generates a system of two linear equations in the unknown coefficients $a$ and $b$. Solve this system.






                                  share|cite|improve this answer









                                  $endgroup$


















                                    0












                                    $begingroup$

                                    By the Cayley-Hamilton theorem, any analytic function $f$ of an $ntimes n$ matrix can be expressed as a polynomial $p(A)$ of degree at most $n-1$. So, a cube root $B$ of $A$ can be expressed in the form $aI+bA$ for some to-be-determined coefficients $a$ and $b$. Now, if $lambda$ is an eigenvalue of $A$, then we also have that $f(lambda)=p(lambda)$. For this problem, this second property generates a system of two linear equations in the unknown coefficients $a$ and $b$. Solve this system.






                                    share|cite|improve this answer









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      By the Cayley-Hamilton theorem, any analytic function $f$ of an $ntimes n$ matrix can be expressed as a polynomial $p(A)$ of degree at most $n-1$. So, a cube root $B$ of $A$ can be expressed in the form $aI+bA$ for some to-be-determined coefficients $a$ and $b$. Now, if $lambda$ is an eigenvalue of $A$, then we also have that $f(lambda)=p(lambda)$. For this problem, this second property generates a system of two linear equations in the unknown coefficients $a$ and $b$. Solve this system.






                                      share|cite|improve this answer









                                      $endgroup$



                                      By the Cayley-Hamilton theorem, any analytic function $f$ of an $ntimes n$ matrix can be expressed as a polynomial $p(A)$ of degree at most $n-1$. So, a cube root $B$ of $A$ can be expressed in the form $aI+bA$ for some to-be-determined coefficients $a$ and $b$. Now, if $lambda$ is an eigenvalue of $A$, then we also have that $f(lambda)=p(lambda)$. For this problem, this second property generates a system of two linear equations in the unknown coefficients $a$ and $b$. Solve this system.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Dec 5 '18 at 22:10









                                      amdamd

                                      29.7k21050




                                      29.7k21050






























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