Find $m in mathbb C$ such that $ f(p) = p'(t-m)-p(1)(t+m)^2 + p(0)t^2 $ is isomorphism












3














I have solved this task, but I am not sure If I done this correctly - can you check my way and idea of solving this (calculations are not important)?




Find $m in mathbb C$ such that linear transformation $f in L(mathbb C[x]_2,mathbb C[x]_2)$: $$
f(p) = p'(t-m)-p(1)(t+m)^2 + p(0)t^2$$
is isomorphism




I know that: $ dim Y = dim im(f)$ and $ dim X = dim im(f)$ but X=Y so $ dim im(f) = 3 $



Firstly I calculate formula for $f(p)$ where $ p(t) = at^2+bt+c$
$$f(p) = ... = a(-2m-m^2-t^2) + \b(-t^2-2tm-m^2+1) + \c(-2tm-m^2)$$



so (after factor by $(-1)$)
$$f(p) = span([1,0,m^2+2m]^T,[1,2m,m^2-1]^T[0,2m,m^2]^T) $$
But I know that $ dim im(f) = 3 $ so columns must be linearly independent so I do RREF and I get $$([1,0,0]^T,[0,1,0]^T,[0,0,1]^T) $$ for $ m in mathbb C setminus left{ 0,-4right} $ otherwise $$[1,0,m^2+2m]^T,[1,2m,m^2-1]^T[0,2m,m^2]^T $$ are linearly dependent.
ps: I am adding tag solution verification but system changes it to proof-verification :(






linear-algebra proof-verification







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    3














    I have solved this task, but I am not sure If I done this correctly - can you check my way and idea of solving this (calculations are not important)?




    Find $m in mathbb C$ such that linear transformation $f in L(mathbb C[x]_2,mathbb C[x]_2)$: $$
    f(p) = p'(t-m)-p(1)(t+m)^2 + p(0)t^2$$
    is isomorphism




    I know that: $ dim Y = dim im(f)$ and $ dim X = dim im(f)$ but X=Y so $ dim im(f) = 3 $



    Firstly I calculate formula for $f(p)$ where $ p(t) = at^2+bt+c$
    $$f(p) = ... = a(-2m-m^2-t^2) + \b(-t^2-2tm-m^2+1) + \c(-2tm-m^2)$$



    so (after factor by $(-1)$)
    $$f(p) = span([1,0,m^2+2m]^T,[1,2m,m^2-1]^T[0,2m,m^2]^T) $$
    But I know that $ dim im(f) = 3 $ so columns must be linearly independent so I do RREF and I get $$([1,0,0]^T,[0,1,0]^T,[0,0,1]^T) $$ for $ m in mathbb C setminus left{ 0,-4right} $ otherwise $$[1,0,m^2+2m]^T,[1,2m,m^2-1]^T[0,2m,m^2]^T $$ are linearly dependent.
    ps: I am adding tag solution verification but system changes it to proof-verification :(






    linear-algebra proof-verification







    share|cite|improve this question



























      3












      3








      3







      I have solved this task, but I am not sure If I done this correctly - can you check my way and idea of solving this (calculations are not important)?




      Find $m in mathbb C$ such that linear transformation $f in L(mathbb C[x]_2,mathbb C[x]_2)$: $$
      f(p) = p'(t-m)-p(1)(t+m)^2 + p(0)t^2$$
      is isomorphism




      I know that: $ dim Y = dim im(f)$ and $ dim X = dim im(f)$ but X=Y so $ dim im(f) = 3 $



      Firstly I calculate formula for $f(p)$ where $ p(t) = at^2+bt+c$
      $$f(p) = ... = a(-2m-m^2-t^2) + \b(-t^2-2tm-m^2+1) + \c(-2tm-m^2)$$



      so (after factor by $(-1)$)
      $$f(p) = span([1,0,m^2+2m]^T,[1,2m,m^2-1]^T[0,2m,m^2]^T) $$
      But I know that $ dim im(f) = 3 $ so columns must be linearly independent so I do RREF and I get $$([1,0,0]^T,[0,1,0]^T,[0,0,1]^T) $$ for $ m in mathbb C setminus left{ 0,-4right} $ otherwise $$[1,0,m^2+2m]^T,[1,2m,m^2-1]^T[0,2m,m^2]^T $$ are linearly dependent.
      ps: I am adding tag solution verification but system changes it to proof-verification :(






      linear-algebra proof-verification







      share|cite|improve this question















      I have solved this task, but I am not sure If I done this correctly - can you check my way and idea of solving this (calculations are not important)?




      Find $m in mathbb C$ such that linear transformation $f in L(mathbb C[x]_2,mathbb C[x]_2)$: $$
      f(p) = p'(t-m)-p(1)(t+m)^2 + p(0)t^2$$
      is isomorphism




      I know that: $ dim Y = dim im(f)$ and $ dim X = dim im(f)$ but X=Y so $ dim im(f) = 3 $



      Firstly I calculate formula for $f(p)$ where $ p(t) = at^2+bt+c$
      $$f(p) = ... = a(-2m-m^2-t^2) + \b(-t^2-2tm-m^2+1) + \c(-2tm-m^2)$$



      so (after factor by $(-1)$)
      $$f(p) = span([1,0,m^2+2m]^T,[1,2m,m^2-1]^T[0,2m,m^2]^T) $$
      But I know that $ dim im(f) = 3 $ so columns must be linearly independent so I do RREF and I get $$([1,0,0]^T,[0,1,0]^T,[0,0,1]^T) $$ for $ m in mathbb C setminus left{ 0,-4right} $ otherwise $$[1,0,m^2+2m]^T,[1,2m,m^2-1]^T[0,2m,m^2]^T $$ are linearly dependent.
      ps: I am adding tag solution verification but system changes it to proof-verification :(






      linear-algebra proof-verification




      linear-algebra proof-verification






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      edited Jan 5 at 21:35









      greedoid

      38.6k114797




      38.6k114797










      asked Jan 4 at 23:06









      VirtualUserVirtualUser

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          We need only to find for wich $m$ given linear trasformation is injective i.e. $ker(f) = {0}$, so for which $m$, the equation $f(p)=0$ forces $p=0$.



          Write $p(x) = ax^2+bx+c$, then $p'(x)= 2ax+b$. Since we have $$p'(t-m)-p(1)(t+m)^2 + p(0)t^2 =0;;;;forall t$$



          we have $$-2am+b-m^2p(1)+(2a-2mp(1))t+(c-p(1))t^2=0$$




          So $c=p(1)=a+b+cimplies a=-b$ and



          $a = m(a+b+c)implies a=mc$ and



          $m^2(a+b+c)+2am=b implies cm^2+2am+a=0implies 3am+a=0implies boxed{m=-{1over 3}}$ or $a=0$. But later is true if $mc=0$ so $boxed{m=0}$ (or $c=0$).






          share|cite|improve this answer































            1














            I'm getting the same result as @greedoid. Namely, $f$ is an isomorphism if and only if $m ne 0, -frac13$.



            Just calculate the images of the basis $1, t, t^2$:



            $$f(1) = -2mt-m^2$$
            $$f(t) = 1-t^2-2mt-m^2$$
            $$f(t^2) = 2t-2m-t^2-2mt-m^2$$



            Now $f$ is an isomorphism if and only if ${f(1), f(t), f(t)^2}$ is linearly indepedendent, which is equivalent to
            $${-f(1), f(t)-f(1), f(t)^2-f(t)} = {m^2+2mt,1-t^2,2t-2m-1}$$
            being linearly independent.



            Since $deg(1-t^2) = 2$, the only way this set can be linearly dependent is if $m^2+2mt = lambda(2t-2m-1)$ for some $lambda$, which gives $lambda = m$ and $m(3m+1) = 0$, or $m in left{0, -frac13right}$.






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              We need only to find for wich $m$ given linear trasformation is injective i.e. $ker(f) = {0}$, so for which $m$, the equation $f(p)=0$ forces $p=0$.



              Write $p(x) = ax^2+bx+c$, then $p'(x)= 2ax+b$. Since we have $$p'(t-m)-p(1)(t+m)^2 + p(0)t^2 =0;;;;forall t$$



              we have $$-2am+b-m^2p(1)+(2a-2mp(1))t+(c-p(1))t^2=0$$




              So $c=p(1)=a+b+cimplies a=-b$ and



              $a = m(a+b+c)implies a=mc$ and



              $m^2(a+b+c)+2am=b implies cm^2+2am+a=0implies 3am+a=0implies boxed{m=-{1over 3}}$ or $a=0$. But later is true if $mc=0$ so $boxed{m=0}$ (or $c=0$).






              share|cite|improve this answer




























                2














                We need only to find for wich $m$ given linear trasformation is injective i.e. $ker(f) = {0}$, so for which $m$, the equation $f(p)=0$ forces $p=0$.



                Write $p(x) = ax^2+bx+c$, then $p'(x)= 2ax+b$. Since we have $$p'(t-m)-p(1)(t+m)^2 + p(0)t^2 =0;;;;forall t$$



                we have $$-2am+b-m^2p(1)+(2a-2mp(1))t+(c-p(1))t^2=0$$




                So $c=p(1)=a+b+cimplies a=-b$ and



                $a = m(a+b+c)implies a=mc$ and



                $m^2(a+b+c)+2am=b implies cm^2+2am+a=0implies 3am+a=0implies boxed{m=-{1over 3}}$ or $a=0$. But later is true if $mc=0$ so $boxed{m=0}$ (or $c=0$).






                share|cite|improve this answer


























                  2












                  2








                  2






                  We need only to find for wich $m$ given linear trasformation is injective i.e. $ker(f) = {0}$, so for which $m$, the equation $f(p)=0$ forces $p=0$.



                  Write $p(x) = ax^2+bx+c$, then $p'(x)= 2ax+b$. Since we have $$p'(t-m)-p(1)(t+m)^2 + p(0)t^2 =0;;;;forall t$$



                  we have $$-2am+b-m^2p(1)+(2a-2mp(1))t+(c-p(1))t^2=0$$




                  So $c=p(1)=a+b+cimplies a=-b$ and



                  $a = m(a+b+c)implies a=mc$ and



                  $m^2(a+b+c)+2am=b implies cm^2+2am+a=0implies 3am+a=0implies boxed{m=-{1over 3}}$ or $a=0$. But later is true if $mc=0$ so $boxed{m=0}$ (or $c=0$).






                  share|cite|improve this answer














                  We need only to find for wich $m$ given linear trasformation is injective i.e. $ker(f) = {0}$, so for which $m$, the equation $f(p)=0$ forces $p=0$.



                  Write $p(x) = ax^2+bx+c$, then $p'(x)= 2ax+b$. Since we have $$p'(t-m)-p(1)(t+m)^2 + p(0)t^2 =0;;;;forall t$$



                  we have $$-2am+b-m^2p(1)+(2a-2mp(1))t+(c-p(1))t^2=0$$




                  So $c=p(1)=a+b+cimplies a=-b$ and



                  $a = m(a+b+c)implies a=mc$ and



                  $m^2(a+b+c)+2am=b implies cm^2+2am+a=0implies 3am+a=0implies boxed{m=-{1over 3}}$ or $a=0$. But later is true if $mc=0$ so $boxed{m=0}$ (or $c=0$).







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 5 at 21:45

























                  answered Jan 5 at 21:27









                  greedoidgreedoid

                  38.6k114797




                  38.6k114797























                      1














                      I'm getting the same result as @greedoid. Namely, $f$ is an isomorphism if and only if $m ne 0, -frac13$.



                      Just calculate the images of the basis $1, t, t^2$:



                      $$f(1) = -2mt-m^2$$
                      $$f(t) = 1-t^2-2mt-m^2$$
                      $$f(t^2) = 2t-2m-t^2-2mt-m^2$$



                      Now $f$ is an isomorphism if and only if ${f(1), f(t), f(t)^2}$ is linearly indepedendent, which is equivalent to
                      $${-f(1), f(t)-f(1), f(t)^2-f(t)} = {m^2+2mt,1-t^2,2t-2m-1}$$
                      being linearly independent.



                      Since $deg(1-t^2) = 2$, the only way this set can be linearly dependent is if $m^2+2mt = lambda(2t-2m-1)$ for some $lambda$, which gives $lambda = m$ and $m(3m+1) = 0$, or $m in left{0, -frac13right}$.






                      share|cite|improve this answer


























                        1














                        I'm getting the same result as @greedoid. Namely, $f$ is an isomorphism if and only if $m ne 0, -frac13$.



                        Just calculate the images of the basis $1, t, t^2$:



                        $$f(1) = -2mt-m^2$$
                        $$f(t) = 1-t^2-2mt-m^2$$
                        $$f(t^2) = 2t-2m-t^2-2mt-m^2$$



                        Now $f$ is an isomorphism if and only if ${f(1), f(t), f(t)^2}$ is linearly indepedendent, which is equivalent to
                        $${-f(1), f(t)-f(1), f(t)^2-f(t)} = {m^2+2mt,1-t^2,2t-2m-1}$$
                        being linearly independent.



                        Since $deg(1-t^2) = 2$, the only way this set can be linearly dependent is if $m^2+2mt = lambda(2t-2m-1)$ for some $lambda$, which gives $lambda = m$ and $m(3m+1) = 0$, or $m in left{0, -frac13right}$.






                        share|cite|improve this answer
























                          1












                          1








                          1






                          I'm getting the same result as @greedoid. Namely, $f$ is an isomorphism if and only if $m ne 0, -frac13$.



                          Just calculate the images of the basis $1, t, t^2$:



                          $$f(1) = -2mt-m^2$$
                          $$f(t) = 1-t^2-2mt-m^2$$
                          $$f(t^2) = 2t-2m-t^2-2mt-m^2$$



                          Now $f$ is an isomorphism if and only if ${f(1), f(t), f(t)^2}$ is linearly indepedendent, which is equivalent to
                          $${-f(1), f(t)-f(1), f(t)^2-f(t)} = {m^2+2mt,1-t^2,2t-2m-1}$$
                          being linearly independent.



                          Since $deg(1-t^2) = 2$, the only way this set can be linearly dependent is if $m^2+2mt = lambda(2t-2m-1)$ for some $lambda$, which gives $lambda = m$ and $m(3m+1) = 0$, or $m in left{0, -frac13right}$.






                          share|cite|improve this answer












                          I'm getting the same result as @greedoid. Namely, $f$ is an isomorphism if and only if $m ne 0, -frac13$.



                          Just calculate the images of the basis $1, t, t^2$:



                          $$f(1) = -2mt-m^2$$
                          $$f(t) = 1-t^2-2mt-m^2$$
                          $$f(t^2) = 2t-2m-t^2-2mt-m^2$$



                          Now $f$ is an isomorphism if and only if ${f(1), f(t), f(t)^2}$ is linearly indepedendent, which is equivalent to
                          $${-f(1), f(t)-f(1), f(t)^2-f(t)} = {m^2+2mt,1-t^2,2t-2m-1}$$
                          being linearly independent.



                          Since $deg(1-t^2) = 2$, the only way this set can be linearly dependent is if $m^2+2mt = lambda(2t-2m-1)$ for some $lambda$, which gives $lambda = m$ and $m(3m+1) = 0$, or $m in left{0, -frac13right}$.







                          share|cite|improve this answer












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                          answered Jan 5 at 21:40









                          mechanodroidmechanodroid

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                          27k62446






























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