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







share|cite|improve this question





























    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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 5 at 21:35









      greedoid

      38.6k114797




      38.6k114797










      asked Jan 4 at 23:06









      VirtualUserVirtualUser

      57712




      57712






















          2 Answers
          2






          active

          oldest

          votes


















          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































            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





















              Your Answer





              StackExchange.ifUsing("editor", function () {
              return StackExchange.using("mathjaxEditing", function () {
              StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
              StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
              });
              });
              }, "mathjax-editing");

              StackExchange.ready(function() {
              var channelOptions = {
              tags: "".split(" "),
              id: "69"
              };
              initTagRenderer("".split(" "), "".split(" "), channelOptions);

              StackExchange.using("externalEditor", function() {
              // Have to fire editor after snippets, if snippets enabled
              if (StackExchange.settings.snippets.snippetsEnabled) {
              StackExchange.using("snippets", function() {
              createEditor();
              });
              }
              else {
              createEditor();
              }
              });

              function createEditor() {
              StackExchange.prepareEditor({
              heartbeatType: 'answer',
              autoActivateHeartbeat: false,
              convertImagesToLinks: true,
              noModals: true,
              showLowRepImageUploadWarning: true,
              reputationToPostImages: 10,
              bindNavPrevention: true,
              postfix: "",
              imageUploader: {
              brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
              contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
              allowUrls: true
              },
              noCode: true, onDemand: true,
              discardSelector: ".discard-answer"
              ,immediatelyShowMarkdownHelp:true
              });


              }
              });














              draft saved

              draft discarded


















              StackExchange.ready(
              function () {
              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3062219%2ffind-m-in-mathbb-c-such-that-fp-pt-m-p1tm2-p0t2-is-iso%23new-answer', 'question_page');
              }
              );

              Post as a guest















              Required, but never shown

























              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              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














                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












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 5 at 21:40









                          mechanodroidmechanodroid

                          27k62446




                          27k62446






























                              draft saved

                              draft discarded




















































                              Thanks for contributing an answer to Mathematics Stack Exchange!


                              • Please be sure to answer the question. Provide details and share your research!

                              But avoid …



                              • Asking for help, clarification, or responding to other answers.

                              • Making statements based on opinion; back them up with references or personal experience.


                              Use MathJax to format equations. MathJax reference.


                              To learn more, see our tips on writing great answers.




                              draft saved


                              draft discarded














                              StackExchange.ready(
                              function () {
                              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3062219%2ffind-m-in-mathbb-c-such-that-fp-pt-m-p1tm2-p0t2-is-iso%23new-answer', 'question_page');
                              }
                              );

                              Post as a guest















                              Required, but never shown





















































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown

































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown







                              -

                              Popular posts from this blog

                              An IMO inspired problem

                              Image
                              14 4 This problem from IMO 1988 is said to be one of the most elegant ones in functional equations. Problem : The function $f$ is defined on the set of all positive integers as follows: begin{align} f(1) = 1, f(3) &= 3, f(2n) = f(n), \ f(4n+1) &= 2f(2n+1) - f(n), \ f(4n+3) &= 3 f(2n+1) - 2f(n) end{align} Find the number of $n$ with $f(n) = n, 1 leq n leq 1988$. The main idea towards the solution is realizing that these conditions stand for the fact that $f(n)$ just reverses the digits in the binary representation of the number $n$. So, essentially the solution is finding the number of binary pallindromes $leq 1988_{10}$. My question is the following : How to reformulate the problem so that $f(n)$ reverses the digits of $n$ in its ternary representation. Or even better, can we reformulate it for a
                              Read more

                              Management

                              Image
                              Management From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Manager" redirects here. For other uses, see Management (disambiguation) and Manager (disambiguation). An organization chart for the United States Coast Guard shows the hierarchy of managerial roles in that organization. Business administration Management of a business Accounting Management accounting Financial accounting Financial audit Business entities Cooperative Corporation Limited liability company Partnership Sole proprietorship State-owned enterprise Corporate governance Annual general meeting Board of directors Supervisory board Advisory board Audit committee Corporate law Commercial law Constitutional documents Contract Corporate crime Corporate liability Insolvency law International trade law Mergers and acquisitions Economics Commodity Pu
                              Read more

                              Investment

                              Image
                              Investment From Wikipedia, the free encyclopedia Jump to navigation Jump to search This article is about investment in finance. For investment in macroeconomics, see Investment (macroeconomics). For other uses, see Investment (disambiguation). "Invest" redirects here. For the term in meteorology, see Invest (meteorology). This article needs additional citations for verification . Please help improve this article by adding citations to reliable sources. Unsourced material may be challenged and removed. ( February 2017 ) (Learn how and when to remove this template message) In general, to invest is to distribute money in the expectation of some benefit in the future – for example, investment in durable goods, in real estate by the service industry, in factories for manufacturing, in product development, and in research and development. However, this article focuses specifically on investment in financial
                              Read more