Find the change-of-coordinate matrices












0














Let F be the field with 4 elements, F={0,1,x,y}. Let $V=P_3(F)$ be the space of polynomials p(t) of degree ≤3. (We can't use x as the polynomial variable since it is already used as an element of the field F; so we use t instead.)



a) Find the Lagrange interpolation polynomials $p_i(t)∈V, i=0,1,2,3$ corresponding to



$c_0=0, c_1=1, c_2=x, c_3=y$.



b) Consider the two bases of V given by



β={$1,t,t^2,t^3$}$, γ={p_0(t),p_1(t),p_2(t),p_3(t)}$
Find the change-of-coordinate matrices



$Q=[IV]^γ_β, P=[IV]^β_γ$.



part a) i did$$begin{bmatrix}0frac{(x-1)(x-2)(x-3)}{(0-1)(0-2)(0-3)} +1frac{(x-0)(x-2)(x-3)}{(1-1)(1-2)(1-3)} +xfrac{(x-0)(x-1)(x-3)}{(2-1)(2-2)(2-3)}+yfrac{(x-0)(x-1)(x-2)}{(3-1)(3-2)(3-3)} end{bmatrix}.$$
i have no idea abot part b, anyone can help






linear-algebra







share|cite|improve this question



























    0














    Let F be the field with 4 elements, F={0,1,x,y}. Let $V=P_3(F)$ be the space of polynomials p(t) of degree ≤3. (We can't use x as the polynomial variable since it is already used as an element of the field F; so we use t instead.)



    a) Find the Lagrange interpolation polynomials $p_i(t)∈V, i=0,1,2,3$ corresponding to



    $c_0=0, c_1=1, c_2=x, c_3=y$.



    b) Consider the two bases of V given by



    β={$1,t,t^2,t^3$}$, γ={p_0(t),p_1(t),p_2(t),p_3(t)}$
    Find the change-of-coordinate matrices



    $Q=[IV]^γ_β, P=[IV]^β_γ$.



    part a) i did$$begin{bmatrix}0frac{(x-1)(x-2)(x-3)}{(0-1)(0-2)(0-3)} +1frac{(x-0)(x-2)(x-3)}{(1-1)(1-2)(1-3)} +xfrac{(x-0)(x-1)(x-3)}{(2-1)(2-2)(2-3)}+yfrac{(x-0)(x-1)(x-2)}{(3-1)(3-2)(3-3)} end{bmatrix}.$$
    i have no idea abot part b, anyone can help






    linear-algebra







    share|cite|improve this question

























      0












      0








      0


      1





      Let F be the field with 4 elements, F={0,1,x,y}. Let $V=P_3(F)$ be the space of polynomials p(t) of degree ≤3. (We can't use x as the polynomial variable since it is already used as an element of the field F; so we use t instead.)



      a) Find the Lagrange interpolation polynomials $p_i(t)∈V, i=0,1,2,3$ corresponding to



      $c_0=0, c_1=1, c_2=x, c_3=y$.



      b) Consider the two bases of V given by



      β={$1,t,t^2,t^3$}$, γ={p_0(t),p_1(t),p_2(t),p_3(t)}$
      Find the change-of-coordinate matrices



      $Q=[IV]^γ_β, P=[IV]^β_γ$.



      part a) i did$$begin{bmatrix}0frac{(x-1)(x-2)(x-3)}{(0-1)(0-2)(0-3)} +1frac{(x-0)(x-2)(x-3)}{(1-1)(1-2)(1-3)} +xfrac{(x-0)(x-1)(x-3)}{(2-1)(2-2)(2-3)}+yfrac{(x-0)(x-1)(x-2)}{(3-1)(3-2)(3-3)} end{bmatrix}.$$
      i have no idea abot part b, anyone can help






      linear-algebra







      share|cite|improve this question













      Let F be the field with 4 elements, F={0,1,x,y}. Let $V=P_3(F)$ be the space of polynomials p(t) of degree ≤3. (We can't use x as the polynomial variable since it is already used as an element of the field F; so we use t instead.)



      a) Find the Lagrange interpolation polynomials $p_i(t)∈V, i=0,1,2,3$ corresponding to



      $c_0=0, c_1=1, c_2=x, c_3=y$.



      b) Consider the two bases of V given by



      β={$1,t,t^2,t^3$}$, γ={p_0(t),p_1(t),p_2(t),p_3(t)}$
      Find the change-of-coordinate matrices



      $Q=[IV]^γ_β, P=[IV]^β_γ$.



      part a) i did$$begin{bmatrix}0frac{(x-1)(x-2)(x-3)}{(0-1)(0-2)(0-3)} +1frac{(x-0)(x-2)(x-3)}{(1-1)(1-2)(1-3)} +xfrac{(x-0)(x-1)(x-3)}{(2-1)(2-2)(2-3)}+yfrac{(x-0)(x-1)(x-2)}{(3-1)(3-2)(3-3)} end{bmatrix}.$$
      i have no idea abot part b, anyone can help






      linear-algebra




      linear-algebra






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 22 '18 at 2:03









      DORCTDORCT

      407




      407






















          1 Answer
          1






          active

          oldest

          votes


















          0














          Note that as indicated by the problem, $t$ should be used for polynomial variable, not $x$ (since we already use $x$ for element in the field $F$).



          I believe the problems in part (a) asks for four polynomials $p_i$ which form a (Lagrange) basis of $V$. In particular, $p_i(t)=displaystyleprod_{j=1,jne i}^3 frac{t-c_j}{c_i-c_j}$. For example, $p_1(t)=dfrac{(t-0)(t-x)(t-y)}{(1-0)(1-x)(1-y)}$ since $c_1=1$. (If you don't believe that these four are indeed a basis of $V$, you can try to prove it yourself or you can read the proof at here). I don't think your answer for part (a) is correct.



          For (b), the transition matrix from $beta= (1,t,t^2,t^3)$ to $beta'=(p_0(x),p_1(x),p_2(x),p_3(x))$ is exactly the Vandermonde matrix
          $P_{beta to beta'}= begin{pmatrix} 1 & 0 & 0 & 0 \
          1 & 1 & 1 & 1 \
          1 & x & x^2 & x^3 \
          1 & y & y^2 & y^3
          end{pmatrix}$          . In particular if a polynomial $f(t)=a_0+a_1t+a_2t^2+a_3t^3 in V$
          and $P_{betato beta'} (a_0,a_1,a_2,a_3)^T=(b_0,b_1,b_2,b_3)$ then
          $f(t)=b_0p_0(x)+b_1p_1(x)+b_2p_2(x)+b_3p_3(x)$.



          In fact, $boxed{b_i=f(a_i) text{ for
          all } 0 le i le 3}$           (hint: what is $p_j(a_i)$ for each $0 le i,j le 3$),
          which is the main reason why the above statement is true.






          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%2f3008652%2ffind-the-change-of-coordinate-matrices%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            0














            Note that as indicated by the problem, $t$ should be used for polynomial variable, not $x$ (since we already use $x$ for element in the field $F$).



            I believe the problems in part (a) asks for four polynomials $p_i$ which form a (Lagrange) basis of $V$. In particular, $p_i(t)=displaystyleprod_{j=1,jne i}^3 frac{t-c_j}{c_i-c_j}$. For example, $p_1(t)=dfrac{(t-0)(t-x)(t-y)}{(1-0)(1-x)(1-y)}$ since $c_1=1$. (If you don't believe that these four are indeed a basis of $V$, you can try to prove it yourself or you can read the proof at here). I don't think your answer for part (a) is correct.



            For (b), the transition matrix from $beta= (1,t,t^2,t^3)$ to $beta'=(p_0(x),p_1(x),p_2(x),p_3(x))$ is exactly the Vandermonde matrix
            $P_{beta to beta'}= begin{pmatrix} 1 & 0 & 0 & 0 \
            1 & 1 & 1 & 1 \
            1 & x & x^2 & x^3 \
            1 & y & y^2 & y^3
            end{pmatrix}$            . In particular if a polynomial $f(t)=a_0+a_1t+a_2t^2+a_3t^3 in V$
            and $P_{betato beta'} (a_0,a_1,a_2,a_3)^T=(b_0,b_1,b_2,b_3)$ then
            $f(t)=b_0p_0(x)+b_1p_1(x)+b_2p_2(x)+b_3p_3(x)$.



            In fact, $boxed{b_i=f(a_i) text{ for
            all } 0 le i le 3}$             (hint: what is $p_j(a_i)$ for each $0 le i,j le 3$),
            which is the main reason why the above statement is true.






            share|cite|improve this answer




























              0














              Note that as indicated by the problem, $t$ should be used for polynomial variable, not $x$ (since we already use $x$ for element in the field $F$).



              I believe the problems in part (a) asks for four polynomials $p_i$ which form a (Lagrange) basis of $V$. In particular, $p_i(t)=displaystyleprod_{j=1,jne i}^3 frac{t-c_j}{c_i-c_j}$. For example, $p_1(t)=dfrac{(t-0)(t-x)(t-y)}{(1-0)(1-x)(1-y)}$ since $c_1=1$. (If you don't believe that these four are indeed a basis of $V$, you can try to prove it yourself or you can read the proof at here). I don't think your answer for part (a) is correct.



              For (b), the transition matrix from $beta= (1,t,t^2,t^3)$ to $beta'=(p_0(x),p_1(x),p_2(x),p_3(x))$ is exactly the Vandermonde matrix
              $P_{beta to beta'}= begin{pmatrix} 1 & 0 & 0 & 0 \
              1 & 1 & 1 & 1 \
              1 & x & x^2 & x^3 \
              1 & y & y^2 & y^3
              end{pmatrix}$              . In particular if a polynomial $f(t)=a_0+a_1t+a_2t^2+a_3t^3 in V$
              and $P_{betato beta'} (a_0,a_1,a_2,a_3)^T=(b_0,b_1,b_2,b_3)$ then
              $f(t)=b_0p_0(x)+b_1p_1(x)+b_2p_2(x)+b_3p_3(x)$.



              In fact, $boxed{b_i=f(a_i) text{ for
              all } 0 le i le 3}$               (hint: what is $p_j(a_i)$ for each $0 le i,j le 3$),
              which is the main reason why the above statement is true.






              share|cite|improve this answer


























                0












                0








                0






                Note that as indicated by the problem, $t$ should be used for polynomial variable, not $x$ (since we already use $x$ for element in the field $F$).



                I believe the problems in part (a) asks for four polynomials $p_i$ which form a (Lagrange) basis of $V$. In particular, $p_i(t)=displaystyleprod_{j=1,jne i}^3 frac{t-c_j}{c_i-c_j}$. For example, $p_1(t)=dfrac{(t-0)(t-x)(t-y)}{(1-0)(1-x)(1-y)}$ since $c_1=1$. (If you don't believe that these four are indeed a basis of $V$, you can try to prove it yourself or you can read the proof at here). I don't think your answer for part (a) is correct.



                For (b), the transition matrix from $beta= (1,t,t^2,t^3)$ to $beta'=(p_0(x),p_1(x),p_2(x),p_3(x))$ is exactly the Vandermonde matrix
                $P_{beta to beta'}= begin{pmatrix} 1 & 0 & 0 & 0 \
                1 & 1 & 1 & 1 \
                1 & x & x^2 & x^3 \
                1 & y & y^2 & y^3
                end{pmatrix}$                . In particular if a polynomial $f(t)=a_0+a_1t+a_2t^2+a_3t^3 in V$
                and $P_{betato beta'} (a_0,a_1,a_2,a_3)^T=(b_0,b_1,b_2,b_3)$ then
                $f(t)=b_0p_0(x)+b_1p_1(x)+b_2p_2(x)+b_3p_3(x)$.



                In fact, $boxed{b_i=f(a_i) text{ for
                all } 0 le i le 3}$                 (hint: what is $p_j(a_i)$ for each $0 le i,j le 3$),
                which is the main reason why the above statement is true.






                share|cite|improve this answer














                Note that as indicated by the problem, $t$ should be used for polynomial variable, not $x$ (since we already use $x$ for element in the field $F$).



                I believe the problems in part (a) asks for four polynomials $p_i$ which form a (Lagrange) basis of $V$. In particular, $p_i(t)=displaystyleprod_{j=1,jne i}^3 frac{t-c_j}{c_i-c_j}$. For example, $p_1(t)=dfrac{(t-0)(t-x)(t-y)}{(1-0)(1-x)(1-y)}$ since $c_1=1$. (If you don't believe that these four are indeed a basis of $V$, you can try to prove it yourself or you can read the proof at here). I don't think your answer for part (a) is correct.



                For (b), the transition matrix from $beta= (1,t,t^2,t^3)$ to $beta'=(p_0(x),p_1(x),p_2(x),p_3(x))$ is exactly the Vandermonde matrix
                $P_{beta to beta'}= begin{pmatrix} 1 & 0 & 0 & 0 \
                1 & 1 & 1 & 1 \
                1 & x & x^2 & x^3 \
                1 & y & y^2 & y^3
                end{pmatrix}$                . In particular if a polynomial $f(t)=a_0+a_1t+a_2t^2+a_3t^3 in V$
                and $P_{betato beta'} (a_0,a_1,a_2,a_3)^T=(b_0,b_1,b_2,b_3)$ then
                $f(t)=b_0p_0(x)+b_1p_1(x)+b_2p_2(x)+b_3p_3(x)$.



                In fact, $boxed{b_i=f(a_i) text{ for
                all } 0 le i le 3}$                 (hint: what is $p_j(a_i)$ for each $0 le i,j le 3$),
                which is the main reason why the above statement is true.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Nov 23 '18 at 10:20

























                answered Nov 23 '18 at 7:05









                TenguTengu

                2,6321921




                2,6321921






























                    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.





                    Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                    Please pay close attention to the following guidance:


                    • 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.


                    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%2f3008652%2ffind-the-change-of-coordinate-matrices%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

                    'app-layout' is not a known element: how to share Component with different Modules

                    Image
                    0 My application was working fine until I decided to go with lazy loading: So, my shared component looks like this: import { Component, Renderer2 } from '@angular/core'; export interface FormModel { captcha?: string; } @Component({ selector: 'app-layout', templateUrl: './app-layout.component.html', styleUrls: ['./app-layout.css'] }) export class FullLayoutComponent { } app-layout.component.ts I am using this component in my Another component say school-home.component.html like this: <app-layout> </app-layout> and it was working fine until I created a Module school-home.module.ts like this: import { NgModule } from '@angular/core'; import { CommonModule } from '@angular/common'; import { SchoolhomeRoutingModule ...
                    Read more

                    android studio warns about leanback feature tag usage required on manifest while using Unity exported app?

                    Image
                    .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 This is a simple project showing cubes on Unity, exported to android project and imported on Android studio. On maifest file it showed following warning: Expecting <uses-feature android:name="android.software.leanback" android:required="false" /> tag I have no information about it. And also on default emptyview android project does not use/mention this feature. I know it is required by TV's and such but why its not added by default? Why is this? And Should I concern about it? android unity3d ...
                    Read more

                    WPF add header to Image with URL pettitions [duplicate]

                    Image
                    up vote 0 down vote favorite This question already has an answer here: Image source URI requiring authorization 3 answers Add header to all* HTTP requests on a machine 1 answer first of all sorry for my english. Recently I added some security in the backend of my project, and now I need to send a validation code(apiKey) inside the headers to retrieve the images. I have a problem with ALL images in my project that I had attached in my UI. For example, before I had: <Image Height="90" Source="{Binding Video.Thumbnail}"/> Now I need to change my code to create a CUSTOM petition add the apiKey in the header:...
                    Read more