Linear operator find function












0












$begingroup$


Let $T(x,y)=(x,3x-y)$ be a linear operator on $R^2$ and let $f(x)=x^2-4$. Find $f(T)$



I'm very new to this and I'm stuck on what it's asking for. I'm guessing I make a matrix from the transformation and then use $rref$. I'm not sure where to go from there. Any guidance is appreciated.



I took matrix A and squared it and it returned the identity matrix. I then subtracted 4 times the identity matrix and got just (-3x, -3y). Can someone clarify what I'm doing incorrectly. Thanks






linear-algebra transformation







share|cite|improve this question











$endgroup$












  • $begingroup$
    Hi, and welcome to MSE. You'll get more help if you show evidence of your own efforts.
    $endgroup$
    – user334732
    Dec 17 '18 at 19:24










  • $begingroup$
    Thanks i have added to my question.
    $endgroup$
    – Jim Jones
    Dec 18 '18 at 14:58
















0












$begingroup$


Let $T(x,y)=(x,3x-y)$ be a linear operator on $R^2$ and let $f(x)=x^2-4$. Find $f(T)$



I'm very new to this and I'm stuck on what it's asking for. I'm guessing I make a matrix from the transformation and then use $rref$. I'm not sure where to go from there. Any guidance is appreciated.



I took matrix A and squared it and it returned the identity matrix. I then subtracted 4 times the identity matrix and got just (-3x, -3y). Can someone clarify what I'm doing incorrectly. Thanks






linear-algebra transformation







share|cite|improve this question











$endgroup$












  • $begingroup$
    Hi, and welcome to MSE. You'll get more help if you show evidence of your own efforts.
    $endgroup$
    – user334732
    Dec 17 '18 at 19:24










  • $begingroup$
    Thanks i have added to my question.
    $endgroup$
    – Jim Jones
    Dec 18 '18 at 14:58














0












0








0





$begingroup$


Let $T(x,y)=(x,3x-y)$ be a linear operator on $R^2$ and let $f(x)=x^2-4$. Find $f(T)$



I'm very new to this and I'm stuck on what it's asking for. I'm guessing I make a matrix from the transformation and then use $rref$. I'm not sure where to go from there. Any guidance is appreciated.



I took matrix A and squared it and it returned the identity matrix. I then subtracted 4 times the identity matrix and got just (-3x, -3y). Can someone clarify what I'm doing incorrectly. Thanks






linear-algebra transformation







share|cite|improve this question











$endgroup$




Let $T(x,y)=(x,3x-y)$ be a linear operator on $R^2$ and let $f(x)=x^2-4$. Find $f(T)$



I'm very new to this and I'm stuck on what it's asking for. I'm guessing I make a matrix from the transformation and then use $rref$. I'm not sure where to go from there. Any guidance is appreciated.



I took matrix A and squared it and it returned the identity matrix. I then subtracted 4 times the identity matrix and got just (-3x, -3y). Can someone clarify what I'm doing incorrectly. Thanks






linear-algebra transformation




linear-algebra transformation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 18 '18 at 14:57







Jim Jones

















asked Dec 17 '18 at 14:36









Jim JonesJim Jones

12




12












  • $begingroup$
    Hi, and welcome to MSE. You'll get more help if you show evidence of your own efforts.
    $endgroup$
    – user334732
    Dec 17 '18 at 19:24










  • $begingroup$
    Thanks i have added to my question.
    $endgroup$
    – Jim Jones
    Dec 18 '18 at 14:58


















  • $begingroup$
    Hi, and welcome to MSE. You'll get more help if you show evidence of your own efforts.
    $endgroup$
    – user334732
    Dec 17 '18 at 19:24










  • $begingroup$
    Thanks i have added to my question.
    $endgroup$
    – Jim Jones
    Dec 18 '18 at 14:58
















$begingroup$
Hi, and welcome to MSE. You'll get more help if you show evidence of your own efforts.
$endgroup$
– user334732
Dec 17 '18 at 19:24




$begingroup$
Hi, and welcome to MSE. You'll get more help if you show evidence of your own efforts.
$endgroup$
– user334732
Dec 17 '18 at 19:24












$begingroup$
Thanks i have added to my question.
$endgroup$
– Jim Jones
Dec 18 '18 at 14:58




$begingroup$
Thanks i have added to my question.
$endgroup$
– Jim Jones
Dec 18 '18 at 14:58










2 Answers
2






active

oldest

votes


















0












$begingroup$

$T$ is realized by the matrix



$M =begin{bmatrix} 1 & 0 \ 3 & -1 \ end{bmatrix}$.



Then compute $M^2-4I$. Can you proceed ?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks. I"m getting (-3x, -3y) by doing this but above says it should be (-3x, 6x-5y) . What am i missing?
    $endgroup$
    – Jim Jones
    Dec 17 '18 at 17:23



















0












$begingroup$

Alternatively, since $f(x)= x^2- 4$, $f(T)(vec{x})= T^2(vec{x})+ 4vec{x}$ where $T^2$ is the composition of T with itself: $f(t)= T(T(vec{x})+ 4vec{x}$. $T(vec{x})= T(begin{bmatrix}x \ y end{bmatrix})= begin{bmatrix}x \ 3x- yend{bmatrix}$ so $T^2(vec{x})= begin{bmatrix}x \ 3x+ (3x- y)end{bmatrix}= begin{bmatrix}x \ 6x- yend{bmatrix}$.



So $fleft(Tleft(begin{bmatrix}x \ y end{bmatrix}right)right)= begin{bmatrix}x \ 6x- yend{bmatrix}- begin{bmatrix}4x \ 4yend{bmatrix}= begin{bmatrix}-3x \ 6x- 5yend{bmatrix}$.






share|cite|improve this answer









$endgroup$













    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%2f3044024%2flinear-operator-find-function%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









    0












    $begingroup$

    $T$ is realized by the matrix



    $M =begin{bmatrix} 1 & 0 \ 3 & -1 \ end{bmatrix}$.



    Then compute $M^2-4I$. Can you proceed ?






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thanks. I"m getting (-3x, -3y) by doing this but above says it should be (-3x, 6x-5y) . What am i missing?
      $endgroup$
      – Jim Jones
      Dec 17 '18 at 17:23
















    0












    $begingroup$

    $T$ is realized by the matrix



    $M =begin{bmatrix} 1 & 0 \ 3 & -1 \ end{bmatrix}$.



    Then compute $M^2-4I$. Can you proceed ?






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thanks. I"m getting (-3x, -3y) by doing this but above says it should be (-3x, 6x-5y) . What am i missing?
      $endgroup$
      – Jim Jones
      Dec 17 '18 at 17:23














    0












    0








    0





    $begingroup$

    $T$ is realized by the matrix



    $M =begin{bmatrix} 1 & 0 \ 3 & -1 \ end{bmatrix}$.



    Then compute $M^2-4I$. Can you proceed ?






    share|cite|improve this answer









    $endgroup$



    $T$ is realized by the matrix



    $M =begin{bmatrix} 1 & 0 \ 3 & -1 \ end{bmatrix}$.



    Then compute $M^2-4I$. Can you proceed ?







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 17 '18 at 14:41









    FredFred

    46.9k1848




    46.9k1848












    • $begingroup$
      Thanks. I"m getting (-3x, -3y) by doing this but above says it should be (-3x, 6x-5y) . What am i missing?
      $endgroup$
      – Jim Jones
      Dec 17 '18 at 17:23


















    • $begingroup$
      Thanks. I"m getting (-3x, -3y) by doing this but above says it should be (-3x, 6x-5y) . What am i missing?
      $endgroup$
      – Jim Jones
      Dec 17 '18 at 17:23
















    $begingroup$
    Thanks. I"m getting (-3x, -3y) by doing this but above says it should be (-3x, 6x-5y) . What am i missing?
    $endgroup$
    – Jim Jones
    Dec 17 '18 at 17:23




    $begingroup$
    Thanks. I"m getting (-3x, -3y) by doing this but above says it should be (-3x, 6x-5y) . What am i missing?
    $endgroup$
    – Jim Jones
    Dec 17 '18 at 17:23











    0












    $begingroup$

    Alternatively, since $f(x)= x^2- 4$, $f(T)(vec{x})= T^2(vec{x})+ 4vec{x}$ where $T^2$ is the composition of T with itself: $f(t)= T(T(vec{x})+ 4vec{x}$. $T(vec{x})= T(begin{bmatrix}x \ y end{bmatrix})= begin{bmatrix}x \ 3x- yend{bmatrix}$ so $T^2(vec{x})= begin{bmatrix}x \ 3x+ (3x- y)end{bmatrix}= begin{bmatrix}x \ 6x- yend{bmatrix}$.



    So $fleft(Tleft(begin{bmatrix}x \ y end{bmatrix}right)right)= begin{bmatrix}x \ 6x- yend{bmatrix}- begin{bmatrix}4x \ 4yend{bmatrix}= begin{bmatrix}-3x \ 6x- 5yend{bmatrix}$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Alternatively, since $f(x)= x^2- 4$, $f(T)(vec{x})= T^2(vec{x})+ 4vec{x}$ where $T^2$ is the composition of T with itself: $f(t)= T(T(vec{x})+ 4vec{x}$. $T(vec{x})= T(begin{bmatrix}x \ y end{bmatrix})= begin{bmatrix}x \ 3x- yend{bmatrix}$ so $T^2(vec{x})= begin{bmatrix}x \ 3x+ (3x- y)end{bmatrix}= begin{bmatrix}x \ 6x- yend{bmatrix}$.



      So $fleft(Tleft(begin{bmatrix}x \ y end{bmatrix}right)right)= begin{bmatrix}x \ 6x- yend{bmatrix}- begin{bmatrix}4x \ 4yend{bmatrix}= begin{bmatrix}-3x \ 6x- 5yend{bmatrix}$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Alternatively, since $f(x)= x^2- 4$, $f(T)(vec{x})= T^2(vec{x})+ 4vec{x}$ where $T^2$ is the composition of T with itself: $f(t)= T(T(vec{x})+ 4vec{x}$. $T(vec{x})= T(begin{bmatrix}x \ y end{bmatrix})= begin{bmatrix}x \ 3x- yend{bmatrix}$ so $T^2(vec{x})= begin{bmatrix}x \ 3x+ (3x- y)end{bmatrix}= begin{bmatrix}x \ 6x- yend{bmatrix}$.



        So $fleft(Tleft(begin{bmatrix}x \ y end{bmatrix}right)right)= begin{bmatrix}x \ 6x- yend{bmatrix}- begin{bmatrix}4x \ 4yend{bmatrix}= begin{bmatrix}-3x \ 6x- 5yend{bmatrix}$.






        share|cite|improve this answer









        $endgroup$



        Alternatively, since $f(x)= x^2- 4$, $f(T)(vec{x})= T^2(vec{x})+ 4vec{x}$ where $T^2$ is the composition of T with itself: $f(t)= T(T(vec{x})+ 4vec{x}$. $T(vec{x})= T(begin{bmatrix}x \ y end{bmatrix})= begin{bmatrix}x \ 3x- yend{bmatrix}$ so $T^2(vec{x})= begin{bmatrix}x \ 3x+ (3x- y)end{bmatrix}= begin{bmatrix}x \ 6x- yend{bmatrix}$.



        So $fleft(Tleft(begin{bmatrix}x \ y end{bmatrix}right)right)= begin{bmatrix}x \ 6x- yend{bmatrix}- begin{bmatrix}4x \ 4yend{bmatrix}= begin{bmatrix}-3x \ 6x- 5yend{bmatrix}$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 17 '18 at 15:09









        user247327user247327

        11.1k1515




        11.1k1515






























            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%2f3044024%2flinear-operator-find-function%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

            Wiesbaden

            Image
            Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is
            Read more

            Marschland

            Image
            Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
            Read more

            Dieringhausen

            Image
            Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
            Read more