Show $(ev_alpha)_{alpha∈mathbb{R}}$ is linearly independent












0















Show $(ev_alpha)_{alpha∈mathbb{R}}$ is linearly independent



$(ev_alpha): mathbb{R}[X]rightarrowmathbb{R}, fmapsto f(alpha)$




So I need to show, maybe with a counter-example, that $lambda=0$ in $sumlambda ev_alpha=0$



How can I do it?






linear-algebra







share|cite|improve this question




















  • 1




    It's going to be difficult: how come you want to give a counterexample of something that you want to prove? PS: You need to replace $lambda$ by $lambda_alpha$.
    – John B
    Nov 21 at 20:51
















0















Show $(ev_alpha)_{alpha∈mathbb{R}}$ is linearly independent



$(ev_alpha): mathbb{R}[X]rightarrowmathbb{R}, fmapsto f(alpha)$




So I need to show, maybe with a counter-example, that $lambda=0$ in $sumlambda ev_alpha=0$



How can I do it?






linear-algebra







share|cite|improve this question




















  • 1




    It's going to be difficult: how come you want to give a counterexample of something that you want to prove? PS: You need to replace $lambda$ by $lambda_alpha$.
    – John B
    Nov 21 at 20:51














0












0








0


1






Show $(ev_alpha)_{alpha∈mathbb{R}}$ is linearly independent



$(ev_alpha): mathbb{R}[X]rightarrowmathbb{R}, fmapsto f(alpha)$




So I need to show, maybe with a counter-example, that $lambda=0$ in $sumlambda ev_alpha=0$



How can I do it?






linear-algebra







share|cite|improve this question
















Show $(ev_alpha)_{alpha∈mathbb{R}}$ is linearly independent



$(ev_alpha): mathbb{R}[X]rightarrowmathbb{R}, fmapsto f(alpha)$




So I need to show, maybe with a counter-example, that $lambda=0$ in $sumlambda ev_alpha=0$



How can I do it?






linear-algebra




linear-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 21 at 20:49









John B

12.2k51740




12.2k51740










asked Nov 21 at 20:47









Dada

6810




6810








  • 1




    It's going to be difficult: how come you want to give a counterexample of something that you want to prove? PS: You need to replace $lambda$ by $lambda_alpha$.
    – John B
    Nov 21 at 20:51














  • 1




    It's going to be difficult: how come you want to give a counterexample of something that you want to prove? PS: You need to replace $lambda$ by $lambda_alpha$.
    – John B
    Nov 21 at 20:51








1




1




It's going to be difficult: how come you want to give a counterexample of something that you want to prove? PS: You need to replace $lambda$ by $lambda_alpha$.
– John B
Nov 21 at 20:51




It's going to be difficult: how come you want to give a counterexample of something that you want to prove? PS: You need to replace $lambda$ by $lambda_alpha$.
– John B
Nov 21 at 20:51










1 Answer
1






active

oldest

votes


















2














An uncountable set is linearly independent if and only if every finite subset is linearly independent.



So suppose $U subset mathbb{R}$ is finite and there exists $lambda_alpha$ for each $alpha in U$ such that $ displaystyle sum_{alpha in U} lambda_alpha ev_alpha = 0$.



Then by definition of $ev_alpha$ we have $displaystyle sum_{alpha in U} lambda_alpha f(alpha) = 0$ for any $f in mathbb{R}[X]$.



Choose some $alpha' in U$ and define $displaystyle f_{alpha'}(x) =prod_{alpha in U - { alpha' }}(x-alpha)$.



Then by construction we have $f_{alpha'}(alpha') neq 0$, but $f_{alpha'}(alpha) = 0$ for all $alpha in U - {alpha'}$.



Then $displaystyle 0= sum_{alpha in U} lambda_alpha f_{alpha'}(alpha) = lambda_{alpha'}f_{alpha'}(alpha')$ and since $f_{alpha'}(alpha')neq 0$ we have $lambda_{alpha'}=0$.



Since our choice of $alpha'$ was arbitrary, we have $lambda_alpha=0$ for all $alpha in U$.



Thus the finite subset $(ev_alpha)_{alpha in U}$ is linearly independent.



And since our choice of $U$ was arbitrary, every finite subset is linearly independent.



Therefore $(ev_alpha)_{alpha in mathbb{R}}$ is linearly independent.






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%2f3008331%2fshow-ev-alpha-alpha%25e2%2588%2588-mathbbr-is-linearly-independent%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









    2














    An uncountable set is linearly independent if and only if every finite subset is linearly independent.



    So suppose $U subset mathbb{R}$ is finite and there exists $lambda_alpha$ for each $alpha in U$ such that $ displaystyle sum_{alpha in U} lambda_alpha ev_alpha = 0$.



    Then by definition of $ev_alpha$ we have $displaystyle sum_{alpha in U} lambda_alpha f(alpha) = 0$ for any $f in mathbb{R}[X]$.



    Choose some $alpha' in U$ and define $displaystyle f_{alpha'}(x) =prod_{alpha in U - { alpha' }}(x-alpha)$.



    Then by construction we have $f_{alpha'}(alpha') neq 0$, but $f_{alpha'}(alpha) = 0$ for all $alpha in U - {alpha'}$.



    Then $displaystyle 0= sum_{alpha in U} lambda_alpha f_{alpha'}(alpha) = lambda_{alpha'}f_{alpha'}(alpha')$ and since $f_{alpha'}(alpha')neq 0$ we have $lambda_{alpha'}=0$.



    Since our choice of $alpha'$ was arbitrary, we have $lambda_alpha=0$ for all $alpha in U$.



    Thus the finite subset $(ev_alpha)_{alpha in U}$ is linearly independent.



    And since our choice of $U$ was arbitrary, every finite subset is linearly independent.



    Therefore $(ev_alpha)_{alpha in mathbb{R}}$ is linearly independent.






    share|cite|improve this answer


























      2














      An uncountable set is linearly independent if and only if every finite subset is linearly independent.



      So suppose $U subset mathbb{R}$ is finite and there exists $lambda_alpha$ for each $alpha in U$ such that $ displaystyle sum_{alpha in U} lambda_alpha ev_alpha = 0$.



      Then by definition of $ev_alpha$ we have $displaystyle sum_{alpha in U} lambda_alpha f(alpha) = 0$ for any $f in mathbb{R}[X]$.



      Choose some $alpha' in U$ and define $displaystyle f_{alpha'}(x) =prod_{alpha in U - { alpha' }}(x-alpha)$.



      Then by construction we have $f_{alpha'}(alpha') neq 0$, but $f_{alpha'}(alpha) = 0$ for all $alpha in U - {alpha'}$.



      Then $displaystyle 0= sum_{alpha in U} lambda_alpha f_{alpha'}(alpha) = lambda_{alpha'}f_{alpha'}(alpha')$ and since $f_{alpha'}(alpha')neq 0$ we have $lambda_{alpha'}=0$.



      Since our choice of $alpha'$ was arbitrary, we have $lambda_alpha=0$ for all $alpha in U$.



      Thus the finite subset $(ev_alpha)_{alpha in U}$ is linearly independent.



      And since our choice of $U$ was arbitrary, every finite subset is linearly independent.



      Therefore $(ev_alpha)_{alpha in mathbb{R}}$ is linearly independent.






      share|cite|improve this answer
























        2












        2








        2






        An uncountable set is linearly independent if and only if every finite subset is linearly independent.



        So suppose $U subset mathbb{R}$ is finite and there exists $lambda_alpha$ for each $alpha in U$ such that $ displaystyle sum_{alpha in U} lambda_alpha ev_alpha = 0$.



        Then by definition of $ev_alpha$ we have $displaystyle sum_{alpha in U} lambda_alpha f(alpha) = 0$ for any $f in mathbb{R}[X]$.



        Choose some $alpha' in U$ and define $displaystyle f_{alpha'}(x) =prod_{alpha in U - { alpha' }}(x-alpha)$.



        Then by construction we have $f_{alpha'}(alpha') neq 0$, but $f_{alpha'}(alpha) = 0$ for all $alpha in U - {alpha'}$.



        Then $displaystyle 0= sum_{alpha in U} lambda_alpha f_{alpha'}(alpha) = lambda_{alpha'}f_{alpha'}(alpha')$ and since $f_{alpha'}(alpha')neq 0$ we have $lambda_{alpha'}=0$.



        Since our choice of $alpha'$ was arbitrary, we have $lambda_alpha=0$ for all $alpha in U$.



        Thus the finite subset $(ev_alpha)_{alpha in U}$ is linearly independent.



        And since our choice of $U$ was arbitrary, every finite subset is linearly independent.



        Therefore $(ev_alpha)_{alpha in mathbb{R}}$ is linearly independent.






        share|cite|improve this answer












        An uncountable set is linearly independent if and only if every finite subset is linearly independent.



        So suppose $U subset mathbb{R}$ is finite and there exists $lambda_alpha$ for each $alpha in U$ such that $ displaystyle sum_{alpha in U} lambda_alpha ev_alpha = 0$.



        Then by definition of $ev_alpha$ we have $displaystyle sum_{alpha in U} lambda_alpha f(alpha) = 0$ for any $f in mathbb{R}[X]$.



        Choose some $alpha' in U$ and define $displaystyle f_{alpha'}(x) =prod_{alpha in U - { alpha' }}(x-alpha)$.



        Then by construction we have $f_{alpha'}(alpha') neq 0$, but $f_{alpha'}(alpha) = 0$ for all $alpha in U - {alpha'}$.



        Then $displaystyle 0= sum_{alpha in U} lambda_alpha f_{alpha'}(alpha) = lambda_{alpha'}f_{alpha'}(alpha')$ and since $f_{alpha'}(alpha')neq 0$ we have $lambda_{alpha'}=0$.



        Since our choice of $alpha'$ was arbitrary, we have $lambda_alpha=0$ for all $alpha in U$.



        Thus the finite subset $(ev_alpha)_{alpha in U}$ is linearly independent.



        And since our choice of $U$ was arbitrary, every finite subset is linearly independent.



        Therefore $(ev_alpha)_{alpha in mathbb{R}}$ is linearly independent.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 27 at 23:29









        sfmiller940

        3615




        3615






























            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%2f3008331%2fshow-ev-alpha-alpha%25e2%2588%2588-mathbbr-is-linearly-independent%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

            Plaza Victoria

            Image
            Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
            Read more

            Puebla de Zaragoza

            Image
            Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
            Read more

            Musa

            Image
            Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
            Read more