What does this notation mean? How can an isomorphism be a function?












1












$begingroup$


enter image description here



I'm extremely confused with this notation.



Could someone translate the second and third line of this example into something a little more declarative or understandable for me?



On the second line it is given that there is an isomorphism between the space of bounded linear transformations from R^2 to R^3 and the space of 3x2 matrices. This means that there exists a bijective linear transformation between the two spaces. How can the isomorphism be used in a function declaration? How can I interpret the matrix given below in terms of that definition?



Thanks so much










share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    What else is an isomorphism going to be, other than a function? Check your definitions!
    $endgroup$
    – user3482749
    Jan 2 at 21:38






  • 2




    $begingroup$
    I'm not sure without seeing the full statement, but I think it should probably read "the function $A in mathcal{B}(mathbb{R}^2,mathbb{R}^3) cong M_{3,2}(mathbb{R})$ given by ...", i.e. the function given by the matrix is not the isomorphism itself.
    $endgroup$
    – Riley
    Jan 3 at 0:31


















1












$begingroup$


enter image description here



I'm extremely confused with this notation.



Could someone translate the second and third line of this example into something a little more declarative or understandable for me?



On the second line it is given that there is an isomorphism between the space of bounded linear transformations from R^2 to R^3 and the space of 3x2 matrices. This means that there exists a bijective linear transformation between the two spaces. How can the isomorphism be used in a function declaration? How can I interpret the matrix given below in terms of that definition?



Thanks so much










share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    What else is an isomorphism going to be, other than a function? Check your definitions!
    $endgroup$
    – user3482749
    Jan 2 at 21:38






  • 2




    $begingroup$
    I'm not sure without seeing the full statement, but I think it should probably read "the function $A in mathcal{B}(mathbb{R}^2,mathbb{R}^3) cong M_{3,2}(mathbb{R})$ given by ...", i.e. the function given by the matrix is not the isomorphism itself.
    $endgroup$
    – Riley
    Jan 3 at 0:31
















1












1








1





$begingroup$


enter image description here



I'm extremely confused with this notation.



Could someone translate the second and third line of this example into something a little more declarative or understandable for me?



On the second line it is given that there is an isomorphism between the space of bounded linear transformations from R^2 to R^3 and the space of 3x2 matrices. This means that there exists a bijective linear transformation between the two spaces. How can the isomorphism be used in a function declaration? How can I interpret the matrix given below in terms of that definition?



Thanks so much










share|cite|improve this question









$endgroup$




enter image description here



I'm extremely confused with this notation.



Could someone translate the second and third line of this example into something a little more declarative or understandable for me?



On the second line it is given that there is an isomorphism between the space of bounded linear transformations from R^2 to R^3 and the space of 3x2 matrices. This means that there exists a bijective linear transformation between the two spaces. How can the isomorphism be used in a function declaration? How can I interpret the matrix given below in terms of that definition?



Thanks so much







real-analysis functional-analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 2 at 21:36









Kohler FryerKohler Fryer

1102




1102








  • 2




    $begingroup$
    What else is an isomorphism going to be, other than a function? Check your definitions!
    $endgroup$
    – user3482749
    Jan 2 at 21:38






  • 2




    $begingroup$
    I'm not sure without seeing the full statement, but I think it should probably read "the function $A in mathcal{B}(mathbb{R}^2,mathbb{R}^3) cong M_{3,2}(mathbb{R})$ given by ...", i.e. the function given by the matrix is not the isomorphism itself.
    $endgroup$
    – Riley
    Jan 3 at 0:31
















  • 2




    $begingroup$
    What else is an isomorphism going to be, other than a function? Check your definitions!
    $endgroup$
    – user3482749
    Jan 2 at 21:38






  • 2




    $begingroup$
    I'm not sure without seeing the full statement, but I think it should probably read "the function $A in mathcal{B}(mathbb{R}^2,mathbb{R}^3) cong M_{3,2}(mathbb{R})$ given by ...", i.e. the function given by the matrix is not the isomorphism itself.
    $endgroup$
    – Riley
    Jan 3 at 0:31










2




2




$begingroup$
What else is an isomorphism going to be, other than a function? Check your definitions!
$endgroup$
– user3482749
Jan 2 at 21:38




$begingroup$
What else is an isomorphism going to be, other than a function? Check your definitions!
$endgroup$
– user3482749
Jan 2 at 21:38




2




2




$begingroup$
I'm not sure without seeing the full statement, but I think it should probably read "the function $A in mathcal{B}(mathbb{R}^2,mathbb{R}^3) cong M_{3,2}(mathbb{R})$ given by ...", i.e. the function given by the matrix is not the isomorphism itself.
$endgroup$
– Riley
Jan 3 at 0:31






$begingroup$
I'm not sure without seeing the full statement, but I think it should probably read "the function $A in mathcal{B}(mathbb{R}^2,mathbb{R}^3) cong M_{3,2}(mathbb{R})$ given by ...", i.e. the function given by the matrix is not the isomorphism itself.
$endgroup$
– Riley
Jan 3 at 0:31












1 Answer
1






active

oldest

votes


















0












$begingroup$

It is a standard fact of linear algebra that any linear map $A: >Xto Y$ between finite dimensional vector spaces with chosen bases $({bf e}_k)_{1leq kleq n}$, resp. $({bf f}_i)_{1leq ileq m}$, has a uniquely determined $(mtimes n)$-matrix $[A]$ describing this map "computation wise". For each $(i,k)in[m]times[n]$ the element $a_{ik}$ of this matrix is the $i^{rm th}$ coordinate of the vector $A{bf e}_k$:
$$A{bf e}_k=sum_{i=1}^m a_{ik},{bf f}_i .$$



This general principle is now invoked when dealing with a differentiable function $$f:>(x,y)mapsto(x^3,xy,y^2)in{mathbb R}^3 .$$ This $f$ has a derivative $$df({bf p}): >T_{bf p}mapsto T_{f({bf p})}tag{1}$$ at the point ${bf p}:=(2,3)$.
This derivative is a map $(1)$, and therefore has a matrix $[df({bf p})]in M_{3,2}({mathbb R})$. It turns out that the matrix elements are given by the partial derivatives of $f$ at ${bf p}$:
$$[df({bf p})]=left[matrix{{partial f_1overpartial x}&{partial f_1overpartial y}cr
{partial f_2overpartial x}&{partial f_2overpartial y}cr {partial f_3overpartial x}&{partial f_3overpartial y}cr}right]_{bf p} =left[matrix{3x^2&0cr
y&xcr 0&2ycr}right]_{(2,3)}=left[matrix{12&0cr 3&2cr0&6cr}right] .$$






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%2f3060003%2fwhat-does-this-notation-mean-how-can-an-isomorphism-be-a-function%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












    $begingroup$

    It is a standard fact of linear algebra that any linear map $A: >Xto Y$ between finite dimensional vector spaces with chosen bases $({bf e}_k)_{1leq kleq n}$, resp. $({bf f}_i)_{1leq ileq m}$, has a uniquely determined $(mtimes n)$-matrix $[A]$ describing this map "computation wise". For each $(i,k)in[m]times[n]$ the element $a_{ik}$ of this matrix is the $i^{rm th}$ coordinate of the vector $A{bf e}_k$:
    $$A{bf e}_k=sum_{i=1}^m a_{ik},{bf f}_i .$$



    This general principle is now invoked when dealing with a differentiable function $$f:>(x,y)mapsto(x^3,xy,y^2)in{mathbb R}^3 .$$ This $f$ has a derivative $$df({bf p}): >T_{bf p}mapsto T_{f({bf p})}tag{1}$$ at the point ${bf p}:=(2,3)$.
    This derivative is a map $(1)$, and therefore has a matrix $[df({bf p})]in M_{3,2}({mathbb R})$. It turns out that the matrix elements are given by the partial derivatives of $f$ at ${bf p}$:
    $$[df({bf p})]=left[matrix{{partial f_1overpartial x}&{partial f_1overpartial y}cr
    {partial f_2overpartial x}&{partial f_2overpartial y}cr {partial f_3overpartial x}&{partial f_3overpartial y}cr}right]_{bf p} =left[matrix{3x^2&0cr
    y&xcr 0&2ycr}right]_{(2,3)}=left[matrix{12&0cr 3&2cr0&6cr}right] .$$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      It is a standard fact of linear algebra that any linear map $A: >Xto Y$ between finite dimensional vector spaces with chosen bases $({bf e}_k)_{1leq kleq n}$, resp. $({bf f}_i)_{1leq ileq m}$, has a uniquely determined $(mtimes n)$-matrix $[A]$ describing this map "computation wise". For each $(i,k)in[m]times[n]$ the element $a_{ik}$ of this matrix is the $i^{rm th}$ coordinate of the vector $A{bf e}_k$:
      $$A{bf e}_k=sum_{i=1}^m a_{ik},{bf f}_i .$$



      This general principle is now invoked when dealing with a differentiable function $$f:>(x,y)mapsto(x^3,xy,y^2)in{mathbb R}^3 .$$ This $f$ has a derivative $$df({bf p}): >T_{bf p}mapsto T_{f({bf p})}tag{1}$$ at the point ${bf p}:=(2,3)$.
      This derivative is a map $(1)$, and therefore has a matrix $[df({bf p})]in M_{3,2}({mathbb R})$. It turns out that the matrix elements are given by the partial derivatives of $f$ at ${bf p}$:
      $$[df({bf p})]=left[matrix{{partial f_1overpartial x}&{partial f_1overpartial y}cr
      {partial f_2overpartial x}&{partial f_2overpartial y}cr {partial f_3overpartial x}&{partial f_3overpartial y}cr}right]_{bf p} =left[matrix{3x^2&0cr
      y&xcr 0&2ycr}right]_{(2,3)}=left[matrix{12&0cr 3&2cr0&6cr}right] .$$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        It is a standard fact of linear algebra that any linear map $A: >Xto Y$ between finite dimensional vector spaces with chosen bases $({bf e}_k)_{1leq kleq n}$, resp. $({bf f}_i)_{1leq ileq m}$, has a uniquely determined $(mtimes n)$-matrix $[A]$ describing this map "computation wise". For each $(i,k)in[m]times[n]$ the element $a_{ik}$ of this matrix is the $i^{rm th}$ coordinate of the vector $A{bf e}_k$:
        $$A{bf e}_k=sum_{i=1}^m a_{ik},{bf f}_i .$$



        This general principle is now invoked when dealing with a differentiable function $$f:>(x,y)mapsto(x^3,xy,y^2)in{mathbb R}^3 .$$ This $f$ has a derivative $$df({bf p}): >T_{bf p}mapsto T_{f({bf p})}tag{1}$$ at the point ${bf p}:=(2,3)$.
        This derivative is a map $(1)$, and therefore has a matrix $[df({bf p})]in M_{3,2}({mathbb R})$. It turns out that the matrix elements are given by the partial derivatives of $f$ at ${bf p}$:
        $$[df({bf p})]=left[matrix{{partial f_1overpartial x}&{partial f_1overpartial y}cr
        {partial f_2overpartial x}&{partial f_2overpartial y}cr {partial f_3overpartial x}&{partial f_3overpartial y}cr}right]_{bf p} =left[matrix{3x^2&0cr
        y&xcr 0&2ycr}right]_{(2,3)}=left[matrix{12&0cr 3&2cr0&6cr}right] .$$






        share|cite|improve this answer









        $endgroup$



        It is a standard fact of linear algebra that any linear map $A: >Xto Y$ between finite dimensional vector spaces with chosen bases $({bf e}_k)_{1leq kleq n}$, resp. $({bf f}_i)_{1leq ileq m}$, has a uniquely determined $(mtimes n)$-matrix $[A]$ describing this map "computation wise". For each $(i,k)in[m]times[n]$ the element $a_{ik}$ of this matrix is the $i^{rm th}$ coordinate of the vector $A{bf e}_k$:
        $$A{bf e}_k=sum_{i=1}^m a_{ik},{bf f}_i .$$



        This general principle is now invoked when dealing with a differentiable function $$f:>(x,y)mapsto(x^3,xy,y^2)in{mathbb R}^3 .$$ This $f$ has a derivative $$df({bf p}): >T_{bf p}mapsto T_{f({bf p})}tag{1}$$ at the point ${bf p}:=(2,3)$.
        This derivative is a map $(1)$, and therefore has a matrix $[df({bf p})]in M_{3,2}({mathbb R})$. It turns out that the matrix elements are given by the partial derivatives of $f$ at ${bf p}$:
        $$[df({bf p})]=left[matrix{{partial f_1overpartial x}&{partial f_1overpartial y}cr
        {partial f_2overpartial x}&{partial f_2overpartial y}cr {partial f_3overpartial x}&{partial f_3overpartial y}cr}right]_{bf p} =left[matrix{3x^2&0cr
        y&xcr 0&2ycr}right]_{(2,3)}=left[matrix{12&0cr 3&2cr0&6cr}right] .$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 3 at 10:49









        Christian BlatterChristian Blatter

        173k7113326




        173k7113326






























            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%2f3060003%2fwhat-does-this-notation-mean-how-can-an-isomorphism-be-a-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

            Human spaceflight

            Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

            張江高科駅