changes of coordinates, confusing book language.












1












$begingroup$


A book i'm reading says consider the "dictionary" $x = u + a$ and $y = v + b$ between the $x,y$ and $u,v$ coordinate systems, where the $u,v$ coordinate axes are obtained by translating the $x,y$ $a$ units to the right, and $b$ units upward.



It makes sense geometrically that given the $(x,y)$ coordinates of a point $P$ of the plane, the "dictionary" above allows us to find the $(u,v)$ coordinates of the same point, and vice versa.



Then the book wants to illustrate that sometimes the origin and axes direction remain unchanged but the unit of length may differ. And so they say suppose we have the following "dictionary" between $x,y$ and $u,v$ planes:



$u = 2x, v = 3y$. Where they depict the point $(1,0)$ in the $u,v$ plane being in the position of the point $(2,0)$ in the $x,y$ plane. So the step sizes in the $u,v$ plane are larger.



After many hours I think I figured out that the use of the word dictionary here (unlike the first example for some mysterious reason?) is misleading.



The transformation above does not allow us to feed in $(x,y)$ coordinates of a point $P$ and determine the $(u,v)$ coordinates of the same point. It seems to instead tell us how to obtain the $u-v$ plane basis from the $x-y$ one. And it seems that the dictionary between the two system is given by $u = frac{1}{2}x$, and $v = frac{1}{3}y$.



Is my interpretation correct? Would you agree that the description of dictionary for the second example (but not the first) was misleading? Since it seemed to give a transformation, and not the change of basis formulas.










share|cite|improve this question









$endgroup$












  • $begingroup$
    The image of $(2,0)$ under $(x,y)mapsto(2x,3y)$ is $(4,0)$, not $(1,0)$. The step sizes in the $uv$ plane are half that in the $xy$ plane
    $endgroup$
    – Shubham Johri
    Jan 16 at 8:18












  • $begingroup$
    I maybe wrong but I think the usage of the word 'dicitionary' here is non-technical and fairly abstract. It just means a rule of correspondence, or a transformation, between points in the $xy$ and $uv$ planes
    $endgroup$
    – Shubham Johri
    Jan 16 at 8:24










  • $begingroup$
    @ShubhamJohri thats what I would think (about the stepsizes in $uv$ plane being half) based on the formulas, but the picture they drew indicated the opposite.
    $endgroup$
    – trynalearn
    Jan 16 at 8:28
















1












$begingroup$


A book i'm reading says consider the "dictionary" $x = u + a$ and $y = v + b$ between the $x,y$ and $u,v$ coordinate systems, where the $u,v$ coordinate axes are obtained by translating the $x,y$ $a$ units to the right, and $b$ units upward.



It makes sense geometrically that given the $(x,y)$ coordinates of a point $P$ of the plane, the "dictionary" above allows us to find the $(u,v)$ coordinates of the same point, and vice versa.



Then the book wants to illustrate that sometimes the origin and axes direction remain unchanged but the unit of length may differ. And so they say suppose we have the following "dictionary" between $x,y$ and $u,v$ planes:



$u = 2x, v = 3y$. Where they depict the point $(1,0)$ in the $u,v$ plane being in the position of the point $(2,0)$ in the $x,y$ plane. So the step sizes in the $u,v$ plane are larger.



After many hours I think I figured out that the use of the word dictionary here (unlike the first example for some mysterious reason?) is misleading.



The transformation above does not allow us to feed in $(x,y)$ coordinates of a point $P$ and determine the $(u,v)$ coordinates of the same point. It seems to instead tell us how to obtain the $u-v$ plane basis from the $x-y$ one. And it seems that the dictionary between the two system is given by $u = frac{1}{2}x$, and $v = frac{1}{3}y$.



Is my interpretation correct? Would you agree that the description of dictionary for the second example (but not the first) was misleading? Since it seemed to give a transformation, and not the change of basis formulas.










share|cite|improve this question









$endgroup$












  • $begingroup$
    The image of $(2,0)$ under $(x,y)mapsto(2x,3y)$ is $(4,0)$, not $(1,0)$. The step sizes in the $uv$ plane are half that in the $xy$ plane
    $endgroup$
    – Shubham Johri
    Jan 16 at 8:18












  • $begingroup$
    I maybe wrong but I think the usage of the word 'dicitionary' here is non-technical and fairly abstract. It just means a rule of correspondence, or a transformation, between points in the $xy$ and $uv$ planes
    $endgroup$
    – Shubham Johri
    Jan 16 at 8:24










  • $begingroup$
    @ShubhamJohri thats what I would think (about the stepsizes in $uv$ plane being half) based on the formulas, but the picture they drew indicated the opposite.
    $endgroup$
    – trynalearn
    Jan 16 at 8:28














1












1








1





$begingroup$


A book i'm reading says consider the "dictionary" $x = u + a$ and $y = v + b$ between the $x,y$ and $u,v$ coordinate systems, where the $u,v$ coordinate axes are obtained by translating the $x,y$ $a$ units to the right, and $b$ units upward.



It makes sense geometrically that given the $(x,y)$ coordinates of a point $P$ of the plane, the "dictionary" above allows us to find the $(u,v)$ coordinates of the same point, and vice versa.



Then the book wants to illustrate that sometimes the origin and axes direction remain unchanged but the unit of length may differ. And so they say suppose we have the following "dictionary" between $x,y$ and $u,v$ planes:



$u = 2x, v = 3y$. Where they depict the point $(1,0)$ in the $u,v$ plane being in the position of the point $(2,0)$ in the $x,y$ plane. So the step sizes in the $u,v$ plane are larger.



After many hours I think I figured out that the use of the word dictionary here (unlike the first example for some mysterious reason?) is misleading.



The transformation above does not allow us to feed in $(x,y)$ coordinates of a point $P$ and determine the $(u,v)$ coordinates of the same point. It seems to instead tell us how to obtain the $u-v$ plane basis from the $x-y$ one. And it seems that the dictionary between the two system is given by $u = frac{1}{2}x$, and $v = frac{1}{3}y$.



Is my interpretation correct? Would you agree that the description of dictionary for the second example (but not the first) was misleading? Since it seemed to give a transformation, and not the change of basis formulas.










share|cite|improve this question









$endgroup$




A book i'm reading says consider the "dictionary" $x = u + a$ and $y = v + b$ between the $x,y$ and $u,v$ coordinate systems, where the $u,v$ coordinate axes are obtained by translating the $x,y$ $a$ units to the right, and $b$ units upward.



It makes sense geometrically that given the $(x,y)$ coordinates of a point $P$ of the plane, the "dictionary" above allows us to find the $(u,v)$ coordinates of the same point, and vice versa.



Then the book wants to illustrate that sometimes the origin and axes direction remain unchanged but the unit of length may differ. And so they say suppose we have the following "dictionary" between $x,y$ and $u,v$ planes:



$u = 2x, v = 3y$. Where they depict the point $(1,0)$ in the $u,v$ plane being in the position of the point $(2,0)$ in the $x,y$ plane. So the step sizes in the $u,v$ plane are larger.



After many hours I think I figured out that the use of the word dictionary here (unlike the first example for some mysterious reason?) is misleading.



The transformation above does not allow us to feed in $(x,y)$ coordinates of a point $P$ and determine the $(u,v)$ coordinates of the same point. It seems to instead tell us how to obtain the $u-v$ plane basis from the $x-y$ one. And it seems that the dictionary between the two system is given by $u = frac{1}{2}x$, and $v = frac{1}{3}y$.



Is my interpretation correct? Would you agree that the description of dictionary for the second example (but not the first) was misleading? Since it seemed to give a transformation, and not the change of basis formulas.







linear-algebra change-of-basis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 16 at 8:10









trynalearntrynalearn

662314




662314












  • $begingroup$
    The image of $(2,0)$ under $(x,y)mapsto(2x,3y)$ is $(4,0)$, not $(1,0)$. The step sizes in the $uv$ plane are half that in the $xy$ plane
    $endgroup$
    – Shubham Johri
    Jan 16 at 8:18












  • $begingroup$
    I maybe wrong but I think the usage of the word 'dicitionary' here is non-technical and fairly abstract. It just means a rule of correspondence, or a transformation, between points in the $xy$ and $uv$ planes
    $endgroup$
    – Shubham Johri
    Jan 16 at 8:24










  • $begingroup$
    @ShubhamJohri thats what I would think (about the stepsizes in $uv$ plane being half) based on the formulas, but the picture they drew indicated the opposite.
    $endgroup$
    – trynalearn
    Jan 16 at 8:28


















  • $begingroup$
    The image of $(2,0)$ under $(x,y)mapsto(2x,3y)$ is $(4,0)$, not $(1,0)$. The step sizes in the $uv$ plane are half that in the $xy$ plane
    $endgroup$
    – Shubham Johri
    Jan 16 at 8:18












  • $begingroup$
    I maybe wrong but I think the usage of the word 'dicitionary' here is non-technical and fairly abstract. It just means a rule of correspondence, or a transformation, between points in the $xy$ and $uv$ planes
    $endgroup$
    – Shubham Johri
    Jan 16 at 8:24










  • $begingroup$
    @ShubhamJohri thats what I would think (about the stepsizes in $uv$ plane being half) based on the formulas, but the picture they drew indicated the opposite.
    $endgroup$
    – trynalearn
    Jan 16 at 8:28
















$begingroup$
The image of $(2,0)$ under $(x,y)mapsto(2x,3y)$ is $(4,0)$, not $(1,0)$. The step sizes in the $uv$ plane are half that in the $xy$ plane
$endgroup$
– Shubham Johri
Jan 16 at 8:18






$begingroup$
The image of $(2,0)$ under $(x,y)mapsto(2x,3y)$ is $(4,0)$, not $(1,0)$. The step sizes in the $uv$ plane are half that in the $xy$ plane
$endgroup$
– Shubham Johri
Jan 16 at 8:18














$begingroup$
I maybe wrong but I think the usage of the word 'dicitionary' here is non-technical and fairly abstract. It just means a rule of correspondence, or a transformation, between points in the $xy$ and $uv$ planes
$endgroup$
– Shubham Johri
Jan 16 at 8:24




$begingroup$
I maybe wrong but I think the usage of the word 'dicitionary' here is non-technical and fairly abstract. It just means a rule of correspondence, or a transformation, between points in the $xy$ and $uv$ planes
$endgroup$
– Shubham Johri
Jan 16 at 8:24












$begingroup$
@ShubhamJohri thats what I would think (about the stepsizes in $uv$ plane being half) based on the formulas, but the picture they drew indicated the opposite.
$endgroup$
– trynalearn
Jan 16 at 8:28




$begingroup$
@ShubhamJohri thats what I would think (about the stepsizes in $uv$ plane being half) based on the formulas, but the picture they drew indicated the opposite.
$endgroup$
– trynalearn
Jan 16 at 8:28










1 Answer
1






active

oldest

votes


















1












$begingroup$

What you seem to have stumbled upon is the fact that basis vectors and vector components transform co-variantly and contra-variantly respectively.



I will use Einstein summation notation, let me know if it's confusing for you.
I will also use the following notation: $e_i$ are basis vectors, $v^i$ are vector components, $A_i^j$ are elements of a matrix. This is the standard tensor notation.



A linear transformation tells us how to find the new basis vectors from the old basis vectors:
$$ tilde{e}_i = A_i^j e_j $$
In your case, $A_1^1 = 2, A_1^2 = 0, A_2^1=0, A_2^2=3$. Now a vector $v$ can be expressed in both bases:
$$ v = tilde{v^i}tilde{e_i} = v^j e_j $$
How do we get the new vector components $tilde{v^i}$?
$$ tilde{v^i}tilde{e_i} = tilde{v^i}A_i^j e_j $$
That is, by comparing the two forms:
$$ tilde{v^i}A_i^j = v^j $$
This is a matrix equation, so we can multiply both sides by the matrix $(A^{-1})$, which is defined as the inverse matrix of $A$. In your case, $(A^{-1})_1^1 = 1/2, (A^{-1})_2^1 = 0, (A^{-1})_1^2 = 0, (A^{-1})_2^2 = 1/3$.
$$ tilde{v^i} = v^j (A^{-1})_j^i$$



So the basis vectors transform using $A_i^j$ matrix (we say they are covariant) while the vector components transform using the $(A^{-1})_j^i$ matrix (we say they are contravariant). This is why the transformation is still a dictionary but it tells you different things when you look at vector components or basis vectors.






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%2f3075451%2fchanges-of-coordinates-confusing-book-language%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









    1












    $begingroup$

    What you seem to have stumbled upon is the fact that basis vectors and vector components transform co-variantly and contra-variantly respectively.



    I will use Einstein summation notation, let me know if it's confusing for you.
    I will also use the following notation: $e_i$ are basis vectors, $v^i$ are vector components, $A_i^j$ are elements of a matrix. This is the standard tensor notation.



    A linear transformation tells us how to find the new basis vectors from the old basis vectors:
    $$ tilde{e}_i = A_i^j e_j $$
    In your case, $A_1^1 = 2, A_1^2 = 0, A_2^1=0, A_2^2=3$. Now a vector $v$ can be expressed in both bases:
    $$ v = tilde{v^i}tilde{e_i} = v^j e_j $$
    How do we get the new vector components $tilde{v^i}$?
    $$ tilde{v^i}tilde{e_i} = tilde{v^i}A_i^j e_j $$
    That is, by comparing the two forms:
    $$ tilde{v^i}A_i^j = v^j $$
    This is a matrix equation, so we can multiply both sides by the matrix $(A^{-1})$, which is defined as the inverse matrix of $A$. In your case, $(A^{-1})_1^1 = 1/2, (A^{-1})_2^1 = 0, (A^{-1})_1^2 = 0, (A^{-1})_2^2 = 1/3$.
    $$ tilde{v^i} = v^j (A^{-1})_j^i$$



    So the basis vectors transform using $A_i^j$ matrix (we say they are covariant) while the vector components transform using the $(A^{-1})_j^i$ matrix (we say they are contravariant). This is why the transformation is still a dictionary but it tells you different things when you look at vector components or basis vectors.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      What you seem to have stumbled upon is the fact that basis vectors and vector components transform co-variantly and contra-variantly respectively.



      I will use Einstein summation notation, let me know if it's confusing for you.
      I will also use the following notation: $e_i$ are basis vectors, $v^i$ are vector components, $A_i^j$ are elements of a matrix. This is the standard tensor notation.



      A linear transformation tells us how to find the new basis vectors from the old basis vectors:
      $$ tilde{e}_i = A_i^j e_j $$
      In your case, $A_1^1 = 2, A_1^2 = 0, A_2^1=0, A_2^2=3$. Now a vector $v$ can be expressed in both bases:
      $$ v = tilde{v^i}tilde{e_i} = v^j e_j $$
      How do we get the new vector components $tilde{v^i}$?
      $$ tilde{v^i}tilde{e_i} = tilde{v^i}A_i^j e_j $$
      That is, by comparing the two forms:
      $$ tilde{v^i}A_i^j = v^j $$
      This is a matrix equation, so we can multiply both sides by the matrix $(A^{-1})$, which is defined as the inverse matrix of $A$. In your case, $(A^{-1})_1^1 = 1/2, (A^{-1})_2^1 = 0, (A^{-1})_1^2 = 0, (A^{-1})_2^2 = 1/3$.
      $$ tilde{v^i} = v^j (A^{-1})_j^i$$



      So the basis vectors transform using $A_i^j$ matrix (we say they are covariant) while the vector components transform using the $(A^{-1})_j^i$ matrix (we say they are contravariant). This is why the transformation is still a dictionary but it tells you different things when you look at vector components or basis vectors.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        What you seem to have stumbled upon is the fact that basis vectors and vector components transform co-variantly and contra-variantly respectively.



        I will use Einstein summation notation, let me know if it's confusing for you.
        I will also use the following notation: $e_i$ are basis vectors, $v^i$ are vector components, $A_i^j$ are elements of a matrix. This is the standard tensor notation.



        A linear transformation tells us how to find the new basis vectors from the old basis vectors:
        $$ tilde{e}_i = A_i^j e_j $$
        In your case, $A_1^1 = 2, A_1^2 = 0, A_2^1=0, A_2^2=3$. Now a vector $v$ can be expressed in both bases:
        $$ v = tilde{v^i}tilde{e_i} = v^j e_j $$
        How do we get the new vector components $tilde{v^i}$?
        $$ tilde{v^i}tilde{e_i} = tilde{v^i}A_i^j e_j $$
        That is, by comparing the two forms:
        $$ tilde{v^i}A_i^j = v^j $$
        This is a matrix equation, so we can multiply both sides by the matrix $(A^{-1})$, which is defined as the inverse matrix of $A$. In your case, $(A^{-1})_1^1 = 1/2, (A^{-1})_2^1 = 0, (A^{-1})_1^2 = 0, (A^{-1})_2^2 = 1/3$.
        $$ tilde{v^i} = v^j (A^{-1})_j^i$$



        So the basis vectors transform using $A_i^j$ matrix (we say they are covariant) while the vector components transform using the $(A^{-1})_j^i$ matrix (we say they are contravariant). This is why the transformation is still a dictionary but it tells you different things when you look at vector components or basis vectors.






        share|cite|improve this answer









        $endgroup$



        What you seem to have stumbled upon is the fact that basis vectors and vector components transform co-variantly and contra-variantly respectively.



        I will use Einstein summation notation, let me know if it's confusing for you.
        I will also use the following notation: $e_i$ are basis vectors, $v^i$ are vector components, $A_i^j$ are elements of a matrix. This is the standard tensor notation.



        A linear transformation tells us how to find the new basis vectors from the old basis vectors:
        $$ tilde{e}_i = A_i^j e_j $$
        In your case, $A_1^1 = 2, A_1^2 = 0, A_2^1=0, A_2^2=3$. Now a vector $v$ can be expressed in both bases:
        $$ v = tilde{v^i}tilde{e_i} = v^j e_j $$
        How do we get the new vector components $tilde{v^i}$?
        $$ tilde{v^i}tilde{e_i} = tilde{v^i}A_i^j e_j $$
        That is, by comparing the two forms:
        $$ tilde{v^i}A_i^j = v^j $$
        This is a matrix equation, so we can multiply both sides by the matrix $(A^{-1})$, which is defined as the inverse matrix of $A$. In your case, $(A^{-1})_1^1 = 1/2, (A^{-1})_2^1 = 0, (A^{-1})_1^2 = 0, (A^{-1})_2^2 = 1/3$.
        $$ tilde{v^i} = v^j (A^{-1})_j^i$$



        So the basis vectors transform using $A_i^j$ matrix (we say they are covariant) while the vector components transform using the $(A^{-1})_j^i$ matrix (we say they are contravariant). This is why the transformation is still a dictionary but it tells you different things when you look at vector components or basis vectors.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 16 at 8:38









        Niki Di GianoNiki Di Giano

        961211




        961211






























            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%2f3075451%2fchanges-of-coordinates-confusing-book-language%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

            Mario Kart Wii

            The Binding of Isaac: Rebirth/Afterbirth

            What does “Dominus providebit” mean?