what does it mean by determinant of Jacobian matrix = 0?












5












$begingroup$


I have an example:
$$ u={x+yover 1-xy} $$
$$ v = tan^{-1}(x)+tan^{-1}(y) $$
So by calculating the determinant of the Jacobian matrix I get zero. Does it mean there is no functional relationship between u and v? What does $|J|=0$ mean?






calculus linear-algebra jacobian







share|cite|improve this question











$endgroup$

















    5












    $begingroup$


    I have an example:
    $$ u={x+yover 1-xy} $$
    $$ v = tan^{-1}(x)+tan^{-1}(y) $$
    So by calculating the determinant of the Jacobian matrix I get zero. Does it mean there is no functional relationship between u and v? What does $|J|=0$ mean?






    calculus linear-algebra jacobian







    share|cite|improve this question











    $endgroup$















      5












      5








      5





      $begingroup$


      I have an example:
      $$ u={x+yover 1-xy} $$
      $$ v = tan^{-1}(x)+tan^{-1}(y) $$
      So by calculating the determinant of the Jacobian matrix I get zero. Does it mean there is no functional relationship between u and v? What does $|J|=0$ mean?






      calculus linear-algebra jacobian







      share|cite|improve this question











      $endgroup$




      I have an example:
      $$ u={x+yover 1-xy} $$
      $$ v = tan^{-1}(x)+tan^{-1}(y) $$
      So by calculating the determinant of the Jacobian matrix I get zero. Does it mean there is no functional relationship between u and v? What does $|J|=0$ mean?






      calculus linear-algebra jacobian




      calculus linear-algebra jacobian






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 2 hours ago









      El Pasta

      44615




      44615










      asked 4 hours ago









      Yibei HeYibei He

      1758




      1758






















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          By the Inverse Function Theorem, it means that your function is not locally invertible at any point where the Jacobian has determinant $0$. The function in question is the map $f: D to mathbb{R}^2$ defined by $f(x,y) = (u(x,y), v(x,y))$ in your notation, where $D$ is the appropriate domain given your coordinate functions.



          This is the higher-dimensional analogue to the behavior we see for $f: mathbb{R} to mathbb{R}$ by $f(x)=x^2$ about the origin. There is no neighborhood $(-epsilon, epsilon)$ around $0$ for which the restriction of $f$ to that neighborhood is one-to-one: there's always a little piece that will make it fail the horizontal line test.



          As an aside, $|J|=0$ indicating that your function is not locally invertible also means you cannot use the resulting transformation to perform multivariable substitution in integrals.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            That's not true. The Inverse Function Theorem is an "if", not an "if and only if": it says nothing about what happens when the Jacobian has determinant $0$. The function might still be locally invertible. For example, in the case $n=1$ the function $f(x) = x^3$ has derivative $0$ at $x=0$ but is locally invertible there.
            $endgroup$
            – Robert Israel
            2 hours ago










          • $begingroup$
            Oh, that's a good point. Is there a fix for this?
            $endgroup$
            – Randall
            2 hours ago










          • $begingroup$
            OP: you should probably un-accept this. This isn't a good answer.
            $endgroup$
            – Randall
            2 hours ago












          • $begingroup$
            What you can say is that, if the function is locally invertible, the inverse function is not differentiable at the point in question. This follows from the (multivariate) chain rule.
            $endgroup$
            – Robert Israel
            2 hours ago



















          2












          $begingroup$

          In this case, there is a functional relationship between $u$ and $v$: in fact $u = tan(v)$. Thus the transformation $(x,y) to (u,v)$ is not invertible: there is no way to get $x$ and $y$ back from $u$ and $v$.






          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%2f3077789%2fwhat-does-it-mean-by-determinant-of-jacobian-matrix-0%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









            2












            $begingroup$

            By the Inverse Function Theorem, it means that your function is not locally invertible at any point where the Jacobian has determinant $0$. The function in question is the map $f: D to mathbb{R}^2$ defined by $f(x,y) = (u(x,y), v(x,y))$ in your notation, where $D$ is the appropriate domain given your coordinate functions.



            This is the higher-dimensional analogue to the behavior we see for $f: mathbb{R} to mathbb{R}$ by $f(x)=x^2$ about the origin. There is no neighborhood $(-epsilon, epsilon)$ around $0$ for which the restriction of $f$ to that neighborhood is one-to-one: there's always a little piece that will make it fail the horizontal line test.



            As an aside, $|J|=0$ indicating that your function is not locally invertible also means you cannot use the resulting transformation to perform multivariable substitution in integrals.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              That's not true. The Inverse Function Theorem is an "if", not an "if and only if": it says nothing about what happens when the Jacobian has determinant $0$. The function might still be locally invertible. For example, in the case $n=1$ the function $f(x) = x^3$ has derivative $0$ at $x=0$ but is locally invertible there.
              $endgroup$
              – Robert Israel
              2 hours ago










            • $begingroup$
              Oh, that's a good point. Is there a fix for this?
              $endgroup$
              – Randall
              2 hours ago










            • $begingroup$
              OP: you should probably un-accept this. This isn't a good answer.
              $endgroup$
              – Randall
              2 hours ago












            • $begingroup$
              What you can say is that, if the function is locally invertible, the inverse function is not differentiable at the point in question. This follows from the (multivariate) chain rule.
              $endgroup$
              – Robert Israel
              2 hours ago
















            2












            $begingroup$

            By the Inverse Function Theorem, it means that your function is not locally invertible at any point where the Jacobian has determinant $0$. The function in question is the map $f: D to mathbb{R}^2$ defined by $f(x,y) = (u(x,y), v(x,y))$ in your notation, where $D$ is the appropriate domain given your coordinate functions.



            This is the higher-dimensional analogue to the behavior we see for $f: mathbb{R} to mathbb{R}$ by $f(x)=x^2$ about the origin. There is no neighborhood $(-epsilon, epsilon)$ around $0$ for which the restriction of $f$ to that neighborhood is one-to-one: there's always a little piece that will make it fail the horizontal line test.



            As an aside, $|J|=0$ indicating that your function is not locally invertible also means you cannot use the resulting transformation to perform multivariable substitution in integrals.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              That's not true. The Inverse Function Theorem is an "if", not an "if and only if": it says nothing about what happens when the Jacobian has determinant $0$. The function might still be locally invertible. For example, in the case $n=1$ the function $f(x) = x^3$ has derivative $0$ at $x=0$ but is locally invertible there.
              $endgroup$
              – Robert Israel
              2 hours ago










            • $begingroup$
              Oh, that's a good point. Is there a fix for this?
              $endgroup$
              – Randall
              2 hours ago










            • $begingroup$
              OP: you should probably un-accept this. This isn't a good answer.
              $endgroup$
              – Randall
              2 hours ago












            • $begingroup$
              What you can say is that, if the function is locally invertible, the inverse function is not differentiable at the point in question. This follows from the (multivariate) chain rule.
              $endgroup$
              – Robert Israel
              2 hours ago














            2












            2








            2





            $begingroup$

            By the Inverse Function Theorem, it means that your function is not locally invertible at any point where the Jacobian has determinant $0$. The function in question is the map $f: D to mathbb{R}^2$ defined by $f(x,y) = (u(x,y), v(x,y))$ in your notation, where $D$ is the appropriate domain given your coordinate functions.



            This is the higher-dimensional analogue to the behavior we see for $f: mathbb{R} to mathbb{R}$ by $f(x)=x^2$ about the origin. There is no neighborhood $(-epsilon, epsilon)$ around $0$ for which the restriction of $f$ to that neighborhood is one-to-one: there's always a little piece that will make it fail the horizontal line test.



            As an aside, $|J|=0$ indicating that your function is not locally invertible also means you cannot use the resulting transformation to perform multivariable substitution in integrals.






            share|cite|improve this answer











            $endgroup$



            By the Inverse Function Theorem, it means that your function is not locally invertible at any point where the Jacobian has determinant $0$. The function in question is the map $f: D to mathbb{R}^2$ defined by $f(x,y) = (u(x,y), v(x,y))$ in your notation, where $D$ is the appropriate domain given your coordinate functions.



            This is the higher-dimensional analogue to the behavior we see for $f: mathbb{R} to mathbb{R}$ by $f(x)=x^2$ about the origin. There is no neighborhood $(-epsilon, epsilon)$ around $0$ for which the restriction of $f$ to that neighborhood is one-to-one: there's always a little piece that will make it fail the horizontal line test.



            As an aside, $|J|=0$ indicating that your function is not locally invertible also means you cannot use the resulting transformation to perform multivariable substitution in integrals.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 3 hours ago

























            answered 3 hours ago









            RandallRandall

            9,50911230




            9,50911230












            • $begingroup$
              That's not true. The Inverse Function Theorem is an "if", not an "if and only if": it says nothing about what happens when the Jacobian has determinant $0$. The function might still be locally invertible. For example, in the case $n=1$ the function $f(x) = x^3$ has derivative $0$ at $x=0$ but is locally invertible there.
              $endgroup$
              – Robert Israel
              2 hours ago










            • $begingroup$
              Oh, that's a good point. Is there a fix for this?
              $endgroup$
              – Randall
              2 hours ago










            • $begingroup$
              OP: you should probably un-accept this. This isn't a good answer.
              $endgroup$
              – Randall
              2 hours ago












            • $begingroup$
              What you can say is that, if the function is locally invertible, the inverse function is not differentiable at the point in question. This follows from the (multivariate) chain rule.
              $endgroup$
              – Robert Israel
              2 hours ago


















            • $begingroup$
              That's not true. The Inverse Function Theorem is an "if", not an "if and only if": it says nothing about what happens when the Jacobian has determinant $0$. The function might still be locally invertible. For example, in the case $n=1$ the function $f(x) = x^3$ has derivative $0$ at $x=0$ but is locally invertible there.
              $endgroup$
              – Robert Israel
              2 hours ago










            • $begingroup$
              Oh, that's a good point. Is there a fix for this?
              $endgroup$
              – Randall
              2 hours ago










            • $begingroup$
              OP: you should probably un-accept this. This isn't a good answer.
              $endgroup$
              – Randall
              2 hours ago












            • $begingroup$
              What you can say is that, if the function is locally invertible, the inverse function is not differentiable at the point in question. This follows from the (multivariate) chain rule.
              $endgroup$
              – Robert Israel
              2 hours ago
















            $begingroup$
            That's not true. The Inverse Function Theorem is an "if", not an "if and only if": it says nothing about what happens when the Jacobian has determinant $0$. The function might still be locally invertible. For example, in the case $n=1$ the function $f(x) = x^3$ has derivative $0$ at $x=0$ but is locally invertible there.
            $endgroup$
            – Robert Israel
            2 hours ago




            $begingroup$
            That's not true. The Inverse Function Theorem is an "if", not an "if and only if": it says nothing about what happens when the Jacobian has determinant $0$. The function might still be locally invertible. For example, in the case $n=1$ the function $f(x) = x^3$ has derivative $0$ at $x=0$ but is locally invertible there.
            $endgroup$
            – Robert Israel
            2 hours ago












            $begingroup$
            Oh, that's a good point. Is there a fix for this?
            $endgroup$
            – Randall
            2 hours ago




            $begingroup$
            Oh, that's a good point. Is there a fix for this?
            $endgroup$
            – Randall
            2 hours ago












            $begingroup$
            OP: you should probably un-accept this. This isn't a good answer.
            $endgroup$
            – Randall
            2 hours ago






            $begingroup$
            OP: you should probably un-accept this. This isn't a good answer.
            $endgroup$
            – Randall
            2 hours ago














            $begingroup$
            What you can say is that, if the function is locally invertible, the inverse function is not differentiable at the point in question. This follows from the (multivariate) chain rule.
            $endgroup$
            – Robert Israel
            2 hours ago




            $begingroup$
            What you can say is that, if the function is locally invertible, the inverse function is not differentiable at the point in question. This follows from the (multivariate) chain rule.
            $endgroup$
            – Robert Israel
            2 hours ago











            2












            $begingroup$

            In this case, there is a functional relationship between $u$ and $v$: in fact $u = tan(v)$. Thus the transformation $(x,y) to (u,v)$ is not invertible: there is no way to get $x$ and $y$ back from $u$ and $v$.






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              In this case, there is a functional relationship between $u$ and $v$: in fact $u = tan(v)$. Thus the transformation $(x,y) to (u,v)$ is not invertible: there is no way to get $x$ and $y$ back from $u$ and $v$.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                In this case, there is a functional relationship between $u$ and $v$: in fact $u = tan(v)$. Thus the transformation $(x,y) to (u,v)$ is not invertible: there is no way to get $x$ and $y$ back from $u$ and $v$.






                share|cite|improve this answer









                $endgroup$



                In this case, there is a functional relationship between $u$ and $v$: in fact $u = tan(v)$. Thus the transformation $(x,y) to (u,v)$ is not invertible: there is no way to get $x$ and $y$ back from $u$ and $v$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 2 hours ago









                Robert IsraelRobert Israel

                320k23209459




                320k23209459






























                    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%2f3077789%2fwhat-does-it-mean-by-determinant-of-jacobian-matrix-0%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

                    Knooppunt Holsloot

                    Image
                    Knooppunt Holsloot Kenmerken Wegen A37 × N34 Type knooppunt klaverblad Portaal     Verkeer & Vervoer Knooppunt Holsloot is een verkeersknooppunt op de kruising van de autosnelweg A37 en de N34 in Drenthe, tussen Emmen en Coevorden. Oorspronkelijk was deze aansluiting een afritconstructie van het type Haarlemmermeeraansluiting, zoals bij Knooppunt Hooipolder. Sinds 2002 is het knooppunt omgebouwd tot een volledig klaverblad. Als noviteit werd hier voor het eerst gebruikgemaakt van dimbare verlichting in plaats van de normale vaste tijden waarop de verlichting aan en uit gaat, wat een grote kostenbesparing opleverde. [1] Dat Holsloot als volwaardig knooppunt is uitgevoerd, waarbij alle verkeersstromen en afbuigingen ononderbroken kunnen doorgaan, is uitzonderlijk. Het betreft hier namelijk een aansluiting tussen een autosnelweg en niet-snelweg. Een vergelijkbare situatie treft men nabij Hoensbroek bij het knooppunt Ten Esschen.     Aa
                    Read more

                    Altaar (religie)

                    Image
                    Deel van de serie over kerkelijk gerei waaronder paramenten en liturgisch vaatwerk gebruikt in de liturgie Liturgisch vaatwerk Vasa sacra Miskelk · Pateen Kelklepeltje Ciborie · Monstrans Pyxis · Custodiale Vasa non sacra Ampullen · Wijwatervat Olievaatje Wierookvat · Ablutievat Paramenten Amict · Albe · Baarkleed · Cingel Tuniek · Dalmatiek · Fanon Kazuifel · Manipel · Stola Gremiale · Benedictievelum Mijter Koorkledij Rochet · Superplie Koorkap · Cappa magna Kovel Kelkgerei Bursa · Kelkvelum Ciborievelum Kerklinnen Corporale · Kelkdoekje Lavabodoekje · Palla Altaardwaal Kerkinterieur Altaar · Ambo Biechtstoel · Communiebank Doksaal · Doopvont Faldistorium Godslamp · Hoogaltaar Heilig Kruisaltaar Katheder · Preekstoel Sedilia Tabernakel · Volksaltaar Liturgische boeken Altaarmissaal Benedictionale · Brevier Evangeliarium · Evangelistarium Graduale · Kyriale Lectionarium Psalter · Rituaal Sacramentarium · Volksmissaal Overige Flambou
                    Read more

                    Gregoriusmis

                    Image
                    Een afbeelding van de Gregoriusmis uit een getijdenboek gemaakt in de zuidelijke Nederlanden tussen 1500 en 1525 Gregoriusmis, Meester van het Akense Altaar, ca. 1500-1525. Collectie Museum Catharijneconvent De Gregoriusmis of de Mis van Sint Gregorius is een motief in katholieke kunst, dat het eerst verscheen in de late Middeleeuwen en dat nog steeds werd gebruikt tijdens de contrareformatie. Paus Gregorius I (ca. 540-604) is afgebeeld, de H. Mis opdragend, terwijl Christus als de Man van Smarten verschijnt op het altaar dat voor Sint Gregorius staat. Volgens de legende had St. Gregorius gebeden om een teken voor een betwijfelaar van de transsubstantiatieleer, waarop de lijdende Christus tijdens het opdragen van de Mis verscheen om hem te overtuigen en waarbij zijn bloed in de altaarkelk liep. Vaak zijn ook de passiewerktuigen afgebeeld. Zie de categorie Mass of Saint Gregory van Wikimedia Commons voor mediabestanden over dit onderwerp. This page is onl
                    Read more