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













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      share|cite|improve this question








      edited 2 hours ago









      El Pasta

      44615




      44615










      asked 4 hours ago









      Yibei HeYibei He

      1758




      1758






















          2 Answers
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          $begingroup$

          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
            3 hours ago










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










          • $begingroup$
            OP: you should probably un-accept this. This isn't a good answer.
            $endgroup$
            – Randall
            3 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
            3 hours ago



















          3












          $begingroup$

          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









          $endgroup$













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            2 Answers
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            active

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            2 Answers
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            active

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            active

            oldest

            votes






            active

            oldest

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            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
              3 hours ago










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










            • $begingroup$
              OP: you should probably un-accept this. This isn't a good answer.
              $endgroup$
              – Randall
              3 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
              3 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
              3 hours ago










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










            • $begingroup$
              OP: you should probably un-accept this. This isn't a good answer.
              $endgroup$
              – Randall
              3 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
              3 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 4 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
              3 hours ago










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










            • $begingroup$
              OP: you should probably un-accept this. This isn't a good answer.
              $endgroup$
              – Randall
              3 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
              3 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
              3 hours ago










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










            • $begingroup$
              OP: you should probably un-accept this. This isn't a good answer.
              $endgroup$
              – Randall
              3 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
              3 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
            3 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
            3 hours ago












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




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












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






            $begingroup$
            OP: you should probably un-accept this. This isn't a good answer.
            $endgroup$
            – Randall
            3 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
            3 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
            3 hours ago











            3












            $begingroup$

            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









            $endgroup$


















              3












              $begingroup$

              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









              $endgroup$
















                3












                3








                3





                $begingroup$

                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 3 hours ago









                Robert IsraelRobert Israel

                320k23209459




                320k23209459






























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