# The Jacobian of the inverse of a transformation

The Jacobian of the inverse of a transformation is the inverse of the Jacobian of that transformation

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In this post, we would like to illustrate the meaning of

the Jacobian of the inverse of a transformation = the inverse of the Jacobian of that transformation

by proving a special case.

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Consider a transformation $\mathscr{T}: \bar{x}^i=\bar{x}^i (x^1,x^2)$, which is an one-to-one mapping from unbarred $x^i$‘s to barred $\bar{x}^i$ coordinates, where $i=1, 2$.

By definition, the Jacobian matrix J of $\mathscr{T}$ is $J= \begin{pmatrix} \displaystyle{\frac{\partial \bar{x}^1}{\partial x^1}} & \displaystyle{\frac{\partial \bar{x}^1}{\partial x^2}} \\ \displaystyle{\frac{\partial \bar{x}^2}{\partial x^1}} & \displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}} \end{pmatrix}$

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Now we consider the the inverse of the transformation $\mathscr{T}$: $\mathscr{T}^{-1}: x^i=x^i(\bar{x}^1,\bar{x}^2)$

By definition, the Jacobian matrix $\bar{J}$ of this inverse transformation, $\mathscr{T}^{-1}$, is $\bar{J}= \begin{pmatrix} \displaystyle{\frac{\partial x^1}{\partial \bar{x}^1}} & \displaystyle{\frac{\partial x^1}{\partial \bar{x}^2}} \\ \displaystyle{\frac{\partial x^2}{\partial \bar{x}^1}} & \displaystyle{\frac{\partial x^2}{\partial \bar{x}^2}} \end{pmatrix}$

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On the other hand, the inverse of Jacobian $J$ of the original transformation $\mathscr{T}$ is $J^{-1}=\displaystyle{\frac{1}{ \begin{vmatrix} \displaystyle{\frac{\partial \bar{x}^1}{\partial x^1}} & \displaystyle{\frac{\partial \bar{x}^1}{\partial x^2}} \\ \displaystyle{\frac{\partial \bar{x}^2}{\partial x^1}} & \displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}} \end{vmatrix} }} \begin{pmatrix} \displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}} & \displaystyle{-\frac{\partial \bar{x}^1}{\partial x^2}} \\ \displaystyle{-\frac{\partial \bar{x}^2}{\partial x^1}} & \displaystyle{\frac{\partial \bar{x}^1}{\partial x^1}} \end{pmatrix}$

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If $\bar{J} = J^{-1}$, their (1, 1)-elementd should be equation: $\displaystyle{\frac{\partial x^1}{\partial \bar{x}^1}}\stackrel{?}{=}\displaystyle{\frac{1}{\displaystyle{\frac{\partial \bar{x}^1}{\partial x^1}}\displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}}-\displaystyle{\frac{\partial \bar{x}^1}{\partial x^2}}\displaystyle{\frac{\partial \bar{x}^2}{\partial x^1}} }} \bigg( \displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}} \bigg)$

Let’s try to prove that.

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Consider equations $\bar{x}^1 = \bar{x}^1(x^1,x^2)$ $\bar{x}^2 = \bar{x}^2(x^1,x^2)$

Differentiate both sides of each equation with respect to $\bar{x}^1$, we have: $A := 1=\displaystyle{\frac{\partial \bar{x}^1}{\partial \bar{x}^1}=\frac{\partial \bar{x}^1}{\partial x^1}\frac{\partial x^1}{\partial \bar{x}^1}+\frac{\partial \bar{x}^1}{\partial x^2}\frac{\partial x^2}{\partial \bar{x}^1}}$ $B := 0 = \displaystyle{\frac{\partial \bar{x}^2}{\partial \bar{x}^1}=\frac{\partial \bar{x}^2}{\partial x^1}\frac{\partial x^1}{\partial \bar{x}^1}+\frac{\partial \bar{x}^2}{\partial x^2}\frac{\partial x^2}{\partial \bar{x}^1}}$

. $A \times \displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}}:~~~~~C := \displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}=\frac{\partial \bar{x}^1}{\partial x^1}\frac{\partial x^1}{\partial \bar{x}^1}\frac{\partial \bar{x}^2}{\partial x^2}+\frac{\partial \bar{x}^1}{\partial x^2}\frac{\partial x^2}{\partial \bar{x}^1}\frac{\partial \bar{x}^2}{\partial x^2}}$ $B \times \displaystyle{\frac{\partial \bar{x}^1}{\partial x^2}}:~~~~~D := \displaystyle{0=\frac{\partial \bar{x}^2}{\partial x^1}\frac{\partial x^1}{\partial \bar{x}^1}\frac{\partial \bar{x}^1}{\partial x^2}+\frac{\partial \bar{x}^2}{\partial x^2}\frac{\partial x^2}{\partial \bar{x}^1}\frac{\partial \bar{x}^1}{\partial x^2}}$

. $D-C:$ $\displaystyle{ \frac{\partial \bar{x}^2}{\partial x^2}= \bigg( \frac{\partial \bar{x}^1}{\partial x^1}\frac{\partial \bar{x}^2}{\partial x^2} - \frac{\partial \bar{x}^2}{\partial x^1}\frac{\partial \bar{x}^1}{\partial x^2}\bigg) \frac{\partial x^1}{\partial \bar{x}^1}}$,

results $\displaystyle{ \frac{\partial x^1}{\partial \bar{x}^1}}=\frac{\displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}}}{\displaystyle{\frac{\partial \bar{x}^1}{\partial x^1}\frac{\partial \bar{x}^2}{\partial x^2} - \frac{\partial \bar{x}^1}{\partial x^2}\frac{\partial \bar{x}^2}{\partial x^1}}}$

— Me@2018-08-09 09:49:51 PM

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2018.08.09 Thursday (c) All rights reserved by ACHK