Show that the product of the Jacobian and the inverse Jacobian is 1

Question

I have seen the following fact in a textbook, but am having trouble proving it. If the Jacobian ("stretch factor" for change-of-variables) is given by

$\left | \frac{\partial (x,y)}{\partial (u,v)} \right | = \begin{vmatrix} \frac{\partial x}{\partial u}& \frac{\partial y}{\partial u}\\ \frac{\partial x}{\partial v}& \frac{\partial y}{\partial v} \end{vmatrix}$, with $x=x(u,v)$ and $y=y(u,v)$,

then

$\left | \frac{\partial (x,y)}{\partial (u,v)} \right | \left | \frac{\partial (u,v)}{\partial (x,y)} \right | = 1$.

I am attempting to show this is true by multiplying the matrices and then taking the determinant of the product, but I do not understand how the product becomes equivalent to $\begin{bmatrix} 1& 0\\ 0& 1 \end{bmatrix}$.

Thanks!

Answer 1

We have $$x = x(u,v) \ \ \ \text{and} \ \ \ y = y(u,v)$$ then the chain rule for functions of several variables states that $$ \frac{\partial x}{\partial x} = \frac{\partial x}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial x}{\partial v} \frac{\partial v}{\partial x}$$ But $$\frac{\partial x}{\partial x} = 1$$ I'll use these facts later on. For now,

I believe we are trying to prove $$ \frac{\partial (x,y)}{\partial(u,v)}\frac{\partial(u,v)}{\partial(x,y)} = I$$ where $$I = \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}$$ We'll also prove $$\Bigg \lvert \frac{\partial (x,y)}{\partial(u,v)}\frac{\partial(u,v)}{\partial(x,y)}\Bigg \rvert = 1$$

So we begin:

$$\frac{\partial (x,y)}{\partial(u,v)}\frac{\partial(u,v)}{\partial(x,y)} = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}\end{vmatrix}\begin{vmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}\end{vmatrix}$$

$$ = \det \Bigg (\begin{bmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}\end{bmatrix} \Bigg ) \det \Bigg (\begin{bmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}\end{bmatrix} \Bigg )$$

$$ = \det \Bigg ( \begin{bmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}\end{bmatrix}\begin{bmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}\end{bmatrix} \Bigg )$$

by the identity $$\det(AB) = \det(A) \det(B)$$

so $$ \det \Bigg ( \begin{bmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}\end{bmatrix}\begin{bmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}\end{bmatrix} \Bigg )$$

$$ = \det \Bigg ( \begin{bmatrix} \frac{\partial x}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial x}{\partial v} \frac{\partial v}{\partial x} & \frac{\partial x}{\partial u} \frac{\partial u}{\partial y} + \frac{\partial x}{\partial v} \frac{\partial v}{\partial y} \\ \frac{\partial y}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial y}{\partial v} \frac{\partial v}{\partial x} & \frac{\partial y}{\partial u} \frac{\partial u}{\partial y} + \frac{\partial y}{\partial v} \frac{\partial v}{\partial y}\end{bmatrix} \Bigg )$$

$$ = \det \Bigg ( \begin{bmatrix} \frac{\partial x}{\partial x} & \frac{\partial x}{\partial y} \\ \frac{\partial y}{\partial x} & \frac{\partial x}{\partial x}\end{bmatrix} \Bigg )$$

$$ = \det \Bigg ( \begin{bmatrix} 1 & \frac{\partial x}{\partial y} \\ \frac{\partial y}{\partial x} & 1\end{bmatrix} \Bigg ) $$ by the chain rule for functions of several variables, using what we found above.

Now, since $x$ is not a function of $y$, and $y$ is not a function of $x$, we have $$ \frac{\partial x}{\partial y} = 0 \ \ \ \text{and} \ \ \ \frac{\partial y}{\partial x} = 0$$

so $$ \det \Bigg ( \begin{bmatrix} 1 & \frac{\partial x}{\partial y} \\ \frac{\partial y}{\partial x} & 1\end{bmatrix} \Bigg ) = \det \Bigg (\begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix} \Bigg ) = \det(I)$$

so we've proven that $$ \frac{\partial (x,y)}{\partial(u,v)}\frac{\partial(u,v)}{\partial(x,y)} = I $$

Now, $$ \det(I) = \det \Bigg (\begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix} \Bigg ) = 1 $$

so we've also proven that $$ \Bigg \lvert \frac{\partial (x,y)}{\partial(u,v)}\frac{\partial(u,v)}{\partial(x,y)}\Bigg \rvert = 1 $$ as required

Answer 2

Do the calculation straight forward.

$$ \frac{\partial (x,y)}{\partial(u,v)} = \begin{vmatrix} x_u & x_v \\ y_u & y_v \\ \end{vmatrix} $$ $$ \frac{\partial (x,y)}{\partial(u,v)} = \begin{vmatrix} u_x & u_y \\ v_x & v_y \\ \end{vmatrix} $$ So, we want to prove is, $$ \frac{\partial (x,y)}{\partial(u,v)}\frac{\partial(u,v)}{\partial(x,y)} = x_uy_vu_xv_y + xvy_uu_yv_x - x_uy_vu_yv_x - x_vy_uu_xv_y = 1 $$ Since $x = x(u, v)$, $y = y(u, v)$, Thus, $$dx = x_udu + x_vdv$$ $$dy = y_udu + y_vdv$$ By chain rule, $$\frac{dx}{dx} = x_uu_x + x_vv_x = 1 \qquad (1)$$ $$\frac{dx}{dy} = x_uu_y + x_vv_y = 0 \qquad (2)$$ $$\frac{dy}{dy} = y_uu_y + y_vv_y = 1 \qquad (3)$$ $$\frac{dy}{dx} = y_uu_x + y_vv_x = 0 \qquad (4)$$ Then, showing $$(1) \times (3) - (2) \times (4) = \frac{\partial (x,y)}{\partial(u,v)}\frac{\partial(u,v)}{\partial(x,y)} = x_uy_vu_xv_y + xvy_uu_yv_x - x_uy_vu_yv_x - x_vy_uu_xv_y = 1$$ Q.E.D