# jacobian times its inverse - should be identity

Here's an easy one.

A Jacobian is $\frac{dx^i}{dy^j}$. The inverse is $\frac{dy^j}{dx^k}$. So, in tensor notation,

$\frac{dx^i}{dy^j} \frac{dy^j}{dx^k} = \frac{dx^i}{dx^k} = \delta^i_k$

Now I'll try to this as in matrix form, in two dimensions:

$\left[ \begin{array}{cc} \frac{dx^1}{dy^1} & \frac{dx^1}{dy^2} \\ \frac{dx^2}{dy^1} & \frac{dx^2}{dy^2} \\ \end{array} \right] \cdot \left[ \begin{array}{cc} \frac{dy^1}{dx^1} & \frac{dy^1}{dx^2} \\ \frac{dy^2}{dx^1} & \frac{dy^2}{dx^2} \\ \end{array} \right]$

The 1,2 element of this product is

$\frac{dx^1}{dy^1}\frac{dy^1}{dx^2} + \frac{dx^1}{dy^2}\frac{dy^2}{dx^2} = \frac{dx^1}{dx^2}+\frac{dx^1}{dx^2} = 0$

as required.

But looking at the 1,1 element,

$\frac{dx^1}{dy^1}\frac{dy^1}{dx^1} + \frac{dx^1}{dy^2}\frac{dy^2}{dx^1} = 2 \frac{dx^1}{dx^1} = 2$

which is wrong… but why?

In two dimensions, we have functions $f_1$ and $f_2$ which are both functions of $x_1$ and $x_2$. The Jacobian matrix is given by,

$$J=\left( \begin{matrix} \partial_1 f_1 & \partial_2 f_1 \\ \partial_1 f_2 & \partial_2 f_2 \end{matrix} \right)$$

where $\partial_n := \partial/\partial x_n$. Computing the inverse of the Jacobian using the standard formula yields,

$$J^{-1}=\frac{1}{\partial_1f_1 \partial_2f_2-\partial_2f_1\partial_1f_2}\left( \begin{matrix} \partial_2 f_2 & -\partial_2 f_1 \\ -\partial_1 f_2 & \partial_1 f_1 \end{matrix} \right)$$

Multiplying both matrices yields the desired result,

$$JJ^{-1}=J^{-1}J=\left( \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right)$$

You just used the wrong expression for the inverse.

• thank you. I knew I would feel stupid... Commented May 2, 2014 at 11:42
• @beginner: Yeah. But remember if the matrix is purely diagonal, you can take the reciprocal of each element to quickly obtain the inverse, but only in that case.
– JPhy
Commented May 2, 2014 at 12:42