Det of Jacobian becomes 0 when computing Joint density of $V_1=X^2+Y^2$ and $V_2=X^2−Y^2$

Question

I am trying to solve the following.

Let $X, Y$ be independent random variables distributed as Uniform$([−1, 1])$. Give the joint density of $U,V$, where $U=X^2+Y^2$ and $V =X^2 - Y^2$.

I am using change of variables method to solve this.

g₁^-1(U,V) -> (U + V / 2)^1/2 = x

g₂^-1(U,V) -> (U - V / 2)^1/2 = y

Let L = g₁^-1(U,V) and K = g₂^-1(U,V)

When I compute Det of Jacobian matrix, it becomes zero.

$$ \det\begin{pmatrix} \frac{\partial L}{\partial u} & \frac{\partial L}{\partial v}\\ \frac{\partial K}{\partial u} & \frac{\partial K}{\partial v} \end{pmatrix} = \det\begin{pmatrix} \frac{\sqrt 2}{4\sqrt{U + V}} & \frac{\sqrt 2}{4\sqrt{U - V}} \\ \frac{\sqrt 2}{4\sqrt{U + V}} & \frac{\sqrt 2}{4\sqrt{U - V}} \end{pmatrix}= 0 $$

Need help on what I making mistake.

Thanks

Answer 1

$U = X^2 + Y^2, V = X^2 - Y^2$

$x = \pm \sqrt{\frac{u+v}{2}}, y = \pm\sqrt{\frac{u-v}{2}}$

$\det\begin{pmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v}\\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{pmatrix} = \det\begin{pmatrix} \pm\frac{1}{2 \sqrt2\sqrt{u + v}} & \pm\frac{1}{2 \sqrt2\sqrt{u + v}} \\ \pm\frac{1}{2 \sqrt2\sqrt{u - v}} & \mp\frac{1}{2 \sqrt2\sqrt{u - v}} \end{pmatrix}= \pm\frac{1}{4\sqrt{u^2 - v^2}}$

So, $|J| = \frac{1}{4\sqrt{u^2 - v^2}}$