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I want to calcualte the inverse of $f(x,y)=\frac{1}{2}(x^2+y^2, x^2-y^2)$.

My book does it like this:

$f: (x,y) \to \begin{pmatrix}u\\v\end{pmatrix}=\frac{1}{2}\begin{pmatrix}x^2+y^2\\x^2-y^2\end{pmatrix}$

Now they calculate $u+v=...=x^2 \Rightarrow x=\sqrt{u+v}$ and $u-v=..=y^2 \Rightarrow y=\sqrt{u-v}$

Now, I'm wondering, couldn't I solve for x and y, like I'd do for 1-dim functions? Like:

$x=\frac{1}{2}(x^2+y^2) \Rightarrow x_{1,2}=1 \mp \sqrt{1-2y^2} \Rightarrow y_{1,2}=1 \mp \sqrt{1-2x^2}$

I know there are also Theorems for this and what not. But I'm wondering if I could do what I just did. I don't really like the first approach since it's not dependend on $x,y$

xotix
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1 Answers1

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I haven't studied all of the technicalities behind this, but here are my two cents.

Think about how this works in the $\mathbb{R} \to \mathbb{R}$ approach: you want to show that for every $y$, there is only one $x$. In the algebraic procedure, it's a clean $x \leftrightarrow y$ swap, solve for $y$, you get the inverse function.

In the $\mathbb{R}^2 \to \mathbb{R}^2$ case (say $(x, y) \mapsto (z_1, z_2)$), you want to show for every $(z_1, z_2)$, there is only one $(x, y)$. There isn't an easy way to do this, because there's not an easy one-to-one correspondence between one of the variables in $(x, y)$ and another one of the variables in $(z_1, z_2)$. You can't, for example, necessarily say that $x \leftrightarrow z_1$ and $y \leftrightarrow z_2$ are valid swaps, because $z_1$ and $z_2$ are both dependent on $x$ and $y$. Thus, the best you can do is manipulate $z_1$ and $z_2$ until you get $x$ and $y$ in terms of $z_1$ and $z_2$.

Clarinetist
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