# Undo this transformation

I have two variables $$x$$, $$y$$ and calculate the following:

$$a = \frac{x}{\sqrt{x^2+y^2}}$$, $$b = \frac{y}{\sqrt{x^2+y^2}}$$

Using $$a$$ and $$b$$ is there a way I can derive my original $$x$$ and $$y$$?

• No, because they stay the same if you multiply $x$ and $y$ by a constant, ie. replace $x$ by $x*k$ and $y$ by $y*k$. You're missing the constant $k = \sqrt{x^2+y^2}$ , which needs to be known in addition to $a$ and $b$ to recover $x$ and $y$. – user3257842 Feb 23 at 18:01
• note that $a = \cos\theta$ and $b=\sin\theta$, but we have no information on $r$, so the best that we can do to identify a point would be $(\tan^{-1}\frac{b}{a}, r)$. Alternatively, $(x,y) = (ar,br)$ – John Joy Feb 23 at 22:18

No. You can't

For $$x=1,y=0$$ you get $$a=1,b=0$$ also for $$x=2,y=0$$ you still get $$a=1,b=0$$. So given $$a,b$$ there is no way to determine $$x,y$$.

Note however that $$a/b$$ gives you $$x/y$$ (when defined, otherwise it tells you whether $$y$$ is zero or not). This is the best you can do as the pairs $$(x,y)$$ and $$(cx,cy)$$ will give you the same $$a$$ and $$b$$.

This the classical normalization operation :

$$\binom{a}{b} = \frac{1}{\sqrt{x^2+y^2}} \underbrace{\binom{x}{y}}_V=\frac{1}{\|V\|} V$$

transforming a vector into the proportional vector with unit norm ( = length) thus belonging to the unit circle.

It is like a projection onto a straight line (imagine the circle is unrolled). And, like a projection, there exists an infinity of vectors $$(x,y)$$ that have the same vector $$(a,b)$$ with unit length as their image. Thus this transformation has no inverse.

For example $$\binom{0.6}{0.8}$$ is the image of $$\binom{3}{4}$$, $$\binom{6}{8}$$, $$\binom{9}{12}$$,...

By squaring we get $$a^2=\frac{x^2}{x^2+y^2},b^2=\frac{y^2}{x^2+y^2}$$ and we get $$a^2+b^2=\frac{x^2+y^2}{x^2+y^2}=1$$ and you can not compute $$x$$ or $$y$$.