Transformations of Level Sets

1 General Transformation

Theorem 1.

Let $f:\mathbb{R}^{n}\to\mathbb{R}$ define a level set $D=L_{0}(f)=\left\{\boldsymbol{r}\in\mathbb{R}^{n}\;\middle|\;f(\boldsymbol{r})% =0\right\}$ . If $T:\mathbb{R}^{n}\to\mathbb{R}^{n}$ is a bijective (invertible) transformation, then the image set $D^{\prime}=T(D)$ under $T$ is given by:

D^{\prime}=\left\{\boldsymbol{r}\in\mathbb{R}^{n}\;\middle|\;(f\circ T^{-1})(% \boldsymbol{r})=0\right\}

(1)

Informal Proof

By definition of function composition, $(f\circ T^{-1})(\boldsymbol{r})=f(T^{-1}(\boldsymbol{r}))$ .
Letting $\boldsymbol{u}=T^{-1}(\boldsymbol{r})$ , we have $\boldsymbol{r}=T(\boldsymbol{u})$ . Substituting back gives:

	$\displaystyle D^{\prime}$	$\displaystyle=\left\{T(\boldsymbol{u})\;\middle\|\;f(\boldsymbol{u})=0\right\}$
	$\displaystyle D^{\prime}$	$\displaystyle=T(D)$

Proof.

	$\displaystyle\boldsymbol{r}\in T(D)$	$\displaystyle\iff\exists\boldsymbol{u}\in D\quad\text{such that}\quad% \boldsymbol{r}=T(\boldsymbol{u})$
		$\displaystyle\iff\exists\boldsymbol{u}\in\mathbb{R}^{n}\quad\text{such that}% \quad f(\boldsymbol{u})=0\quad\text{and}\quad\boldsymbol{u}=T^{-1}(\boldsymbol% {r})$
		$\displaystyle\iff f(T^{-1}(\boldsymbol{r}))=0$
		$\displaystyle\iff(f\circ T^{-1})(\boldsymbol{r})=0$

Therefore, $D^{\prime}=T(D)=\left\{\boldsymbol{r}\in\mathbb{R}^{n}\;\middle|\;(f\circ T^{-% 1})(\boldsymbol{r})=0\right\}$ . ∎

2 Examples - 2D Level Sets (Level Curves)

Let $\boldsymbol{r}=\begin{pmatrix}x\\ y\end{pmatrix}$ and $D=\left\{\begin{pmatrix}x\\ y\end{pmatrix}\;\middle|\;f(x,y)=0\right\}$ .

2.1 Translation

For a translation vector $\Delta\boldsymbol{r}=\begin{pmatrix}\Delta x\\ \Delta y\end{pmatrix}$ , the translation mapping is $T_{\Delta\boldsymbol{r}}(\boldsymbol{r})=\boldsymbol{r}+\Delta\boldsymbol{r}$ .

	$\displaystyle D_{T}$	$\displaystyle=\left\{\begin{pmatrix}x\\ y\end{pmatrix}\;\middle\|\;f(x-\Delta x,y-\Delta y)=0\right\}$
	$\displaystyle\boldsymbol{u}$	$\displaystyle=\begin{pmatrix}x^{\prime}\\ y^{\prime}\end{pmatrix}=\begin{pmatrix}x-\Delta x\\ y-\Delta y\end{pmatrix}=\boldsymbol{r}-\Delta\boldsymbol{r}=T_{-\Delta% \boldsymbol{r}}(\boldsymbol{r})$
	$\displaystyle\boldsymbol{r}$	$\displaystyle=\boldsymbol{u}+\Delta\boldsymbol{r}=T_{\Delta\boldsymbol{r}}(% \boldsymbol{u})$
	$\displaystyle D_{T}$	$\displaystyle=\left\{T_{\Delta\boldsymbol{r}}\left(\begin{pmatrix}x^{\prime}\\ y^{\prime}\end{pmatrix}\right)\;\middle\|\;f(x^{\prime},y^{\prime})=0\right\}=T% _{\Delta\boldsymbol{r}}(D)$

2.2 Scaling

Let $S=\begin{pmatrix}s_{x}&0\\ 0&s_{y}\end{pmatrix}$ be the scaling matrix, where $L_{S}(\boldsymbol{r})=S\boldsymbol{r}=\begin{pmatrix}s_{x}x\\ s_{y}y\end{pmatrix}$ .

	$\displaystyle D_{S}$	$\displaystyle=\left\{\begin{pmatrix}x\\ y\end{pmatrix}\;\middle\|\;f\left(\frac{x}{s_{x}},\frac{y}{s_{y}}\right)=0\right\}$
	$\displaystyle\boldsymbol{u}$	$\displaystyle=\begin{pmatrix}x^{\prime}\\ y^{\prime}\end{pmatrix}=\begin{pmatrix}x/s_{x}\\ y/s_{y}\end{pmatrix}=S^{-1}\boldsymbol{r}=L_{S}^{-1}(\boldsymbol{r})$
	$\displaystyle\boldsymbol{r}$	$\displaystyle=S\boldsymbol{u}=L_{S}(\boldsymbol{u})$
	$\displaystyle D_{S}$	$\displaystyle=\left\{S\begin{pmatrix}x^{\prime}\\ y^{\prime}\end{pmatrix}\;\middle\|\;f(x^{\prime},y^{\prime})=0\right\}=L_{S}(D)$

2.3 Rotation

Let $R_{\theta}=\begin{pmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{pmatrix}$ . The inverse transformation matrix is:

R_{\theta}^{-1}=R_{-\theta}=\begin{pmatrix}\cos\theta&\sin\theta\\ -\sin\theta&\cos\theta\end{pmatrix}

Applying $R_{\theta}^{-1}$ to $\boldsymbol{r}$ yields:

L_{R_{\theta}^{-1}}(\boldsymbol{r})=R_{\theta}^{-1}\begin{pmatrix}x\\ y\end{pmatrix}=\begin{pmatrix}x\cos\theta+y\sin\theta\\ -x\sin\theta+y\cos\theta\end{pmatrix}

Thus, the rotated set $D_{R}$ is given by:

	$\displaystyle D_{R}$	$\displaystyle=\left\{\begin{pmatrix}x\\ y\end{pmatrix}\;\middle\|\;f(x\cos\theta+y\sin\theta,-x\sin\theta+y\cos\theta)=% 0\right\}$
	$\displaystyle\boldsymbol{u}$	$\displaystyle=\begin{pmatrix}x^{\prime}\\ y^{\prime}\end{pmatrix}=R_{\theta}^{-1}\boldsymbol{r}\iff\boldsymbol{r}=R_{% \theta}\boldsymbol{u}$
	$\displaystyle D_{R}$	$\displaystyle=\left\{R_{\theta}\begin{pmatrix}x^{\prime}\\ y^{\prime}\end{pmatrix}\;\middle\|\;f(x^{\prime},y^{\prime})=0\right\}=L_{R_{% \theta}}(D)$