Consider for simplicity the 2-dimensional space. Define a triangle $T$ on this space that has the vertices $$v_i=(x_i,y_i),\,i=1,2,3$$
Define the reference triangle $\hat T$ as the triangle with vertices $$\hat v_1 = (0,0), \hat v_2 = (1,0), \hat v_3 = (0,1)$$
We can define an affine linear transformation between these triangles by $$\begin{eqnarray}\Phi:\hat T&\to& T\\(\hat x,\hat y)&\mapsto& \begin{bmatrix}x_1\\y_1\end{bmatrix} + \begin{bmatrix}x_2-x_1 & x_3-x_1\\y_2-y_1&y_3-y_1\end{bmatrix}\begin{bmatrix}\hat x\\\hat y\end{bmatrix} =: \begin{bmatrix}x\\y\end{bmatrix}\end{eqnarray}$$
This map transforms the vertices of $\hat T$ to the vertices of $T$.
Now I'm reading, that affine maps DON'T conserve the vector-property of being outer normal. Why is this the case?
I tried for example with $\begin{bmatrix}-1\\0\end{bmatrix}$ which is outer normal to $\hat T$ at $(0,0.5)$, and everything works out fine. The transformed vector is again normal to one side of $T$.
Explanation is appreciated 🙂
Thanks!
EDIT: Here's my computation
Let the triangle $T$ be given by the vertices
$$v_1 = (1,2), v_2 = (3,1), v_3 = (2,4)$$ The affine map reads then
$$\Phi(\hat x,\hat y) = \begin{bmatrix}1\\2\end{bmatrix} + \begin{bmatrix}2&1\\-1&2\end{bmatrix}\begin{bmatrix}\hat x\\\hat y\end{bmatrix}$$
Take the vector $n = \begin{bmatrix}-1\\0\end{bmatrix}$ that's perpendicular the side of $\hat{T}$ connecting $(0,0)$ and $(0,1)$, so it's e.g. an outer normal at $(0,0.5)$. Therefore then I want to map this direction using $\Phi$ I map the points $(0,0.5)$ and $(-1,0.5)$ (because $n$ is the vector connecting them) and then take the difference to again get the direction connecting the mapped points. That is
$$\Phi(-1,0.5)-\Phi(0,0.5) = \begin{bmatrix}-2\\1\end{bmatrix}$$
This vector is perpendicular to the side of T connecting $(1,2)$ and $(2,4)$:
$$\begin{bmatrix}-2&1\end{bmatrix}\begin{bmatrix}1\\-2\end{bmatrix} = 0$$
I mean, it is clear, that the length of the vector is not one, but here it states that the angles don' get transformed.
Best Answer
An affine transform may not preserve orthogonality.
For example, the points $(0,0),(1,0),(0,1)$ form a right angle.
By the transformation $(x,y)\to(x,x+y)$, they become $(0,0),(1,1),(0,1)$. These do no form a right angle.
An affine transform has six degrees of freedom (independent coefficients), and can map any triangle onto any other triangle. All angles can change.
By the way, linear transforms don't preserve orthogonality either.