Affine transformation on a plane

Question

Given is a plane with the points $A = (1, 1, 1) , B = (1, 0, 1), C = (0, 1, 1)$ and their affine transformations $f(A) = (0,0), f(B) = (1, 0), f(C) = (0, 1)$

I must find all points $x$ on the plane that have the affine transformation $f(x) = (1, 1)$.

When I look at the points and their transformations I think that $x=(0,0,1)$ should be a possible solution, but how do I know if there are also other points that get mapped to $(1,1)$? I tried to use the formula $x'=a+Ax$ and find values for $A$ to get the transformation but had no success with that.

Answer 1

Notice that your original plane is $z=1$. So you can ignore the last coordinate. Then you have a transformation: $$\vec y= \begin{bmatrix}a & b\\c & d\end{bmatrix}\vec x+\begin{bmatrix}e\\f\end{bmatrix}$$ Now we plug in your three points (only the first two coordinates) to get the following system of equations: $$a+b+e=0\\c+d+f=0\\a+0+e=1\\c+0+f=0\\0+b+e=0\\0+d+f=1$$ From the first and third you get $b=-1$, from the second and fourth you get $d=0$. Plugging in these values in the last two equations, you have $e=1$ and $f=1$. Finally, $a=0$ and $c=-1$. So now your original transformation is $$\vec y= \begin{bmatrix}0 & -1\\-1 & 0\end{bmatrix}\vec x+\begin{bmatrix}1\\1\end{bmatrix}$$ We want points that transform into $(1,1)$. $$\begin{bmatrix}1\\1\end{bmatrix}= \begin{bmatrix}0 & -1\\-1 & 0\end{bmatrix}\vec x+\begin{bmatrix}1\\1\end{bmatrix}$$ Since the determinant of the matrix is non-zero, you get that the only solution is $\vec x=(0,0)$, or in the original plane $\vec x=(0,0,1)$

Answer 2

Any affine transformation has a linear part: if $f(\mathbf{x})$ is an affine transformation, then as you said $f(\mathbf{x})=\mathbf{a}+L\mathbf{x}$, where $\mathbf{a}$ is a constant vector and $L$ is a linear transformation. Note that in this problem this linear transformation $L$ is from $\mathbb{R}^3$ to $\mathbb{R}^2$. So by dimension considerations it can't be one-to-one; instead, it has to have a nontrivial kernel (null space), and so there are infinitely many points in the preimage of any point either with respect to $L$ or with respect to $f$.

From the data given to you, you know three values of the affine transformation $f$. You need to find all points $D$ such that $f(D)=(1,1)$. You correctly found one of them $(0,0,1)$. The issue now is that there is a whole affine plane of solutions to the equation $f(D)=(1,1)$ given by $D\in(0,0,1)+\ker(L)$. But we can't determine $\ker(L)$ from the given data, because we need to have at least one more data point to do that.

In general: a linear transformation from an $n$-dimensional space is completely determined by its values on $n$ linearly independent vectors; but an affine transformation from an $n$-dimensional space is completely determined by its values at $n+1$ points in general position.