1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 $\vec x=(0,0,1)$

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.