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Let $A$ be a matrix with determinant 1. Then we call a general affine transformation, a transformation of the form \begin{align*} \begin{bmatrix}x'\\y'\end{bmatrix}=A\begin{bmatrix}x\\y\end{bmatrix}+\begin{bmatrix}r\\s\end{bmatrix} \end{align*} Let $p_1,p_2,p_3$ be colinear points. Prove that \begin{align*} \frac{||p_2-p_1||}{||p_2-p_3||} \end{align*} Is preserved under these transformations.

I have been working this proof for some time now. I can break these transformations down into pure translations, rotations, and scalings. Everything works out quite nicely for the translations and the rotations but if I try to prove this using a scaling matrix i.e. a matrix of the form \begin{align*} \begin{bmatrix}a&0\\0&1/a\end{bmatrix} \end{align*}.

Any help is greatly appreciated.

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  • $\begingroup$ Use \| instead of || for the vector norm. $\endgroup$
    – amd
    Commented Feb 26, 2019 at 23:57
  • $\begingroup$ This “general” affine transformation isn’t as general as it could be. We only need $\det A\ne0$. With $\det A=1$, the transformation also preserves areas and orientation. $\endgroup$
    – amd
    Commented Feb 26, 2019 at 23:59

2 Answers 2

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Since $p_1, p_2$ and $p_3$ are colinear, we can write $\vec{v} = p_2 - p_3$ and $p2 - p_1 = \lambda \vec{v}$ whence $$ \frac{||p_2-p_1||}{||p_2-p_3||} = \lambda $$ Now write the components of $v$ as $(v_x, v_y)$; we have $$ ||p_2-p_3|| = (v_x, v_y) \\ ||p_2-p_1|| = (\lambda v_x, \lambda v_y) $$

Now apply (multiply by) a scaling transformation. Although it is adequate to just use your example, it might be illuminating to use a general matrix $$S = \pmatrix{a & b \\ c & d}$$with $ad-bc = 1$. $$ S (p_2-p_3)= \pmatrix{ a \,v_x + b \,v_y \\ c \,v_x + d \,v_y}\\ S (p_2-p_1) = \pmatrix{ a \lambda \,v_x + b \lambda\,v_y \\ c \lambda\,v_x + d \lambda \,v_y} = \lambda S(p2-p3)\\ $$

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You don’t really need to break the transformation down into cases. This is a straightforward consequence of the linearity of $A$ and homogeneity of norms.

Note first that the translation part of the transformation—the addition of $[r,s]^T$—is irrelevant since it cancels when you subtract the images of the points from each other.

The line through $p_2$ and $p_3$ can be parameterized as $(1-\lambda)p_2+\lambda p_3$. Since $p_1$ is colinear with these two points, there’s some real $\lambda$ for which $p_1 = (1-\lambda)p_2+\lambda p_3$, from which $$p_2-p_1 = p_2-((1-\lambda)p_2+\lambda p_3) = \lambda(p_2-p_3)$$ and so $${\|p_2-p_1\|\over\|p_2-p_3\|} = |\lambda|.$$

Now, by linearity, $Ap_1 = (1-\lambda)Ap_2+\lambda Ap_3$, therefore $$p_2'-p_1' = Ap_2-((1-\lambda)Ap_2+\lambda Ap_3) = \lambda(Ap_2-Ap_3)=\lambda(p_2'-p_3').$$ If $\det A\ne0$, then $p_2\ne p_3$ implies $p_2'\ne p_3'$ and the result follows immediately.

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