Conditions for Similarity in an Affine Transformation

Proof

Consider the general affine transformation of the plane:

\[ \begin{cases} x' = a x + b y + c \\[4pt] y' = a' x + b' y + c' \end{cases} \]

Let \( A(x_A,y_A) \) and \( B(x_B,y_B) \) be two arbitrary points. Their images satisfy:

\[ x_B' - x_A' = a(x_B-x_A) + b(y_B-y_A) \] \[ y_B' - y_A' = a'(x_B-x_A) + b'(y_B-y_A) \]

The Euclidean distance between the image points is therefore:

\[ \overline{A'B'} = \sqrt{\big[a(x_B-x_A)+b(y_B-y_A)\big]^2 + \big[a'(x_B-x_A)+b'(y_B-y_A)\big]^2} \]

Squaring both sides yields a more manageable expression:

\[ (\overline{A'B'})^2 = \big[a(x_B-x_A)+b(y_B-y_A)\big]^2 + \big[a'(x_B-x_A)+b'(y_B-y_A)\big]^2 \]

Expanding the two squares gives:

\[ \begin{aligned} (\overline{A'B'})^2 &= a^2(x_B-x_A)^2 + 2ab(x_B-x_A)(y_B-y_A) + b^2(y_B-y_A)^2 \\ &\quad + a'^2(x_B-x_A)^2 + 2a'b'(x_B-x_A)(y_B-y_A) + b'^2(y_B-y_A)^2 \end{aligned} \]

Collecting like terms produces the standard quadratic form associated with the linear part of the affine map:

\[ (\overline{A'B'})^2 = (a^2+a'^2)(x_B-x_A)^2 + (b^2+b'^2)(y_B-y_A)^2 + 2(ab+a'b')(x_B-x_A)(y_B-y_A) \]

The original squared distance is simply:

\[ (\overline{AB})^2 = (x_B-x_A)^2 + (y_B-y_A)^2. \]

We now impose the metric requirement for similarity. By definition:

\[ \overline{A'B'} = k\,\overline{AB} \]

which is equivalent to

\[ (\overline{A'B'})^2 = k^2(\overline{AB})^2. \]

Substituting the expanded expression for \( (\overline{A'B'})^2 \), we obtain:

\[ (a^2+a'^2)(x_B-x_A)^2 + (b^2+b'^2)(y_B-y_A)^2 + 2(ab+a'b')(x_B-x_A)(y_B-y_A) = k^2(\overline{AB})^2. \]

The first condition, \( ab+a'b' = 0 \), eliminates the mixed term entirely:

\[ \require{cancel} (a^2+a'^2)(x_B-x_A)^2 + (b^2+b'^2)(y_B-y_A)^2 + \cancel{2(ab+a'b')(x_B-x_A)(y_B-y_A)} = k^2(\overline{AB})^2. \]

We are left with:

\[ (a^2+a'^2)(x_B-x_A)^2 + (b^2+b'^2)(y_B-y_A)^2 = k^2(\overline{AB})^2. \]

The second condition requires that the two coefficients be equal, that is:

\[ a^2+a'^2 = b^2+b'^2 = k^2. \]

Substituting this common value gives:

\[ k^2(x_B-x_A)^2 + k^2(y_B-y_A)^2 = k^2(\overline{AB})^2. \]

Factoring out \( k^2 \):

\[ k^2\big[(x_B-x_A)^2 + (y_B-y_A)^2\big] = k^2(\overline{AB})^2. \]

Since the expression in brackets is exactly \( (\overline{AB})^2 \), the equality holds identically:

\[ (\overline{A'B'})^2 = k^2(\overline{AB})^2. \]

Taking square roots, we recover the defining property of a similarity transformation:

\[ \overline{A'B'} = k\,\overline{AB}. \]

Thus the affine map preserves all distances up to the constant factor \( k \), and is therefore a similarity transformation with similarity ratio

\[ k = \sqrt{a^2+a'^2} = \sqrt{b^2+b'^2}. \]

The argument is complete.

And so on.