Affine Transformations

Today, however, Fahrenheit has been replaced by the Celsius (and in scientific applications, Kelvin) scale in all but a handful of the world's countries. On the kelvin scale, 0°K is equal to −273.15 °C and the boiling point of water is 373°K, which is 100°C. \[ \mbox{F}^{\circ} = \frac{9}{5}\,\mbox{C}^{\circ} + 32 . \] For converting Fahrenheit (°F) into Celsius (°C) scale, use the formula: \[ \mbox{C}^{\circ} = \frac{5}{9} \left( \mbox{F}^{\circ} - 32^{\circ} \right) . \] Note that both, the transformation from Fahrenheit scale into Celsius scale and its reverse, are not linear maps because they do not preserve zero. Actually, these transformations are affine ones.    ■

Example 2: Let us consider an affine transformation \[ \begin{split} x &\mapsto x+y +1 , \\ y &\mapsto 2\,y - 1 . \end{split} \] It is convenient to rewrite this transformation in matrix/vector form using either column vectors \[ \begin{bmatrix} x \\ y \end{bmatrix} \,\mapsto \, \begin{bmatrix} 1 & 1 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} + \begin{bmatrix} \phantom{-}1 \\ -1 \end{bmatrix} \] or row vectors \[ \begin{bmatrix} x & y \end{bmatrix} \,\mapsto \, \begin{bmatrix} x & y \end{bmatrix} \begin{pmatrix} 1&0 \\ 1&2 \end{pmatrix} + \begin{bmatrix} 1 & -1 \end{bmatrix} . \] It is a “shear” followed by a translation. The effect of this shear on the square (𝑎, b, c, d) is shown in the following figure. The image of this square is the parallelogram.

Now we plot the same figures by keeping grid visible.

Note that we are using the second version of the AffineTransform. It has an argument set which is a list, {m, v}, where m is a linear transform and v is a shift (translate) vector. Note below the TransformationFunction it produces is an augmented matrix in the usual form with the b vector the far right column to the right of the dividing line.

The transformation function performs two operations: The linear transformation associated with the "m" matrix is the first. The second is the translation association with the "b" vector.

Applying the transform to each corner of the rectangle, we get the same transformed vertices images as provided in the narration above

Wolfram has a function, FindGeometricTransform, which essentially reverses the process we just completed. Below we provide Mathematica with just the new corners, the original corners and ask for an Affine Transform matrix. It returns our matrix above.

One might reasonably ask: "What does the last line mean? What does the row containing {0.,0.|1.}, of the TransformationFunction do?" Those questions come after answering the broader question: "What makes a mathematician?" Getting close to these answers takes a little thought about the order of learning things. A mathematician might say the first questions are premature. We will answer them later in this lesson, further down the page you are reading. But people today are impatient. Wolfram Mathematica version 14 offers a "Chat-Enabled Notebook" with a direct "tap" on the wonders of Artificial Intelligence. You can ask one of many Large Language Models your questions about that last line of the Transformation Function. Here is the answer you might get:

The "0 0 1" row in the transformation matrix ensures that when this matrix is multiplied with a point represented in homogeneous coordinates, the 1 in the third component of the point stays a 1. This is necessary to keep the point a point, and not a vector.

In contrast, if the transformation matrix is multiplied with a vector, the 0 in the third component of the vector stays a 0, ensuring that the vector remains a vector and is not translated. So, the "0 0 1" row is essentially a part of the mathematical machinery that allows for points and vectors to be treated differently by affine transformations, particularly translations. It doesn't directly affect the scaling, rotation, shear or translation applied to the points or vectors.

For now, patience is advised as we will revisit this later, in subsection devoted to augmented (affine) matrices.    ■

In general, for any given closed region, the area under an affine transformation A x + b is scaled by det(A). This result is valid for any linear mapping y = A x.

Illustrating this in Mathematica requires some careful definitions and distinguishing a transformation matrix with a zero determinant from one with a non-zero determinant. First, we define a transformation matrix, A, which has a zero determinant

Now we use a transformation matrix, B, with a non-zero determinant.

Part 1:

Part 2:

Part 3: We generate a plot showing the original lines and their transformed versions. The transformed lines are parallel to each other, just like the original lines. First, we define the affine transformation

Part 4: We have two parts. The first part plots a triangle T and its transformed triangle T1 side by side. T is mapped to T1 by the affine transformation.

The second part of item #4 generalized further by plotting an original quadrilateral (as two triangles) and the transformed quadrilateral (also as two triangles, which is no longer a quadrilateral). The original is mapped to the transformed one by the affine transformation.

Part 5: Here we need to illustrate that an affine transformation maps a parallelogram into a parallelogram.

Define a parallelogram *)

Part 6:

Here we need to illustrate that an affine transformation preserves the ratio of lengths. We can do this by defining a line segment and its transformation, then verifying the ratio of lengths before and after the transformation.

Let's consider a simple affine transformation f(x) = A x + b with \[ {\bf A} = \begin{bmatrix} 3&4 \\ -1&2 \end{bmatrix} , \qquad {\bf b} = \begin{bmatrix} 2 \\ -1 \end{bmatrix} , \] and four line segments S₃ = [1,2] of length √5 and S₂ = [2, 4], S₃ = [4, 2], S₄ = [−2, 4] of length √20, which is twice the length of S₁.

After transformation, these will become f(S₁) = [13, 2] and f(S₂) = [24, 5], f(S₃) = [22, 1], f(S₄) = [12, 9].

We plot the original line segments S₁ and S₂ and the transformed line segments f(S₁) and f(S₂). The ratio of the lengths of the transformed line segments is 2, which is the same as the ratio of the lengths of the original line segments.

Part 7: This example illustrates the fact that an Affine Transformation preserves the ratios of areas of polygons.

Define the polygons (that we choose as parallelogram and a triangle):

Part 7A:

Now we extend this part by considering a half unit circle instead of triangle: \[ C = \left\{ (x, y) \in \mathbb{R}^2 \ : \ 0 \le x^2 + y^2 \le 1, \quad x\ge 0, \ y\ge 0 \right\} . \tag{5.7.2}\]

The alert student might compare the range of X and the range of Y above with the ranges of the code for the same variables immediately below. While they are different, it is of no matter as using a different ranges in the code just changes the perimeter white space in the graphic.

The figure above does not provide a true presentation because Mathematica plots it using standard ordering of independent variable. You need to flip it with respect to ordinate (vertical axis) to obtain a true picture.

Now we plot the region after given affine transformation

We improve this plot by adding arrows to show how points are transformed.

In order to visualize an affine plane, we consider a 2D plane ℝ² inside ℝ³. In order to separate points from vectors, we choose two planes, one for points and another one for vectors, as in the picture below. We'll call the green one the vector space V ≌ ℝ² and the blue one as the point plane A. The plane V passes through the origin since it is a vector space, but the blue plane A does not. However, the inhabited set A looks almost exactly the same as V, having the exact same, flat geometry, and in fact A and V are simply translates of one another. This plane A is a classical example of an affine space. Later you will see that any affine space is isomorphic to the affine space generated by V. You will learn in Part 3 that A is a coset of V.

The Wolfram code below produces Figure 4 above.

The left picture shows an attempt to introduce vector structure in the inhabited set A. Let T : A ↦ V be a translation of the point set to the vector space. You may try to define addition of two points as

Recall that a frame in ℝ³ (we restrict ourselves with n = 3 for simplicity) is a pair consisting of the point O (called the origin) and an ordered basis ε = [e₁, e₂, e₃]. For example, the standard frame in ℝ³ has origin O = (0, 0, 0) and the basis of three vectors e₁ = (1, 0, 0) = i, e₂ = (0, 1, 0) = j, and e₃ = (0, 0, 1) = k. The position of a point P is then defined by the “unique vector” from O to P. This approach identifies point P(p₁, p₂, p₃) with corresponding vector \( \displaystyle \quad {\bf x} = \overline{OP} = \left( p_1 , p_2 , p_3 \right) . \)

Hence, in a standard frame of ℝ³, points and vectors are identically represented by triples of real numbers. However, if we choose another frame with different origin Ω = (ω₁, ω₂, ω₃), but with the same basis vectors ε, points and position vectors are no longer identified. This time, the point P = (p₁, p₂, p₃) is defined by by two position vectors: \[ OP = \left( p_1 , p_2 , p_3 \right) \] in frame (O, ε) and \[ \Omega P = \left( p_1 - \omega_1 , p_2 - \omega_2 , p_3 - \omega_3 \right) \] in frame (Ω, ε). This is because \[ OP = O\Omega + \Omega P \qquad \mbox{and} \qquad O\Omega = \left( \omega_1 , \omega_2 , \omega_3 \right) . \] In the second frame (Ω, ε), points and position vectors are no longer identical.    ■

For a tangent plane to a surface S to exist at a point on that surface, it is sufficient for the function that defines the surface to be continuously differentiable at that point. If a surface is defined by a differentiable function z = f(x, y), and P₀ = (x₀, y₀, z₀) is a point on S, then the equation of the tangent plane at P₀ is given by \[ z = f(x_0 , y_0 ) + f_x (x_0 , y_0 ) \left( x- x_0 \right) + f_y (x_0 , y_0 ) \left( y- y_0 \right) . \tag{9.1} \] To see why this formula is correct, let’s first find two lines tangent to the surface S. The equation of the tangent line to the curve that is represented by the intersection of S with the vertical trace given by x = x₀ is z = f(x₀, y₀) + f_y(x₀, y₀)(y − y₀). Similarly, the equation of the tangent line to the curve that is represented by the intersection of S with the vertical trace given by y = y₀ is z = f(x₀, y₀) + f_x(x₀, y₀)(x − x₀). A parallel vector to the first tangent line is a = j + f_y(x₀, y₀)k, a parallel vector to the second tangent line is b = i + f_x(x₀, y₀)k. We can take the cross product of these two vectors \begin{align*} {\bf a} \times {\bf b} &= \left( {\bf j} + f_y (x_0 , y_0 )\,{\bf k} \right) \times \left( {\bf i} + f_x (x_0 , y_0 )\,{\bf k} \right) \\ &= \begin{vmatrix} {\bf i} & {\bf j} & {\bf k} \\ 0&1& f_y (x_0 , y_0 ) \\ 1&0& f_x (x_0 , y_0 ) \end{vmatrix} \\ &= f_x (x_0 , y_0 )\,{\bf i} + f_y (x_0 , y_0 )\,{\bf j} - {\bf k} . \end{align*} This vector is perpendicular to both lines and is therefore perpendicular to the tangent plane. We can use this vector as a normal vector to the tangent plane, which we denote as n = a × b/(∥a × b∥) (of unit length), along with the point P₀ = (x₀, y₀, z₀) in the equation for a plane: \begin{align*} {\bf n} \bullet \left( (x- x_0 )\,{\bf i} + (y - y_0 )\,{\bf j} + (z - f(x_0 , y_0 ))\,{\bf k} \right) &= 0 , \\ \left( f_x (x_0 , y_0 )\, {\bf i} + f_y (x_0 , y_0 )\,{\bf j} - {\bf k} \right) \bullet \left( (x- x_0 )\,{\bf i} + (y - y_0 )\,{\bf j} + (z - f(x_0 , y_0 ))\,{\bf k} \right) &= 0 , \\ f_x (x_0 , y_0 )\left( x - x_0 \right) + f_y (x_0 , y_0 ) \left( y - y_0 \right) - \left( z - f(x_0 , y_0 ) \right) &= 0. \end{align*} Solving this equation for z gives Equation (9.1).

As an example, let us choose a quadratic function f(x, y) = 3 x² − 5 x y − 7 y² + 4 x − 5 y + 1 at point (−2, 1). First, we calculate derivatives \begin{align*} f_x (x,y) &=6\,x - 5\,y + 4, \\ f_x (-2,1) &= -13 , \\ f_y (x,y) &= -5 - 5 x - 14 y , \\ f_y (-2,1) &= -9 . \end{align*} Mathematica confirms calculations above.

The two parabolas you see in the plot are the vertical traces of the surface, which are obtained by intersecting the surface z = f(x,y) with the planes x = x₀ and y = y₀. The equations for these parabolas can be derived by substituting these values into the surface equation.

Given the surface equation:

2. Vertical trace at y = y₀ ==1:
z==f(x,1)==3x^2-5x(1)-7(1)^2+4x-5(1)+1
Simplifying the expression:
z==3x^2-5x-7+4x-5+1
z==3x^2-x-11
So, the equation for the second parabola is:
z==3x^2-x-11

Thus, the equations for the two parabolas are:
1. z==-7 y^2+5y+5
2. z==3x^2-x-11
These are the vertical traces of the surface at x==-2 and y==1, respectively.

Similarly, consider the plane P in ℝ³ through three non-collinear points x, y, z ∈ ℝ³. As in the case of the line, we reach arbitrary points in P from the origin by going to the tip of x, then traveling within the plane some amount in the direction y − x and some other amount in the direction z − x. Algebraically, we have

As a further example, we consider two points P₁ and P₂ in a two-dimensional affine space. The following expression \[ P = P_1 + t \left( P_2 - P_1 \right) \] makes sense because the difference P₁ − P₂ is a vector, and thus so is t(P₁ − P₂). Therefore, P is the sum of a point and a vector which is again a point in the inhabited set. This point P lies on the line going through two other points P₁ and P₂. Note that if 0 ≤ t ≤ 1, then P is somewhere on the line segment joining P₁ and P₂.

Now we consider three points; the following figure shows a combination of these points: \[ P = \alpha_1 P_1 + \alpha_2 P_2 + \alpha_3 P_3 , \qquad \alpha_1 + \alpha_2 + \alpha_3 = 1. \]

For arbitrary two real scalars α, β ∈ ℝ, we have \[ f \left( \alpha {\bf x} + \beta {\bf y} \right) = \alpha\,{\bf A}\,{\bf x} + \beta \,{\bf A}\,{\bf y} + \mathbf{b} , \] for arbitrary column vectors x and y. If we want this affine transformation to preserve linear combination, we need to satisfy the identity \[ f \left( \alpha {\bf x} + \beta {\bf y} \right) = \alpha\,f \left( {\bf x} \right) + \beta\,f \left( {\bf y} \right) \] or \[ \alpha\,{\bf A}\,{\bf x} + \beta \,{\bf A}\,{\bf y} + \mathbf{b} = \alpha \left( {\bf A}\,{\bf x} \right) + \beta \left( {\bf A}\,{\bf y} \right) + \left( \alpha + \beta\right) \mathbf{b} . \] The latter identity is true only when \[ \alpha + \beta = 1. \tag{11.2} \] Therefore, we conclude that the affine transformation (11.1) preserves linear combination of two points only when weights α and β satisfy the condition (11.2).

It is straight forward to verify that a linear combination of n column vectors x₁, x₂, … , x_n with weights c₁, c₂, … , c_n is preserved by affine transformation (11.1) only when these weights satisfy the condition \[ \sum_{i=0}^n c_i = 1 . \]    ■

Note that new p1 is inside the triangle created from the original three points.

Point p2 is also within the triangular outline of the original points.

Note that point p3 does not fall inside the original triangle.

Note that the curve passes through 𝑎 and d, but generally not through b and c. For example, a Bézier curve can be used to specify the velocity over time of an object such as a cursor moving from A to B, rather than simply moving at a fixed number of pixels per step. The Bézier curve can be constructed using the de Casteljau algorithm. Although the algorithm is slower for most architectures when compared with the direct approach, it is more numerically stable. Wolfram implemented de Casteljau's algorithm in a special build-in command: BezierCurve. Here is an illustration from the Mathematica Documentation. Use your cursor to drag the points about to see how it operates.

In particular, if we have two nodes x₀, x₁, the two functions b₀, b₁ : ℝ → ℝ are \[ b_0 (x) = \frac{x_1 - x}{x_1 - x_0} , \qquad b_1 (x) = \frac{x - x_0}{x_1 - x_0} \] These functions form a barycentric basis with respect to x₀ and x₁ that lead to approximation \[ F_1 (x) = \frac{x_1 - x}{x_1 - x_0} \, f\left( x_0 \right) + \frac{x - x_0}{x_1 - x_0} \, f \left( x_1 \right) . \tag{14.2} \] In general, we have \[ F_n (x) = \frac{\sum_{i=0}^n \frac{(-1)^i}{x - x_i} \, f\left( x_i \right)}{\sum_{i=0}^n \frac{(-1)^i}{x - x_i}} . \tag{14.3} \]

As a numerical example, we consider the sine function f(x) = sin(x) and two points x₀ = π/6 and x₁ = 3π/4. The corresponding two points barycentric interpolation reads as \[ F_1 (x) = \frac{3\pi /4 -x}{7\pi /12} \, \frac{1}{2} + \frac{x - \pi /6}{7\pi /12} \, \frac{1}{\sqrt{2}} . \]

Now we turn our attention to three point-interpolation. We again consider sine function and choose three points: x₀ = π/6, x₁ = π/4, and x₂ = 3π/4. So we use formula (14.3) and ask Mathematica for help.

Create module that produces the general barycentric interpolation for three nodes.

We check with Mathematica:

1. Any linear transformation T : V ⇾ U induces an affine mapping of the spaces (V, V, +) ↣ (U, U, +). For it, Df = f.
2. Transformation f : 𝔽ⁿ ↣ 𝔽ⁿ is an affine map for f(x) = A x + b because Df = A.
3. Any translation t_x : A ↣ A, where t_x(P) = P + x, is affine, and D(t_x) = id_V because \[ t_{\bf x} (P) - t_{\bf x} (Q) = \left( P + {\bf x} \right) - \left( Q + {\bf x} \right) = P - Q . \]
4. If f : 𝔸 ⇾ 𝔹 is an affine mapping and y ∈ U, then the mapping t_x ∘ f : 𝔸 ⇾ 𝔹 is affine and D(t_x ∘ f) = D(f). Indeed, \begin{align*} t_{\bf x} \circ f (P) - t_{\bf x} \circ f (Q) &= \left( t_{\bf x} (P) + {\bf x} \right) - \left( t_{\bf x} (Q) + {\bf x} \right) \\ &= f(P) - f(Q) = Df(P-Q) . \end{align*}
5. An affine function f : 𝔸 ↣ 𝔽 is defined as an affine mapping of 𝔸 into (𝔽, 𝔽, +). Thus, f assumes values in 𝔽, while Df is a linear functional on V. Any constant function f is affine: Df = 0.
6. The identity mapping id : 𝔸 ↣ 𝔸 is an affine mapping. Indeed, P − Q = id_V(P − Q). In particular, D(id_A) = id_V.

Example 16: Let us consider the affine transformation S : ℝ³ ↦ ℝ⁴ given by S(x) = A x + b, where \[ \mathbf{A} = \begin{bmatrix} -2& 3& 5 \\ 3& 7& -1 \\ 5& 27& 7 \\ -9& 2& 16 \end{bmatrix}, \qquad {\bf b} = \begin{bmatrix} 3 \\ -6 \\ -12 \\ 15 \end{bmatrix} . \] First, we find solution A u = −b, so S(u) = 0, with the aid of Mathematica.

P = (1, 2) ,
Q = (3, 1) ,
R = (4, 4) ,

Then vectors acting on these points become

u = P − Q = (−2, 1),
v = Q − R = (−1, −3),

Using Mathematica's build-in command, we define an affine transformation with matrix \( \displaystyle \quad {\bf A} = \begin{bmatrix} 1.5 & -1 \\ 2 & 2.5 \end{bmatrix} \quad \) and translation vector b = (2, −1).

1. Part (a):
   Apply the affine transformation to points and vectors
   transformedP=affineFunc[P]; transformedQ=affineFunc[Q];
   Apply the affine transformation to vectors without translation vector v:
   A = {{1.5, -1}, {2, 2.5}}; v = P-Q; A.v
   transformedP - transformedQ
   {-4., -1.5}
   {-4., -1.5}
2. Part (b):
   We set α = 7/3 and calculate
   affineFunc[P + (7/3)*v]
   {-7.83333, 2.5}
   A.(P + 7/3*v) + {2, -1}
   {-7.83333, 2.5}
   Now we calculate the right-hand side:
   affineFunc[P] + (7/3)*A.v
   {-7.83333, 2.5}
3. Part (c):
   Apply the affine transformation to vectors without translation
   DF[vector_]:= A.vector;
   Verify the theorem statement
   lhs=DF[u+v]; rhs=DF[u]+DF[v]; TrueQ[lhs==rhs]
   True
   Graphical Illustration of Part C.

   Points and their transformations

   transformedP = affineFunc[P] + b; transformedQ = affineFunc[Q] + b; transformedR = affineFunc[R] + b;
   Graphics for original points and vectors
   originalGraphics=Graphics[{ Red,PointSize[Large],Point[P],Point[Q],Point[R], Blue,Arrow[{P,Q}],Arrow[{Q,R}],Arrow[{P,R}], Text[P,P,{1,-2}],Text[Q,Q,{-2,-1}],Text[R,R,{1,-1}] }];
   transformedGraphics=Graphics[{ Green,PointSize[Large],Point[transformedP],Point[transformedQ],Point[transformedR], Blue,Arrow[{transformedP,transformedQ}],Arrow[{transformedQ,transformedR}],Arrow[{transformedP,transformedR}], Text[F(P),transformedP,{1,-1}],Text[F(Q),transformedQ,{1,.9}],Text[F(R),transformedR,{1,-1}] }];
   Show both graphics
   Show[originalGraphics,transformedGraphics,PlotRange->All,Axes->True,AxesOrigin->{0,0},GridLines->Automatic]

   

4. Part (d):
   We check the assessment with α = (7/3):
   A.(7/3*v)
   {-9.33333, -3.5}
   and right-hand side is
   (7/3)*A.v
   {-9.33333, -3.5}

Condition (19.2) for vector w shows that singularity of matrix A plays no role in defining affine mappings.    ■

Solution: We use the coordinates to form a linear combination of the vectors in the frame that we then add to the frame’s origin. Because we are adding a vector to a point the result will indeed be a point. \[ Q = -3 \begin{bmatrix} \phantom{-}2 \\ -3 \end{bmatrix} + 2 \begin{bmatrix} 1 \\ 4 \end{bmatrix} + \begin{pmatrix} 6 & 2 \end{pmatrix} = \begin{bmatrix} 2 \\ 19 \end{bmatrix} . \]

Solution: The basis vectors in β can be written as \begin{align*} \begin{bmatrix} 3 \\ 2 \end{bmatrix} &= 0 \cdot \begin{bmatrix} 7 \\ 3 \end{bmatrix} - \begin{bmatrix} -3 \\ -2 \end{bmatrix} + 0 \cdot \left( 3,\ -2 \right) , \\ \begin{bmatrix} 1 \\ 4 \end{bmatrix} &= -2 \cdot \begin{bmatrix} 7 \\ 3 \end{bmatrix} -5\cdot \begin{bmatrix} -3 \\ -2 \end{bmatrix} + 0 \cdot \left( 3,\ -2 \right) , \\ \left( 5, \ 2 \right) &= - \frac{8}{5}\cdot \begin{bmatrix} 7 \\ 3 \end{bmatrix} -\frac{22}{5} \cdot \begin{bmatrix} -3 \\ -2 \end{bmatrix} + 1 \cdot \left( 3,\ -2 \right) , \end{align*}

Suppose that a line L in 3-dimensional Euclidean space is determined by two distinct points x = (x₁, x₂, x₃) and y = (y₁, y₂, t₃). The vector displacement from x to y is nonzero because the points are distinct, and represents the direction of the line. That is, every displacement between points on L is a scalar multiple of d = y – x. If a physical particle of unit mass were to move from x to y, it would have a moment about the origin. The geometric equivalent to this moment, is a vector whose direction is perpendicular to the plane containing L and the origin, and whose length equals twice the area of the triangle formed by the displacement and the origin. Treating the points as displacements from the origin, the moment is the cross product of these vectors, m = x × y, so d • m = 0.

A pair (d, m) identifies uniquely the line, up to a common (nonzero) scalar multiple. That is, the coordinates \[ \left( {\bf d} : {\bf m} \right) = \left( d_1 : d_2 : d_3 : m_1 : m_2 : m_3 \right) \] may be considered homogeneous coordinates for L, in the sense that all pairs (λd : λm) for λ ≠ 0, can be produced by points on L and only L, and any such pair determines a unique line so long as d is not zero and d • m = 0. Furthermore, this approach extends to include points, lines, and a plane "at infinity", in the sense of projective geometry. In addition a point x lies on the line L if and only if x × d = m.

Shoemake's tutorial contains many properties of the Plücker coordinates, of which we mention the following: \[ \mbox{Squared distance from origin to $L$ is} = \frac{m_1^2 + m_2^2 + m_3^2}{d_1^2 + d_2^2 + d_3^2} . \]

For example, x = (1,-3, 2) and y = (5. 2, -3). Their cross product is

If two vectors d = m are the same, then the Plücker coordinates of the line are \[ \left( \mathbf{d} : \mathbf{0} \right) = \left( d_1 : d_2 : d_3 : 0 : 0 : 0 \right) . \] The squared distance of this line from the origin is 0.    ■

Example 23: Let 𝔸 = (A, ℝ², d) be an affine space with frame α = [e₁, e₂, O], where O = (0, 0). Let T : 𝔸 → 𝔸 be defined as T(P) = P + t, where t = [Δx, Δy]. Find the 3 × 3 matrix T that implements this transformation.

Solution: Since only frame α is used, we have \[ \left[ T(\mathbf{e}_1 ) \right]_{\alpha} = [\![ \mathbf{e}_1 ]\!] = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \qquad \mbox{and} \qquad \left[ T(\mathbf{e}_2 ) \right]_{\alpha} = [\![ \mathbf{e}_2 ]\!] = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \] while \[ \left[ T(O) \right]_{\alpha} = \left[ O + \mathbf{t} \right]_{\alpha} = \left[ O \right]_{\alpha} + \left[ {\bf t} \right]_{\alpha} = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} + \begin{bmatrix} \Delta x \\ \Delta y \\ 0 \end{bmatrix} = \begin{bmatrix} \Delta x \\ \Delta y \\ 1 \end{bmatrix} . \] Thus the matrix T is given by \[ \mathbf{T} = \begin{bmatrix} 1 & 0 & \Delta x \\ 0&1& \Delta y \\ 0&0&1 \end{bmatrix} . \]    ■

Example 24: This example illustrates usefulness of affine (augmented) matrices for evaluation of compositions of two or more affine transformations. Once an affine transformation is written in matrix form (by extending the original dimensions), evaluation of compositions of two or more affine transformations is reduced to standard matrix multiplications.

We consider two affine transformations f : 𝔸² ↣ 𝔸³ and g : 𝔸³ ↣ 𝔸⁴, defined by formulas \[ f(\mathbf{x}) = {\bf A}\,\mathbf{x} + \mathbf{b} \quad\mbox{and} \quad g(\mathbf{y}) = {\bf B}\,\mathbf{y} + \mathbf{c} , \] where \[ {\bf A} = \begin{bmatrix} 1&2 \\ -2&1 \\ 3&-2 \end{bmatrix} , \quad {\bf b} = \begin{bmatrix} -2 \\ -3 \\ 1 \end{bmatrix} , \qquad {\bf B} = \begin{bmatrix} 4 & 1 & -2 \\ 2&5&-2 \\ 2&-1&3 \\ 5&-3& 2 \end{bmatrix} , \quad {\bf c} = \begin{bmatrix} 4\\ 3\\ 2 \\ 1 \end{bmatrix} . \] The corresponding augmented (affine) matrices for these transformations are \[ {\bf A}_b = \begin{bmatrix} 1&2 & -2 \\ -2&1 & -3 \\ 3&-2& 1 \\ 0&0&1 \end{bmatrix} , \qquad {\bf B}_c = \begin{bmatrix} 4 & 1 & -2& 4 \\ 2&5&-2&3 \\ 2&-1&3 & 2 \\ 5&-3& 2& 1 \\ 0&0&0&1 \end{bmatrix} . \] Then to determine an augmented matrix corresponding to their conposition g ∘ f, we simply multiply the corresponding augmented matrices: \begin{align*} {\bf B}_c {\bf A}_{b} &= \begin{bmatrix} 4 & 1 & -2& 4 \\ 2&5&-2&3 \\ 2&-1&3 & 2 \\ 5&-3& 2& 1 \\ 0&0&0&1 \end{bmatrix} \cdot \begin{bmatrix} 1&2 & -2 \\ -2&1 & -3 \\ 3&-2& 1 \\ 0&0&1 \end{bmatrix} \\ &= \begin{bmatrix} -4& 13&-9 \\ -14& 13&-18 \\ 13& -3&4 \\ 17& 3&2 \\ 0&0&1\end{bmatrix} , \end{align*} because \[ g \left( f({\bf x}) \right) = {\bf B}\,{\bf A}\,{\bf x} + {\bf B}\,{\bf b} + {\bf c} . \] This identity tells us that the composition g ∘ f is an affine transformation x → C x + d, where \[ {\bf C} = {\bf B}\,{\bf A} , \qquad {\bf d} = {\bf B}\,{\bf b} + {\bf c} . \] So \[ {\bf C} = \begin{bmatrix} -4& 13 \\ -14& 13 \\ 13& -3 \\ 17& 3 \end{bmatrix} , \qquad {\bf d} = \begin{bmatrix} -9\\ -18\\ 4\\ 2 \end{bmatrix} \]

Let us start with two affine mappings that operate on affine spaces of different dimensions. We define them according to formulas: \[ f(\mathbf{x}) = {\bf A}\, \mathbf{x} + \mathbf{b} \qquad\mbox{and} \qquad g(\mathbf{y}) = {\bf B}\, \mathbf{y} + \mathbf{c} , \] where \[ {\bf A} = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} \in \mathbb{R}^{2\times 2} , \qquad {\bf b} = \begin{pmatrix} -2 \\ 3.5 \end{pmatrix} \in \mathbb{R}^{2\times 1} , \] and \[ {\bf B} = \begin{bmatrix} -1 & -2 \\ 2&3 \\ -3& -4 \end{bmatrix} \in \mathbb{R}^{3\times 2}, \qquad {\bf c} = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} \in \mathbb{R}^{3\times 1} . \] Observe that f : 𝔸² ↣ 𝔸² and g : 𝔸² ↣ 𝔸³. So we can define one composition g ∘ f    of these affine mappings, but not f ∘ g---it is undefined.

The composition g ∘ f : 𝔸² ↣ 𝔸³ of these two affine mappings acting on two-dimensional vector v can be evaluated as follows: \begin{align*} g \circ f (\mathbf{v} ) &= g \left( f(\mathbf{v}) \right) = {\bf B} \left( f(\mathbf{v}) \right) + \mathbf{c} \\ &= {\bf B} \left( {\bf A}\, {\bf v} + {\bf b} \right) + \mathbf{c} \\ &= {\bf B} \,{\bf A} \,{\bf v} + {\bf B} \,{\bf b} + \mathbf{c} , \end{align*} which is an affine transformation \[ \mathbb{R}^{2 \times 1} \ni {\bf v} ↣ {\bf C}\,{\bf v} + {\bf d} \in \mathbb{R}^{3 \times 1} , \tag{E25.1} \] with \[ {\bf C} = {\bf B} \,{\bf A} = \begin{bmatrix} -5&-11 \\ 8&18 \\ 5&7 \end{bmatrix} \in \mathbb{R}^{3\times 2} \tag{E25.2} \] and \[ \mathbf{d} = {\bf B} \,{\bf b} +\mathbf{c} = \begin{bmatrix} -4 \\ 5.5 \\ 22 \end{bmatrix} \in \mathbb{R}^{3\times 1} . \tag{E25.3} \]

In order to finish this part, we build augmented matrices corresponding to transformations f, g and their composition g ∘ f \[ {\bf A}_b = \begin{bmatrix} 1 & 3 & -2 \\ 2 & 4& 3.5 \\ 0&0&1 \end{bmatrix} \in \mathbb{R}^{3\times 3} , \qquad {\bf B}_c = \begin{bmatrix} -1&-2& 1 \\ 2&3&-1 \\ -3&-4&2 \\ 0&0&1 \end{bmatrix} \in \mathbb{R}^{4\times 3} , \] Multiplying these matrices, we obtain \[ {\bf B}_c {\bf A}_b = \begin{bmatrix} {\bf B}\,{\bf A} & {\bf B}\,{\bf b} + {\bf c} \\ 0 \cdots 0 & 1 \end{bmatrix} = \begin{bmatrix} -5& -11& -4 \\ 8&18& 5.5 \\ 5&7& 22 \\ 0&0&1 \end{bmatrix} \in \mathbb{R}^{4 \times 3} . \]

Therefore, when vector v = [3.4 −2.1] is considered as an element of the inhabited set, an extra "1" is added to make it three-dimensional: v = [3.4 −2.1 1] Existence of extra component "1" indicates that v is considered as a point in the inhabited set. On the other hand, if the same vector v = [3.4 −2.1] is considered as a vector, we append extra component "0," which shows that this vector belongs to the vector space: v = [3.4 −2.1 0].

Then the augmented matrix C_d can be applied from left to either points or vectors. In computer graphics, this augmented matrix acts on points/vectors written in row format from right. The last component (either "1" or "0") of points/vectors does not effect the outcome. This makes the augmented matrices a universal instrument that can operate on both, points and vectors. \[ {\bf C}_d {\bf v} = \begin{bmatrix} -5& -11& -4 \\ 8&18& 5.5 \\ 5&7& 22 \\ 0&0&1 \end{bmatrix} \begin{pmatrix} 3.4 \\ -2.1 \\ 1 \end{pmatrix} = \begin{pmatrix} 2.1 \\ -5.1 \\ 24.3 \\ 1 \end{pmatrix} . \]

Note that the product A_bB_c is undefined because it has uncompatible dimensions (you cannot multiply 3×3 matrix by 4×3 matrix from right).

Observe that Mathematica automatically generates augmented matrices when command AffineTransform is invoked.

Another example:

Define points P and Q on the plane: \[ P = \left[ 1, 2 \right]^{mathrm T} , \qquad Q = \left[ 3, 4 \right]^{mathrm T} . \]

1. Suppose that an affine space A has two frames \[ \beta = \left( \begin{bmatrix} 2 \\ -2 \end{bmatrix} , \ \begin{bmatrix} 3 \\ 1 \end{bmatrix} , \ (2, -4)\right) \] and \[ \phi = \left( \begin{bmatrix} 3 \\ 1 \end{bmatrix} , \ \begin{bmatrix} 1 \\ 1 \end{bmatrix} , \ (-2, 5)\right) \] Find the change of frame matrix M and use it to compute ⟦Q⟧_β = (5, -3, 1), then find ⟦Q⟧_ϕ.
2. Suppose M is the change of frame matrix that transforms coordinates relative to frame β to coordinates relative to frame ϕ. Prove that M⁻¹ exists.
3. Determine the matrix representation of the affine transformation S : 𝔸 → 𝔸 if 𝔸 = (A, ℝ²) and S(P) = Q where Q = (x+2y, y) if P = (x, y). What type of transformation is this?
4. We call the ith median of the system of points P₁, P₂, … , P_n ∈ A, the inhabited set, segment connecting the point P_i with the center of gravity of the remaining point {P_j : j ≠ i}. Prove that all medians intersect at one point - the center of gravity of points P₁, P₂, … , P_n.

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