Affine Transformations

Example 1: There are a few countries (exclusively by the U.S. and its formerly and presently governed territories, in addition to a few South American and some countries in the Pacific) that still use Fahrenheit (°F) temperature scale while the majority of countries utilize the International System of Units (SI), which includes Celsius (°C) scale. First proposed in 1724 by physicist Daniel Gabriel Fahrenheit (who had also invented the mercury thermometer in 1714), the Fahrenheit temperature scale was used before domination of metric unit system. Celsius scale is named after the Swedish astronomer Anders Celsius (1701–1744), who developed a variant of it in 1742.

Today, however, Fahrenheit has been replaced by the Celsius (and in scientific applications, Kelvin) scale in all but a handful of the world's countries.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.

Then vertical line x = −1 is mapped into the line \[ x \mapsto y , \qquad y \mapsto 2\,y - 1 . \] The images of vertices become \[ \left( -1,-1 \right) \mapsto \left( -1, -3 \right) , \quad \left( -1, 1 \right) \mapsto \left( 1, 1 \right) , \quad \left( 1, 1 \right) \mapsto \left( 3, 1 \right) , \quad \left( 1, -1 \right) \mapsto \left( 1, -3 \right) . \] The next line below is Wolfram code for exporting any graphic to PNG format

Now we plot the same figures by keeping grid visible.

Since Wolfram has a documentation for AffineTransform, we are going to use its command Affine Transform[m,b].

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 lower left coordinate of our rectangle to be transformed is {-1,-1}. We can use our AffineTransform as function, t, and apply that to the lower left coordinate to get the image of that coordinate.

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

Showing both in a grid with different scales

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:

In homogeneous coordinates, a 2D point (x, y) is represented as (x, y, 1) and a 2D vector is represented as (x, y, 0). The reason for this is to allow for translations to be represented as matrix multiplications.

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.    ■

Example 3: Matrix \( \displaystyle \quad \mathbf{A} = \begin{bmatrix} 1 & 0 \\ 0&0 \end{bmatrix} \) maps all points to the x-axis, so it is a projection on this axis. The area of any closed region will become zero. We have det(A) = 0, which verifies that any closed region’s area will be scaled by zero.

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

Define a unit square as a list of points, not as a pair of vectors: In Mathematica, this unitSquare is a graphical object. It has an area

Apply the transformation matrix to the unit square Compute the area of the transformed square This transformation matrix collapses the square into a line, which makes it impossible for Mathematica's Area function to compute the area since a line has no area.

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

B has a positive Determinant Apply transformation matrix B to the unit square The unit square has an area of 1, its transform this time has an area of 2 The entire area below is the transformed area, the darker sub-area is the original unit square.

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Example 4: We consider two affine maps: \[ f \,:\ \mathbb{A}^3 \mapsto \mathbb{A}^2 \qquad \mbox{and} \qquad g\,:\ \mathbb{A}^2 \mapsto \mathbb{A}^2 , \] defined explicitly as \[ f({\bf x}) = {\bf A}\,{\bf x} + {\bf b} = {\bf y} \quad \mbox{and} \qquad g({\bf y}) = {\bf B}\,{\bf y} + {\bf w} , \] where \[ {\bf A} = \begin{bmatrix} 1&2&3 \\ 3&2&1 \end{bmatrix}, \qquad {\bf B} = \begin{bmatrix} 1& -3 \\ 2& -2 \end{bmatrix}, \] and \[ {\bf b} = \begin{bmatrix} -3 \\ \phantom{-}2 \end{bmatrix}, \qquad {\bf w} = \begin{bmatrix} \phantom{-}1 \\ -5 \end{bmatrix} . \] If x = (x₁, x₂, x₃)^T ∈ ℝ^3×1, is an arbitrary column vector, then \[ {\bf y} = {\bf A}\,{\bf x} + {\bf b} = \begin{bmatrix} x_1 + 2\, x_2 + 3\, x_3 \\ 3\, x_1 + 2\, x_2 + x_3 \end{bmatrix} + \begin{bmatrix} -3 \\ \phantom{-}2 \end{bmatrix} . \] Then we apply transformation g and obtain \[ {\bf B} \left( {\bf A}\,{\bf x} + {\bf b} \right) + {\bf w} = {\bf B} \,{\bf A}\,{\bf x} + {\bf B} \, {\bf b} + {\bf w} = \begin{bmatrix} -8\, x_1 - 4\, x_2 \\ -4\, x_1 + 4\, x_3\end{bmatrix} + \begin{bmatrix} -8 \\ -15 \end{bmatrix} . \] We check the answer with Mathematica:    ■

Example 5:

Part 1:

In order to illustrate the first property, we plot a line and compare it with transformed line. For this specific example, the transformation is a translation by the vector b, so the transformed line will be parallel to the original line and offset by the vector b. We consider Eq.(1) when \[ {\bf A} = \begin{bmatrix} 1&-2 \\ 3&-1 \end{bmatrix} , \qquad {\bf b} = \begin{bmatrix} 5 \\ 6 \end{bmatrix} . \]

Part 2:

We consider affine transformation \[ \begin{bmatrix} x \\ y \end{bmatrix} \,\to\, \begin{bmatrix} 2&-1 \\ 1&1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} + \begin{bmatrix} 5 \\ 6 \end{bmatrix} . \] Generated plot shows that this transformation maps lines into lines. Using Mathematica, we define the affine transformation and then plot the original straight line and its output.

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

Now, define some arbitrary parallel lines with slope 2: Applying the affine transformation to the lines, we get Plot the original lines and their transformed versions

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 *)

Define the affine transformation Apply the transformation to the parallelogram Visualize the original and transformed parallelograms

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].

Now we calculate the ratios of lengths of transformed segments (with the aid of Mathematica) and None of these ratios is 2, as you expect. Now we answer why these ratios are different. When vector v [1, 2] is subject to affine transformation, its lenth is ∥A v∥ independently of shift vector b, but not ∥A v + b∥ because shift by any vector (b in our case) does not change its length. Therefore, the lengths of affine transformation f(v) are \begin{align*} \| f(S_1 ) \| &= \| {\bf A}\,[1, 2] \| = \| [11, 3] \| = \sqrt{130} , \\ \| f(S_2 ) \| &= \| {\bf A}\,[2, 4] \| = \| [22, 6] \| = 2\,\sqrt{130} , \\ \| f(S_3 ) \| &= \| {\bf A}\,[4, 2] \| = \| [20, 0] \| = 20 , \\ \| f(S_4 ) \| &= \| {\bf A}\,[-2, 4] \| = \| [10, 10] \| = 10\,\sqrt{2} . \end{align*}

On the other hand,

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):

Define an affine transformation x ↦ A x + b, where \[ {\bf A} = \begin{bmatrix} -3&2 \\ \phantom{-}1&2 \end{bmatrix} , \qquad {\bf b} = \begin{bmatrix} \phantom{-}2 \\ -4 \end{bmatrix} . \tag{5.7.1} \] We apply a build-in command AffineTransform. It is the subroutine that takes two inputs, m and b and outputs an affine transformation function (which we name "t" in the following code). We can then apply "t" to any number of vectors, such as Now we make a module:

Compute the areas before the transformation

Apply the transformation to the polygons. We first determine equations of lines that enclose the parallelogram. For example, the line y = 0 is transferred according to affine transformation \[ \begin{split} X &= -3\,x + 2\,y +2 , \\ Y &= x + 2\,y -4 \end{split} \] leads to the pair \[ \begin{split} X &= -3\,x +2 , \\ Y &= x -4 . \end{split} \] From the latter, we get x = Y + 4. Substitution of x into the former equation, we obtain \[ X = -3 \left( Y+4 \right) +2 \qquad \Longrightarrow \qquad X = -3\,Y + 10 . \] Similarly, we express in new coordinates the equation of lines that embrace the given parallelogram: \begin{align*} y = 0 \ & \mapsto \ Y = -\frac{1}{3}\,X - \frac{10}{3} , \\ y = x \ & \mapsto \ Y = -3\,X + 2 , \\ y = 1 \ & \mapsto \ Y = - \frac{1}{3}\, X - \frac{2}{3} , \\ y = x -1 \ & \mapsto \ Y = -3\,X + 6 . \end{align*} So the corresponding vertices are mapped as follows: \begin{align*} (0, 0) \ & \mapsto \ (2, -4) , \\ (1, 0) \ & \mapsto \ (-1, -3) , \\ (2, 1) \ & \mapsto \ (-2, 0) , \\ (1, 1) \ & \mapsto \ (1, -1) . \end{align*} Apply the transformation to the polygons Compute the areas after the transformation Compare the ratios Plot the original and transformed polygons

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 area of semi-circle of radius 1 is Since the area of the parallelogram through points (0,0), (1, 0), (2, 1), (1,1) is 1, the ratio of areas of semi-circle to the parallelogram becomes \[ \frac{\mbox{area of semicircle}}{\mbox{area of parallelogram}} = \frac{\pi /2}{1} = \frac{\pi}{2} = \approx 1.5708 . \] The equation of the upper boundary of the given semi-circle is \[ \partial C := \sqrt{1 - x^2} = y , \qquad x\ge 0, \ y\ge 0 . \] In order to determine the equations of the boundary of the transformed region, we invoke the affine transformation: \[ \begin{split} X &\mapsto \ -3\,x + 2\,y +2 , \\ Y &\mapsto \ x + 2\,y -4 \end{split} \] because the affine transformation is given in matrix/vector form: \[ \begin{bmatrix} X \\ Y \end{bmatrix} = \begin{bmatrix} -3&2 \\ \phantom{-}1 & 2\end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} + \begin{bmatrix} \phantom{-}2 \\ -4 \end{bmatrix} . \] We can express old variables (x, y) through new ones (X, Y): \begin{align*} x &= \frac{1}{4} \left( 6 - X + Y \right) , \\ y &= \frac{1}{8} \left( 10 + X + 3\, Y \right) . \end{align*} We can express this inverse transformation in succinct form: \[ \begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{8} \begin{bmatrix} -2&2 \\ \phantom{-}1 &3 \end{bmatrix} \begin{bmatrix} X \\ Y \end{bmatrix} + \begin{bmatrix} 3/2 \\ 5/4 \end{bmatrix} \tag{5.7.3} \] because \[ {\bf A}^{-1} = \frac{1}{8} \begin{bmatrix} -2&2 \\ \phantom{-}1 & 3 \end{bmatrix} , \qquad \det{\bf A} = -8 . \] Since the boundary of the semicircle consists of two pieces \begin{align*} y &= \sqrt{1 - x^2} , \quad -1 \le x \le 1, \\ y &= 0 , \quad -1 \le x \le 1, \end{align*} we need to determine equations of these two pieces in new variables. The equation of straight line y = 0 was determined previously (as a part of parallelogram's boundary): \[ Y = - \frac{1}{3} \left( X-2 \right) -4 , \qquad 5 \ge X \ge -1 . \tag{5.7.4} \] Note that the new variable X moves in opposite direction because determinant of matrix A is negative. Equation (5.7.4) follows from affine transformations of endpoints: \begin{align*} (1, 0) &\mapsto\ (-1, -3) , \\ (-1,0) &\mapsto\ (5, -5) . \end{align*} Now we determine the equation of transformed semi-circle. We substitute into the equation of semi-circle \( \displaystyle \quad x^2 + y^2 = 1 , \quad y \ge 0 . \) instead of old variables (x, y) their expressions from Eq.(5.7.3) to obtain \[ 2 \left( 6 -X + Y \right)^2 + \left( 10 + X + 3\,Y \right)^2 = 8^2 , \qquad 5 \ge X \ge -1, \quad -4 \le Y \le -2 . \tag{5.7.5} \]

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.

Finally, we calculate the ratio of transformed areas, semi-circle to parallelogram. This confirms that ratio of areas remain the same upon affine transformations.    ■

Example 6:

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.

Define the vectors Define the starting point for the red arrows on the green plane Define two vectors a and b on the blue plane, not parallel to the\ x, y, or z, and making 110 degrees with each other. Calculate the sum of the two vectors Define the transformation : a translation moving the vectors to the blue plane in the z - direction Define the lateral displacement : a translation moving the\ vectors in the x - y plane. Apply the transformations to the vectors Make cylindrical disks as point markers Show graphics together

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

However, the resulting vector (P(+)Q) does not belong to the inhabited set A. It is impossible to introduce a vector structure into an inhabited set of points---it is not a vector space because it has no algebraic structure. The basic idea of affine space is inherited from physics where forces (vectors) are acting on point objects to move them into another position (point again).

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Example 7: Any finite dimensional vector space V has an affine space structure specified by choosing the inhabited set A = V and letting d be subtraction in the vector space V. We will refer to the affine structure (V, V, d) on a vector space V as the canonical affine structure on V. In particular, the vector space ℝⁿ can be viewed as the affine space (ℝⁿ, ℝⁿ, d), denoted by 𝔸ⁿ. The affine space 𝔸ⁿ is called the real affine space of dimension n.

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.    ■

Example 8: Let us consider the subset A of 𝔸³ consisting of all points (x, y, z) satisfying the equation \[ x^2 + y^2 - z = 0 . \] The set of points A is a paraboloid of revolution, with axis Oz. The surface A can be made into an official affine space by defining the action of addition of points and vectors (which is equivalent to the difference operation d) s : A × ℝ² → A of ℝ² on A defined such that for every point (x, y, x² + y²) on A and any vector v = (v, u) ∈ ℝ², \[ \left( x, y , x^2 + y^2 \right) + \begin{bmatrix} v \\ u \end{bmatrix} = \left( x + v , y + u , (x+v)^2 + (y+u)^2 \right) . \]    ■

Example 9: Let P₀ = (x₀, y₀, z₀) be a point on a surface S ⊂ ℝ³, and let C be any curve passing through P₀ and lying entirely in S. If the tangent lines to all such curves C at P₀ lie in the same plane, then this plane is called the tangent plane to S at P₀.

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.

Then Eq.(9.1) becomes \[ z = 3 -13 \left( x+2 \right) -9 \left( y -1 \right) . \] Here is how Mathematica sees this. First, we define S, the surface function Then we compute partial derivatives Defining Subscript[P, 0] requires our x and y values and the value of the function at those values (z) Next we evaluate the derivatives at P₀. The equation of the tangent plane is We are now in a position to plot. Note that, while the tangent plane is indeed tangent to the surface at P₀, it intercepts the surface in other places. So, we plot the surface and tangent plane using a smaller range for the tangent plane. This adds a green point to show the tangency

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:

The two parabolas are:
1. Vertical trace at x = x₀ ==-2:
z = f(-2,y) == 3(-2)^2-5(-2)y-7y^2+4(-2)-5y+1
Simplifying the expression:
z==12+10y-7y^2-8-5y+1
z==-7 y^2+5y+5
So, the equation for the first parabola is:
z==-7 y^2+5y+5

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.

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Example 10: Let us consider the line L in ℝ³ through the distinct points x, y ∈ ℝ³. Geometrically, we can reach each point on this line by starting at the origin, traveling to the tip of the vector x (viewed as an arrow starting at 0), then following some scalar multiple t(y − x) of the direction vector y − x for the line (where t ∈ ℝ). Algebraically, the line consists of all vectors x + t(y − x) as t ranges over ℝ. Restating this, In other words, L consists of all affine combinations of x and y. Recall that a linear combination of two vectors includes all points \( \displaystyle \quad t\,{\bf y} + s \,{\bf x} , \quad \) where t, s ∈ ℝ, without constraint s = 1 − t.

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

Hence, P is the set of affine combinations of x, y, z.

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: \[ P = \alpha_1 P_1 + \alpha_2 P_2 + \alpha_3 P_3 , \qquad \alpha_1 + \alpha_2 + \alpha_3 = 1. \]

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Example 11: Let us consider the affine transformation applied to column vectors: \[ \mathbb{F}^{n\times 1} \ni \mathbf{x} \longrightarrow f({\bf x}) = {\bf A}\,{\bf x} + \mathbf{b} , \tag{11.1} \] where A is an m-by-n real matrix and b is a given m-column vector. We check whether this (arbitrary) affine transformation transfers affine combinations into another affine combination.

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 \[ \sum_{i=0}^n c_i = 1 . \]    ■

Example 12: Let us consider three points in two-dimensional space ℝ² with respect to some frame: \[ a (0, 0), \qquad b(2, 2), \qquad c (3, 1) . \] Of course, the expressions for points are not written accurately (from mathematician prospective), but lazy people like me use this informal notation. These points should be written as \[ P_1 = O + a , \qquad P_2 = O + b , \qquad P_3 = O + c . \] Using these three points, we determine three other points \[ p_1 = \frac{1}{4}\, a + \frac{1}{4}\, b + \frac{1}{2}\, c , \quad p_2 = \frac{1}{3}\, a + \frac{1}{3}\, b + \frac{1}{3}\, c , \quad p_3 = a - b + c . \] Now we explore Mathematica. First, we define points. Create triangle from these points that are marked now blue, Compute new, large, red points,

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

Note that for p1 we divide b by 4, so we have b/4. Now we modify this point p1 by introducing a coefficient (0,1) multiplier, .5 − t. Then upon changing t, we obtain new barycenters depending on the value of t. Because variable names are re-used below the following animation later in the notebook, the animation will change upon evaluation of later cells.

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

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

Similar to point p1, we make animation by continuously changing position of point p3.

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Example 13: This example demonstrates that an arbitrary polynomial curve can be defined as a set of barycenters of a fixed number of points. For example, let (𝑎, b, c, d) be a sequence of points in 𝔸². Observe that \[ \left( 1- t \right)^3 + 3t \left( 1 - t \right)^2 + 3t^2 \left( 1 - t \right) + t^3 = 1 \] because the sum on the left-hand side is obtained by expanding \[ \left( 1 + (1-t) \right)^3 = \sum_{i=0}^3 \binom{3}{i} t^i \left( 1 - t \right)^{3-i} = 1 \] using the binomial formula. Thus, \[ \left( 1- t \right)^3 a + 3t \left( 1 - t \right)^2 b + 3t^2 \left( 1 - t \right) c + t^3 d \] is a well-defined affine combination. Then, we can define the curve F : 𝔸 → 𝔸² such that \[ F(t) = \left( 1- t \right)^3 a + 3t \left( 1 - t \right)^2 b + 3t^2 \left( 1 - t \right) c + t^3 d \] Such a curve is called a cubic Bézier curve, and (𝑎, b, c, d) are called its control points. The mathematical basis for Bézier curves—the Bernstein polynomials—was established in 1912, but the polynomials were not applied to graphics until some 50 years later when mathematician Paul de Casteljau (1930--2022) in 1959 developed de Casteljau's algorithm, a numerically stable method for evaluating the curves, and became the first to apply them to computer-aided design at French automaker Citroën.

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.

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Example 14: Let us now consider data f₀, f₁, … , f_n corresponding to the nodes x₀, x₁, … , x_n and possibly sampled from some function f : ℝ^m → ℝ, that is, f_i = f(x_i) for i = 0, 1 … , n. The barycentric interpolant of this data is then given by \[ F(x) = \sum_{0 \le i \le n} b_i (x)\, f_i . \tag{14.1} \] The function F : ℝ^m → ℝ interpolates the data f_i at x_i for i = 0, 1 , … , n, and we require that this interpolation should be exact for linear functions: \[ \sum_{0 \le i \le n} b_i (x) = 1 \qquad\mbox{and} \qquad \sum_{0 \le i \le n} b_i (x)\, x_i = x . \] This means that when data f_i at x_i are sampled from a linear polynomial f, the interpolation (14.1) is exact: F(x_i) = f(x_i) for i = 0, 1 , … , n.

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.

\[ F_2 (x) = \frac{\left( 9- 4\sqrt{2} \right) \pi^2 + 24 \left( -2 + \sqrt{2} \right) \pi x + 48 x^2}{10 \pi^2 -48 \pi x + 96 x^2} . \] For Mathematica, using the equations (14.3), we begin by defining the nodes. Next, we define the function values at the nodes Create functions to compute Barycentric weights for the two nodes Define the interpolant for the two nodes. Create the interpolant function for two nodes

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

Define and display the interpolant function for three nodes. However, Mathematica does some simplification. Out[ ] // TableForm =

2 nodes		\( \displaystyle \quad \frac{6 \left( \frac{3\pi}{4} - x \right)}{7\pi} + \frac{6 \sqrt{2} \left( -\frac{\pi}{6} +x \right)}{7\pi} \)
3 nodes		\( \displaystyle \quad \frac{-\frac{1}{\sqrt{2} \left( -\frac{3\pi}{4} + x \right)} + \frac{1}{\sqrt{2} \left( -\frac{\pi}{4} +x \right)} - \frac{1}{2 \left( -\frac{\pi}{6} +x \right)}}{- \frac{1}{-\frac{3\pi}{4} +x} + \frac{1}{-\frac{\pi}{4} +x} - \frac{1}{-\frac{\pi}{6} +x}} \)

We check with Mathematica:

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Example 15:

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.

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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.

From solution given above, we can reconstruct the solution as a line given parametrically \[ L = \left\{ \left( x, y, z \right) \ : \ x = -\frac{38}{23}\, t -\frac{39}{23} , \quad y = \frac{13}{23}\, t - \frac{3}{23} , \quad z = t , \quad \forall t \in \mathbb{R} \right\} . \] We can decompose L into a sum of two components: the first is the line L₀, which passes through origin; the second is a translation by a particular vector v_p. To find the particular vector v_p, notice that all we have to do is set t = 0 in the parametric definition of L given above, which yields \( \displaystyle \quad {\bf v}_p = \left[ -\frac{39}{23} , \ -\frac{3}{23} , \ 0 \right] \quad \) Once we know v_p, the line L₀ is simply the remaining portion of the solution \[ L_0 = \left\{ \left( x, y, z \right) \ : \ x = -\frac{38}{23}\, t , \quad y = \frac{13}{23}\, t , \quad z = t , \quad \forall t \in \mathbb{R} \right\} . \] Clearly, L₀ is a line through the origin, and is thus a subspace of ℝ³. The line L can be realized as a translate of the line L₀ by the particular solution x_p. Now let us plot these two lines along with the particular solution x_p.

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Example 17: We consider a typical affine map: y → A x + b, where A ∈ ℝ^m×n, x ∈ ℝ^n×1, b ∈ ℝ^m×1. For simplicity, we choose m = n = 2 and set

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).

Using this affine transformation with linear part DF(·) = A(·), we check properties of Theorem 6.

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}

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Example 18: Let us consider two sets of points on the plane ℝ²: \[ \begin{split} T_1 &= \left\{ \left( -5, -3 \right) , \ \left( 2, 10 \right) , \ \left( 3, -5 \right) \right\} , \\ T_2 &= \left\{ \left( -4, 1 \right) , \ \left( -3, 11 \right) , \ \left( 1, 9 \right) \right\} . \end{split} \] We want to find an affine transformation f(x) = A x + w that maps points from T₁ into T₂. Writing affine transformation explicitly, we get \[ {\bf A} = \begin{bmatrix} a&b \\ c&d \end{bmatrix} \in \mathbb{R}^{2\times 2} , \qquad {\bf w} = \begin{bmatrix} \alpha \\ \beta \end{bmatrix} \in \mathbb{R}^{2\times 1} . \] In order to have f(T₁) = T₂, we should have \begin{align*} f \left( \begin{bmatrix} -5 \\ -3 \end{bmatrix} \right) &= \begin{bmatrix} -4 \\ \phantom{-}1 \end{bmatrix} \quad \Longrightarrow \quad \begin{split} -5 a --3b &= -4 - \alpha , \\ -5c - 3d &= 1 - \beta ; \end{split} \\ f \left( \begin{bmatrix} 2 \\ 10 \end{bmatrix} \right) &= \begin{bmatrix} -3 \\ 11 \end{bmatrix} \quad \Longrightarrow \quad \begin{split} 2 a -+10b &= -3 -\alpha , \\ 2c + 10 d &= 11 -\beta ; \end{split} \\ f \left( \begin{bmatrix} 3 \\ -5 \end{bmatrix} \right) &= \begin{bmatrix} 1 \\ 9 9 \end{bmatrix} \quad \Longrightarrow \quad \begin{split} 3 a --5b &= 1-\alpha , \\ 3c - 5d &= 9 -\beta . \end{split} \end{align*} Introducing 6-dimensional vector of unknowns X and right-hand side vector Y, \[ {\bf X} = \left[ a\ b \ c\ d\ \alpha \ \beta \right]^{\mathrm T} , \qquad {\bf Y} = \left[-4\ 1 \ -3\ 11\ 1 \ 9 \right]^{\mathrm T} , \] we can write our problem as a matrix/vector equation \[ {\bf M}\, {\bf X} = {\bf Y} , \] where \[ {\bf M} = \begin{bmatrix} -5 & -3 & \phantom{-}0&\phantom{-}0&1&0 \\ \phantom{-}0 &\phantom{-}0&-5&-3&0&1 \\ \phantom{-}2 & 10 & \phantom{-}0 & \phantom{-}0 & 1&0 \\ \phantom{-}0 & \phantom{-}0 & \phantom{-}2&10 & 0&1 \\ \phantom{-}3 & -5 & \phantom{-}0 & \phantom{-}0 & 1&0 \\ \phantom{-}0 & \phantom{-}0& \phantom{-}3 & -5& 0&1 \end{bmatrix}, \quad {\bf Y} = \begin{bmatrix} -4 \\ 1 \\ -3 \\ 11 \\ 1 \\ 9 \end{bmatrix} . \] So we use Mathematica to build matrix and column vector from Eq.(5): The last three Mathematica commands are simply verifications that the vectors (x_k, y_k) determine the corners of triangle T₁ were sent to their corresponding counterparts (z_k, w_k) of T₂.

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Example 19: Let g : ℝ³ ⇾ ℝ³ be a linear transformation that is defined by a singular matrix \[ \mathbf{A} = \begin{bmatrix} \phantom{-}1& \phantom{-}2& 3 \\ \phantom{-}2& -3& 1 \\ -1& \phantom{-}5& 2 \end{bmatrix} \qquad \Longrightarrow \qquad g({\bf x}) = {\bf A}\,{\bf x} . \] Let us consider an affine transformation \[ f(\mathbf{x}) = {\bf A}\,{\bf x} + {\bf w} , \tag{19.1} \] with some vector w to be determined. If we want transformation (19.1) to satisfy the condition f(𝑎) = b for a pair of points 𝑎 ∈ ℝ^3×1, b ∈ ℝ^3×1, we should get the relation \[ f(a) = {\bf A}\,a + {\bf w} = b . \] Hence, vector w must be equal to \[ \mathbf{w} = b - {\bf A}\,a , \tag{19.2} \] the difference of two points. Here we used canonical affine spaces 𝔸 = (ℝ³, +) = 𝔹. Points 𝑎 and b in Eq.(19.2) are identified with some vectors by choosing (fixing) some frames in ℝ³, which is isomorphic to ℝ^3×1.

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

Example 20: Given the frame \[ \phi = \left( \begin{bmatrix} \phantom{-}2 \\ -3 \end{bmatrix} , \ \begin{bmatrix} 1 \\ 4 \end{bmatrix} , \ \begin{pmatrix} 6 & 2 \end{pmatrix} \right) , \] determine the point Q that has the coordinates (-3, 2, 1).

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. \[ Q = -3 \begin{bmatrix} \phantom{-}2 \\ -3 \end{bmatrix} + 2 \begin{bmatrix} 1 \\ 4 \end{bmatrix} + \begin{pmatrix} 6 & 2 \end{pmatrix} = \begin{bmatrix} 2 \\ 19 \end{bmatrix} . \]

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Example 21: Let β and ϕ be two frames for the same affine space such that \[ \beta = \left( \begin{bmatrix} 3 \\ 2 \end{bmatrix} , \ \begin{bmatrix} 1 \\ 4 \end{bmatrix} , \ \left( 5, \ 2 \right) \right) , \qquad \phi = \left( \begin{bmatrix} 7 \\ 3 \end{bmatrix} ,\ \begin{bmatrix} -3 \\ -2 \end{bmatrix} ,\ \left( 3,\ -2 \right) \right) . \] If ⟦Q⟧_β = (-3, 1, 1), then find ⟦Q⟧_ϕ.

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*}

so the change of frame matrix M is \[ \mathbf{M} = \begin{bmatrix} 0&-1& -\frac{8}{5} \\ -2&-5& -\frac{22}{5} \\ 0&0&1 \end{bmatrix} . \] Knowing M we can compute ⟦Q⟧_ϕ: \[ [\![ Q ]\!]_{\phi} = \begin{bmatrix} 0&-1& -\frac{8}{5} \\ -2&-5& -\frac{22}{5} \\ 0&0&1 \end{bmatrix} \begin{bmatrix} -3 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} -\frac{13}{5} \\ - \frac{17}{5} \\ 1 \end{bmatrix} . \]    ■

Example 22: The Plücker coordinates were introduced by the German mathematician and physicist Julius Plücker (1801--1868) in the 19th century. Plücker 3D line coordinates are concise and efficient for numerous chores---they provide faster and simpler code for line representations than if you described lines by a direction vector and a position vector, or as two plane equations, or by two points.

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.

For illustration of Plücker coordinates as displacement-moment (d, m) on the plane, we define first points x and y: Calculate the displacement vector d and moment vector m: Ensure m is in the positive z direction Base point of the moment vector (a point on the z-axis) Now we obtain a graphical illystration:

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

Then \[ \left( {\bf d} : {\bf m} \right) = \left( 4 : 5 : -5 : 5 : 13 : 17 \right) . \] The squared distance of this line from the origin is \[ \frac{5^2 + 13^2 + 17^2}{4^2 + 5^2 + 5^2} = \frac{483}{66} = \frac{161}{22}\approx 7.31818 \approx \left( 2.70521 \right)^2 . \]

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 \[ \mathbf{T} = \begin{bmatrix} 1 & 0 & \Delta x \\ 0&1& \Delta y \\ 0&0&1 \end{bmatrix} = \begin{pmatrix} 1 & 0 &0 \\ 0&1&0 \\ \Delta x& \Delta y&1 \end{pmatrix}^{\mathrm T} . \] Recall that matrix written in parentheses acts on row vectors from right.    ■