Skip to main content

Linear Algebra Notes

🔗

Section 5.4 Linear Transformations in the Plane

🔗

Subsection 5.4.1 Linear Transformations in the Plane

🔗
Transformations from R2 to R2 are transformations of the plane, and so correspond to transformations of vectors (x,y) in the plane. They are represented by 2×2 matrices, of the form,
[abcd]
🔗
For various values of a,b,c,d∈R. The linear transformation is completely determined by these numbers. Then, a vector v=(x,y)∈R2 is transformed by,
[abcd][xy]=x[ac]+y[bd]=[ax+bycx+dy]
The most basic transformation is the identity transformation, which does not change any vector. That is, T(v)=v for every v∈R2. To determine the matrix of this transformation, consider the images T(e1),T(e2). Of course, we should have,
T(e1)=e1=[10]andT(e2)=e2=[01]
Thus, the matrix of T is,
[1001]=I2
which is the 2×2 identity matrix. This is pretty intuitive.
For example, the matrix transformation,
[31−12][xy]=[3x+y−x+2y]
🔗

Subsection 5.4.2 Reflections

🔗
Consider reflections over various lines in the plane. First, consider reflection over the x-axis. You may recall from pre-calculus that reflection a point (x,y) over the x-axis corresponds to the transformation (x,y)→(x,−y). That is, the x-coordinate is unchanged, and the y-coordinate is negated. Thus, the matrix of this transformation satisfies,
[abcd][xy]=[x−y]
[ax+bycx+dy]=[x−y]
🔗
For this equality to be true for all x and y, clearly we need a=1,b=0,c=0,d=−1. Thus, the matrix is given by,
[100−1]
🔗
From a more systematic perspective, we can derive this matrix from the images of the basis vectors. We want this transformation to have the property that,
T([10])=[10]andT([01])=[0−1]
🔗
That is, (1,0)→(1,0) (it is unchanged), and (0,1)→(0,−1). Thus, the matrix for this transformation is,
[100−1]
🔗
and indeed,
[100−1][xy]=[x−y]
🔗
In a similar way, reflection over the y-axis corresponds to the transformation (x,y)→(−x,y), and has the property that,
T([10])=[−10]andT([01])=[01]
[−1001]
Consider the transformation of reflection over the line y=x. From pre-calculus, you may recall this corresponds to the “inverse” of a function or relation, and corresponds to the transformation (x,y)→(y,x) (swapping x and y). Then, the matrix transformation satisfies,
[abcd][xy]=[yx][ax+bycx+dy]=[yx]
Equating entries, we see that a=0,b=1,c=1,d=0. Thus, the matrix is,
[0110]
From the basis vector image approach, we know that reflection over y=x requires that,
T([10])=[01]andT([01])=[10]
which gives the same result.
Consider the transformation of reflection over the line y=−x. Graphically, from some examples, you might deduce that the transformation is (x,y)→(−y,−x). From the basis vector image approach, we need,
T([10])=[0−1]andT([01])=[−10]
Thus, the matrix is given by,
Ry=−x=[0−1−10]
🔗
In summary,
🔗

Subsection 5.4.3 Stretching (Contraction/Expansion) and Dilation

Consider the horizontal transformation (x,y)→(kx,y) which multiplies the x-coordinate by a scalar k>0. If k>1, this is called a horizontal expansion, and if 0<k<1, then this is called a contraction. In terms of matrices, this corresponds to the matrix,
[k001]
Similarly, vertical expansion and contraction is the transformation (x,y)→(x,ky) for k≥0, and is given by,
[0k10]
There is another kind of stretch which can be described as a “radial” stretch, where length are expanded or contracted from the origin. This corresponds to the transformation (x,y)→(rx,ry) for some r>0. This is called a dilation if r>1, and a contraction if 0<r<1. In vector form, this is the transformation T(x)=rx, corresponding to scalar multiplication by r. As a matrix, this corresponds to the matrix,
[r00r]
For example, the matrix
[2002]
corresponds to dilaton by a factor of 2. Notice that,
[2002][xy]=[2x2y]
Notice that dilations are just two consecutive identical stretches, one by r horizontally and another by r vertically. In matrix form,
[r001][0r10]=[r00r]
In addition, we can consider non-uniform dilations, where one axis is stretched more than the other, say by a factor of r in the x-direction and s in the y-direction. This corresponds to the matrix,
[r00s]
Again, this is equivalent to the two stretches,
[r001][0s10]=[r00s]
🔗

Subsection 5.4.4 Rotation Transformations

🔗
Consider the transformation which rotates a point (x,y) about the origin. First, consider counter-clockwise rotation, and first consider a 90∘ counter-clockwise rotation. If you sketch a graph of this transformation for a vector (x,y) in QI, you can infer that the transformation is (x,y)→(−y,x). For vectors in the other 3 quadrants, it is a bit more subtle.
[abcd][xy]=[−yx][ax+bycx+dy]=[−yx]
a=0,b=−1,c=1,d=0,
[0−110]
🔗
From the basis vector image approach, consider the images of the basis vectors. We see that,
T([10])=[01]andT([01])=[−10]
🔗
Rotation 90∘ clockwise (equivalently, 270∘ counter-clockwise) is similar. The transformation is (x,y)→(−x,y), and the images of the basis vectors are,
T([10])=[0−1]andT([01])=[10]
[01−10]
[01−10][xy]=[−xy]
180∘(x,y),(x,y)→(−x,−y).
T([10])=[−10]andT([01])=[0−1]
[−100−1]
[−100−1]
🔗
In summary,
R360∘=R0∘
🔗

Subsection 5.4.5 General Rotations

🔗
Previously, we consided rotations for 90∘,180∘, and 270∘. Next, consider a more general angle. Consider rotation by an angle θ counter-clockwise. To determine the matrix of this transformation, we need to determine its image of the basis vectors. Consider e1. Its terminal point (1,0) is on the unit circle, and any rotation of e1 will also be on the unit circle. Since (1,0) lies on the positive x-axis, almost by definition, the resulting point after rotation counter-clockwise by θ will be the point (cos⁡θ,sin⁡θ). Thus,
T([10])=[cos⁡θsin⁡θ]
🔗
For e2, the rotation by θ takes it to a point on the unit circle corresponding to an angle of θ+90∘. Thus,
T([01])=[cos⁡(θ+90∘)sin⁡(θ+90∘)]
cos⁡(θ+90∘)=−sin⁡θsin⁡(θ+90∘)=cos⁡θ.
T([01])=[−sin⁡θcos⁡θ]
🔗
Putting things together, in summary, ,
🔗
Notice that substituting the particular values of 90∘,180∘,270∘, etc. result in the previously developed rotation matrices.
🔗

Subsection 5.4.6 Linear Transformations Revisited

🔗
The matrix form of a transformation allows us to use matrix multiplication to compose transformations together. Recall that,
B(Ax)=(BA)x
🔗
That is, the matrix for performing the transformation A followed by the transformation B is equivalent to the single transformation BA. From one perspective, BA was defined precisely so that that relation holds.
You may recall this from pre-calculus the definition of an odd function, which is a function f such that f(x)=−f(−x). Graphically, an odd function has rotational symmetry about the origin, in that rotating its graph 180∘ about the origin results in the same graph. You may also recall that the transformation from f(x) to −f(−x) represents a reflection over the x-axis and then reflection over the y-axis. So, equivalently, an odd function is one such that reflecting its graph over both the x-axis and then the y-axis results in the same graph.
In the language of transformations, this is because rotation by 180∘ about the origin is an equivalent transformation as the composition of two reflections. In the language of matrix transformations, this corresponds to the matrix product,
[100−1]⏟Rx[−1001]⏟Ry=[−1001]⏟R180∘
🔗

Subsection 5.4.7 Glide Reflection

🔗
Glide reflection is a reflection and a translation.
<Prev^TopNext>
Authored in PreTeXt
Powered by Runestone