In Exploration exp:shapeTransformation we saw how geometric shapes in the plane can be manipulated with matrix transformations. In this section we will study matrix transformations of the plane in more detail by applying matrix transformations to pixels in photos. We will treat every pixel of a picture as a point or the head of a vector in . When a transformation is applied to each pixel, the output pixel is colored the same color as the input pixel. The figure below shows the result of one such transformation.
Not all transformations are matrix transformations. Practice Problem prob:translation_does_not_work offers a cautionary tale of what happens when we try to find a matrix for a transformation when such a matrix does not exist. Luckily, many familiar transformations, including rotations, scalings, reflections, and shears, are matrix transformations. We will focus our attention on those.
Recall that by Observation obs:imagesOfijk of Matrix Transformations, we can find a matrix of a matrix transformation by examining what the transformation does to the standard basis vectors and .
Consider what this transformation does to the standard unit vectors. We observe that and .
This allows us to construct a candidate for the transformation matrix , by making the images of and the columns of . Thus,
We can now check to see what this matrix does to an arbitrary point . Treating this point as a vector , we compute Thus, this transformation takes point to point . So, the proposed transformation doubles all -coordinates resulting in a vertical stretch by a factor of 2.
In general, a vertical stretch (or compression) leaves unchanged, and scales the vector while preserving its vertical direction. Thus, a vertical stretch (or compression) maps to , and maps to for some positive number . Similarly, a horizontal stretch (or compression) maps to , and maps to .
A matrix transformation induced by \begin{equation} \label{vscale} M_v=\begin{bmatrix} 1 & 0\\ 0 & k \end{bmatrix}, \end{equation} where (), scales objects in the plane vertically by a factor of .
A matrix transformation induced by \begin{equation} \label{hscale} M_h=\begin{bmatrix} k & 0\\ 0 & 1 \end{bmatrix}, \end{equation} where (), scales objects in the plane horizontally by a factor of .
In stating the above formula we stipulated that . If we were to allow to be zero, what would the resulting transformations accomplish? In what way would the resulting matrices be fundamentally different from matrices and ? What would happen if were allowed to be negative? (See Practice Problem prob:k0)
A horizontal shear is a transformation that takes an arbitrary point and maps it to the point . The effect of this transformation is that all points along a fixed horizontal line slide to the left or to the right by a fixed amount. Note that the higher the point is above the -axis, the greater is the magnitude of , resulting in a greater amount of horizontal slide.
Adding a scalar multiple of the component to the component can be accomplished by matrix multiplication. Observe that
A vertical shear is a transformation that takes an arbitrary point and maps it to the point . This too, can be accomplished by matrix multiplication.
A matrix transformation induced by \begin{equation} M_{hs}=\begin{bmatrix} 1 & k\\ 0 & 1 \end{bmatrix} \end{equation} shears the plane horizontally.
A matrix transformation induced by \begin{equation} M_{vs}=\begin{bmatrix} 1 & 0\\ k & 1 \end{bmatrix} \end{equation} shears the plane vertically.
It turns out that rotations about the origin are also matrix transformations. You will have an opportunity to revisit and prove this claim in Challenge Problems for Chapter 6.
In general, we find the rotation matrix by determining the images of vectors and .
When a point is reflected about a line, its image is located on the opposite side of the line and the same distance away from the line as the original point. This is another example of a matrix transformation. (We will prove that this is a matrix transformation in Challenge Problems for Chapter 6.)
For example, the figure below shows the reflection of point about line . Note that the reflection lies on a line through perpendicular to .
Our task is to find the matrix of a reflection of the plane about an arbitrary line through the origin.
We will start with the reflection about the -axis.
So, the matrix that induces the reflection about the -axis is Next, we will consider the reflection about the -axis.
Thus, the matrix that induces the reflection about the -axis is
Now we will turn our attention to transformations that reflect the plane about the line . We will assume that .
Consider the vector and its reflection.
Observe that the head of the image vector, , will lie on the line that passes through and is perpendicular to the line . The equation of this line is given by
\begin{equation} \label{eq:reflectionline} y=-\frac{1}{m}x+\frac{1}{m} \end{equation}
The head of will also lie on the circle with equation
To find the image of we need to determine where the line intersects the circle. Substitution gives us
After a little algebra we get The quadratic formula yields
The solution corresponds to the head of the vector . So, the -component of is . We find the -component of by substituting into Equation eq:reflectionline. Thus, the image of under this reflection is given by
Next we need to find the image of . The head of is located at one of the intersections of line and the circle .
We leave it to the reader to verify that \begin{equation} \label{eq:imageofj} T(\vec{j})=\begin{bmatrix}\frac{2m}{m^2+1}\\\frac{m^2-1}{m^2+1}\end{bmatrix} \end{equation}
This reflection is induced by the matrix
A matrix transformation induced by \begin{equation} \label{eq:reflectionymx} M_{y=mx}=\frac{1}{1+m^2}\begin{bmatrix} 1-m^2 & 2m \\ 2m & m^2-1 \end{bmatrix} \end{equation} reflects the plane about the line .
Note that the eye of the duck in Example ex:reflectedduck is located on the line . The reflection leaves the eye fixed in place. The eye is an example of a fixed point. In Practice Problem prob:fixedpoint you will be asked to show that every point along the line is a fixed point.
If a matrix transformation is followed by another matrix transformation, the resulting transformation can be represented as a product of the two matrices that induce the individual transformations. Thus, if is induced by and is induced by , then is induced by .
Remember that matrix multiplication is not commutative, so the order in which the matrices are multiplied is of utmost importance.
Consider the following example which incorporates a reflection as well as a rotation of vectors.
Using Formula form:rotation, the matrix of the transformation which involves rotating through an angle of is \begin{equation*} M_1 = \begin{bmatrix} \cos \left ( \pi /6\right ) & -\sin \left ( \pi /6\right ) \\ \sin \left ( \pi /6\right ) & \cos \left ( \pi /6\right ) \end{bmatrix} = \begin{bmatrix} \frac{1}{2}\sqrt{3} & -\frac{1}{2} \\ \frac{1}{2} & \frac{1}{2}\sqrt{3} \end{bmatrix} \end{equation*}
As we learned in Exploration init:reflectionxyaxes, the matrix for the transformation which reflects all vectors about the -axis is .
Therefore, the matrix of the linear transformation which first rotates through and then reflects through the -axis is given by \begin{equation*} M = M_2 M_1 = \begin{bmatrix}1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} \frac{1}{2}\sqrt{3} & -\frac{1}{2} \\ \frac{1}{2} & \frac{1}{2}\sqrt{3} \end{bmatrix} = \begin{bmatrix} \frac{1}{2}\sqrt{3} & \frac{1}{2} \\ -\frac{1}{2} & -\frac{1}{2}\sqrt{3} \end{bmatrix} \end{equation*}
Following the rotation with reflection across the -axis will make it point straight down. Sure enough,
Consider the matrix . This matrix induces a transformation that can be broken into two parts: (1) a reflection followed by (2) a shear. Find matrices and that induce the reflection and the shear respectively. Verify that the product of the two matrices is equal to (be careful about the order of the product!).
Let . Use the GeoGebra interactive above to visually examine the transformation induced by . The composition of which transformations is equivalent to the transformation induced by ?
Let . Use the GeoGebra interactive above to visually examine the transformation induced by . The composition of which transformations is equivalent to the transformation induced by ?
To do this, consider transformations and . Transformation is induced by , and is induced by .
The figure on the left illustrates the action of on a single point . First, is reflected about the line , then is reflected about the line .
The figure on the right shows the action of on the same point . The point is first reflected about the line , followed by a reflection about the line . The final images of point under and are clearly different.
Since , we conclude that .
Suppose a 1 by 1 photo of a chipmunk was shifted as shown in the figure.
Suppose we tried to construct a standard matrix for this transformation by making the images of and the columns of . We would obtain Does this matrix describe the transformation? If so, prove it. If not, explain why not.
Example exa:rotationreflection was adapted from Example 5.27 of Ken Kuttler’s A First Course in Linear Algebra. (CC-BY)
Ken Kuttler, A First Course in Linear Algebra, Lyryx 2017, Open Edition, p. 290.
The following images are courtesy of Wikimedia Commons
Adrian Pingstone, A male Mandarin Duck at Slimbridge Wildfowl and Wetlands Centre, Gloucestershire, England. Public Domain
Ansgar Koreng, Facade of ARD-Hauptstadtstudio in Berlin-Mitte. CC-BY 4.0
Christoph Braun, Reflection of St. Michaelis Church in a window of St. Ansgar in Hamburg, Germany. Public Domain.
Daniel Gammert, Red-billed Gulls Chroicocephalus novaehollandiae scopulinus. Brighton Beach, New Zealand. Public Domain
I, Tony Wills, Red billed gull in Wellington Harbour, Wellington, New Zealand. CC-BY
Jackhynes, Lleyn sheep taken with a Sony Digital Camera at 3.2 megapixels in Devon, UK. Public Domain
Erik Davis, Chipmunk. CC-BY
2024-09-11 22:57:31
