Section 5.4 Linear Transformations in the Plane
Subsection 5.4.1 Linear Transformations in the Plane
Transformations from \(\mathbb{R}^2\) to \(\mathbb{R}^2\) are transformations of the plane, and so correspond to transformations of vectors \((x,y)\) in the plane. They are represented by \(2 \times 2\) matrices, of the form,
\begin{equation*}
\begin{bmatrix} a \amp b \\ c \amp d \end{bmatrix}
\end{equation*}
For various values of \(a, b, c, d \in \mathbb{R}\text{.}\) The linear transformation is completely determined by these numbers. Then, a vector \(\vec{v} = (x,y) \in \mathbb{R}^2\) is transformed by,
\begin{equation*}
\begin{bmatrix} a \amp b \\ c \amp d \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = x \begin{bmatrix} a \\ c \end{bmatrix} + y \begin{bmatrix} b \\ d \end{bmatrix} = \begin{bmatrix} ax + by \\ cx + dy \end{bmatrix}
\end{equation*}
The most basic transformation is the identity transformation, which does not change any vector. That is, \(T(\vec{v}) = \vec{v}\) for every \(\vec{v} \in \mathbb{R}^2\text{.}\) To determine the matrix of this transformation, consider the images \(T(\vec{e}_1), T(\vec{e}_2)\text{.}\) Of course, we should have,
\begin{equation*}
T(\vec{e}_1) = \vec{e}_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \qquad \text{and} \qquad T(\vec{e}_2) = \vec{e}_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}
\end{equation*}
Thus, the matrix of \(T\) is,
\begin{equation*}
\begin{bmatrix} 1 \amp 0 \\ 0 \amp 1 \end{bmatrix} = I_2
\end{equation*}
which is the \(2 \times 2\) identity matrix. This is pretty intuitive.Example 5.4.2.
\begin{equation*}
\begin{bmatrix} 3 \amp 1 \\ -1 \amp 2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 3x + y \\ -x + 2y \end{bmatrix}
\end{equation*}
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 \((x,y) \rightarrow (x,-y)\text{.}\) That is, the \(x\)-coordinate is unchanged, and the \(y\)-coordinate is negated. Thus, the matrix of this transformation satisfies,
\begin{equation*}
\begin{bmatrix} a \amp b \\ c \amp d \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} x \\ -y \end{bmatrix}
\end{equation*}
\begin{equation*}
\begin{bmatrix} ax + by \\ cx + dy \end{bmatrix} = \begin{bmatrix} x \\ -y \end{bmatrix}
\end{equation*}
For this equality to be true for all \(x\) and \(y\text{,}\) clearly we need \(a = 1, b = 0, c = 0, d = -1\text{.}\) Thus, the matrix is given by,
\begin{equation*}
\begin{bmatrix} 1 \amp 0 \\ 0 \amp -1 \end{bmatrix}
\end{equation*}
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,
\begin{equation*}
T\brac{\begin{bmatrix} 1 \\ 0 \end{bmatrix}} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \qquad \text{and} \qquad T\brac{\begin{bmatrix} 0 \\ 1 \end{bmatrix}} = \begin{bmatrix} 0 \\ -1 \end{bmatrix}
\end{equation*}
That is, \((1,0) \rightarrow (1,0)\) (it is unchanged), and \((0,1) \rightarrow (0,-1)\text{.}\) Thus, the matrix for this transformation is,
\begin{equation*}
\begin{bmatrix} 1 \amp 0 \\ 0 \amp -1 \end{bmatrix}
\end{equation*}
and indeed,
\begin{equation*}
\begin{bmatrix} 1 \amp 0 \\ 0 \amp -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} x \\ -y \end{bmatrix}
\end{equation*}
In a similar way, reflection over the \(y\)-axis corresponds to the transformation \((x,y) \rightarrow (-x,y)\text{,}\) and has the property that,
\begin{equation*}
T\brac{\begin{bmatrix} 1 \\ 0 \end{bmatrix}} = \begin{bmatrix} -1 \\ 0 \end{bmatrix} \qquad \text{and} \qquad T\brac{\begin{bmatrix} 0 \\ 1 \end{bmatrix}} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}
\end{equation*}
\begin{equation*}
\begin{bmatrix} -1 \amp 0 \\ 0 \amp 1 \end{bmatrix}
\end{equation*}
Example 5.4.3. Reflection over \(y = x\).
Consider the transformation of reflection over the line \(y = x\text{.}\) From pre-calculus, you may recall this corresponds to the “inverse” of a function or relation, and corresponds to the transformation \((x,y) \rightarrow (y,x)\) (swapping \(x\) and \(y\)). Then, the matrix transformation satisfies,
\begin{align*}
\begin{bmatrix} a \amp b \\ c \amp d \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} y \\ x \end{bmatrix}\\
\begin{bmatrix} ax + by \\ cx + dy \end{bmatrix} = \begin{bmatrix} y \\ x \end{bmatrix}
\end{align*}
Equating entries, we see that \(a = 0, b = 1, c = 1, d = 0\text{.}\) Thus, the matrix is,
\begin{equation*}
\begin{bmatrix} 0 \amp 1 \\ 1 \amp 0 \end{bmatrix}
\end{equation*}
From the basis vector image approach, we know that reflection over \(y = x\) requires that,
\begin{equation*}
T\brac{\begin{bmatrix} 1 \\ 0 \end{bmatrix}} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \qquad \text{and} \qquad T\brac{\begin{bmatrix} 0 \\ 1 \end{bmatrix}} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}
\end{equation*}
which gives the same result.Example 5.4.4. Reflection over \(y = -x\).
Consider the transformation of reflection over the line \(y = -x\text{.}\) Graphically, from some examples, you might deduce that the transformation is \((x,y) \rightarrow (-y,-x)\text{.}\) From the basis vector image approach, we need,
\begin{equation*}
T\brac{\begin{bmatrix} 1 \\ 0 \end{bmatrix}} = \begin{bmatrix} 0 \\ -1 \end{bmatrix} \qquad \text{and} \qquad T\brac{\begin{bmatrix} 0 \\ 1 \end{bmatrix}} = \begin{bmatrix} -1 \\ 0 \end{bmatrix}
\end{equation*}
Thus, the matrix is given by,
\begin{equation*}
R_{y=-x} = \begin{bmatrix} 0 \amp -1 \\ -1 \amp 0 \end{bmatrix}
\end{equation*}
In summary,
Theorem 5.4.5.
\begin{equation*}
R_x = \begin{bmatrix} 1 \amp 0 \\ 0 \amp -1 \end{bmatrix} \qquad R_y = \begin{bmatrix} -1 \amp 0 \\ 0 \amp 1 \end{bmatrix} \qquad R_{y=x} = \begin{bmatrix} 0 \amp 1 \\ 1 \amp 0 \end{bmatrix}
\end{equation*}
Subsection 5.4.3 Stretching (Contraction/Expansion) and Dilation
Example 5.4.6. Expansion and contraction.
\begin{equation*}
\begin{bmatrix} k \amp 0 \\ 0 \amp 1 \end{bmatrix}
\end{equation*}
Similarly, vertical expansion and contraction is the transformation \((x,y) \rightarrow (x,ky)\) for \(k \ge 0\text{,}\) and is given by,
\begin{equation*}
\begin{bmatrix} 0 \amp k \\ 1 \amp 0 \end{bmatrix}
\end{equation*}
Example 5.4.7. Dilations.
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 \((x,y) \rightarrow (rx,ry)\) for some \(r > 0\text{.}\) This is called a dilation if \(r > 1\text{,}\) and a contraction if \(0 < r < 1\text{.}\) In vector form, this is the transformation \(T(\vec{x}) = r\vec{x}\text{,}\) corresponding to scalar multiplication by \(r\text{.}\) As a matrix, this corresponds to the matrix,
\begin{equation*}
\begin{bmatrix} r \amp 0 \\ 0 \amp r \end{bmatrix}
\end{equation*}
For example, the matrix
\begin{equation*}
\begin{bmatrix} 2 \amp 0 \\ 0 \amp 2 \end{bmatrix}
\end{equation*}
corresponds to dilaton by a factor of 2. Notice that,
\begin{equation*}
\begin{bmatrix} 2 \amp 0 \\ 0 \amp 2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 2x \\ 2y \end{bmatrix}
\end{equation*}
Notice that dilations are just two consecutive identical stretches, one by \(r\) horizontally and another by \(r\) vertically. In matrix form,
\begin{equation*}
\begin{bmatrix} r \amp 0 \\ 0 \amp 1 \end{bmatrix} \begin{bmatrix} 0 \amp r \\ 1 \amp 0 \end{bmatrix} = \begin{bmatrix} r \amp 0 \\ 0 \amp r \end{bmatrix}
\end{equation*}
Example 5.4.8. Non-uniform dilations.
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,
\begin{equation*}
\begin{bmatrix} r \amp 0 \\ 0 \amp s \end{bmatrix}
\end{equation*}
Again, this is equivalent to the two stretches,
\begin{equation*}
\begin{bmatrix} r \amp 0 \\ 0 \amp 1 \end{bmatrix} \begin{bmatrix} 0 \amp s \\ 1 \amp 0 \end{bmatrix} = \begin{bmatrix} r \amp 0 \\ 0 \amp s \end{bmatrix}
\end{equation*}
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 \(90^{\circ}\) 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 \((x,y) \rightarrow (-y,x)\text{.}\) For vectors in the other 3 quadrants, it is a bit more subtle.
\begin{align*}
\begin{bmatrix} a \amp b \\ c \amp d \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} -y \\ x \end{bmatrix}\\
\begin{bmatrix} ax + by \\ cx + dy \end{bmatrix} = \begin{bmatrix} -y \\ x \end{bmatrix}
\end{align*}
\(a = 0, b = -1, c = 1, d = 0\text{,}\)
\begin{equation*}
\begin{bmatrix} 0 \amp -1 \\ 1 \amp 0 \end{bmatrix}
\end{equation*}
From the basis vector image approach, consider the images of the basis vectors. We see that,
\begin{equation*}
T\brac{\begin{bmatrix} 1 \\ 0 \end{bmatrix}} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \qquad \text{and} \qquad T\brac{\begin{bmatrix} 0 \\ 1 \end{bmatrix}} = \begin{bmatrix} -1 \\ 0 \end{bmatrix}
\end{equation*}
Rotation \(90^{\circ}\) clockwise (equivalently, \(270^{\circ}\) counter-clockwise) is similar. The transformation is \((x,y) \rightarrow (-x,y)\text{,}\) and the images of the basis vectors are,
\begin{equation*}
T\brac{\begin{bmatrix} 1 \\ 0 \end{bmatrix}} = \begin{bmatrix} 0 \\ -1 \end{bmatrix} \qquad \text{and} \qquad T\brac{\begin{bmatrix} 0 \\ 1 \end{bmatrix}} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}
\end{equation*}
\begin{equation*}
\begin{bmatrix} 0 \amp 1 \\ -1 \amp 0 \end{bmatrix}
\end{equation*}
\begin{equation*}
\begin{bmatrix} 0 \amp 1 \\ -1 \amp 0 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} -x \\ y \end{bmatrix}
\end{equation*}
\(180^{\circ}\)\((x,y)\text{,}\)\((x,y) \rightarrow (-x,-y)\text{.}\)
\begin{equation*}
T\brac{\begin{bmatrix} 1 \\ 0 \end{bmatrix}} = \begin{bmatrix} -1 \\ 0 \end{bmatrix} \qquad \text{and} \qquad T\brac{\begin{bmatrix} 0 \\ 1 \end{bmatrix}} = \begin{bmatrix} 0 \\ -1 \end{bmatrix}
\end{equation*}
\begin{equation*}
\begin{bmatrix} -1 \amp 0 \\ 0 \amp -1 \end{bmatrix}
\end{equation*}
\begin{equation*}
\begin{bmatrix} -1 \amp 0 \\ 0 \amp -1 \end{bmatrix}
\end{equation*}
In summary,
Theorem 5.4.9.
The matrices for some rotations include,
\begin{equation*}
R_{90^{\circ}} = \begin{bmatrix} 0 \amp -1 \\ 1 \amp 0 \end{bmatrix} \qquad R_{180^{\circ}} = \begin{bmatrix} -1 \amp 0 \\ 0 \amp -1 \end{bmatrix} \qquad R_{270^{\circ}} = \begin{bmatrix} 0 \amp 1 \\ -1 \amp 0 \end{bmatrix}
\end{equation*}
Subsection 5.4.5 General Rotations
Previously, we consided rotations for \(90^{\circ}, 180^{\circ}\text{,}\) and \(270^{\circ}\text{.}\) Next, consider a more general angle. Consider rotation by an angle \(\theta\) counter-clockwise. To determine the matrix of this transformation, we need to determine its image of the basis vectors. Consider \(\vec{e}_1\text{.}\) Its terminal point \((1,0)\) is on the unit circle, and any rotation of \(\vec{e}_1\) 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 \(\theta\) will be the point \((\cos{\theta}, \sin{\theta})\text{.}\) Thus,
\begin{equation*}
T\brac{\begin{bmatrix} 1 \\ 0 \end{bmatrix}} = \begin{bmatrix} \cos{\theta} \\ \sin{\theta} \end{bmatrix}
\end{equation*}
For \(\vec{e}_2\text{,}\) the rotation by \(\theta\) takes it to a point on the unit circle corresponding to an angle of \(\theta + 90^{\circ}\text{.}\) Thus,
\begin{equation*}
T\brac{\begin{bmatrix} 0 \\ 1 \end{bmatrix}} = \begin{bmatrix} \cos{(\theta + 90^{\circ})} \\ \sin{(\theta + 90^{\circ})} \end{bmatrix}
\end{equation*}
\(\cos{(\theta + 90^{\circ})} = -\sin{\theta}\)\(\sin{(\theta + 90^{\circ})} = \cos{\theta}\text{.}\)
\begin{equation*}
T\brac{\begin{bmatrix} 0 \\ 1 \end{bmatrix}} = \begin{bmatrix} -\sin{\theta} \\ \cos{\theta} \end{bmatrix}
\end{equation*}
Putting things together, in summary, ,
Theorem 5.4.10.
The rotation transformation \(R_{\theta}: \mathbb{R}^2 \rightarrow \mathbb{R}^2\) for rotation by an angle \(\theta\) counter-clockwise has matrix given by,
\begin{equation*}
\boxed{R_{\theta} = \begin{bmatrix} \cos{\theta} \amp -\sin{\theta} \\ \sin{\theta} \amp \cos{\theta} \end{bmatrix}}
\end{equation*}
Notice that substituting the particular values of \(90^{\circ}, 180^{\circ}, 270^{\circ}\text{,}\) 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,
\begin{equation*}
B(A\vec{x}) = (BA) \vec{x}
\end{equation*}
That is, the matrix for performing the transformation \(A\) followed by the transformation \(B\) is equivalent to the single transformation \(BA\text{.}\) From one perspective, \(BA\) was defined precisely so that that relation holds.
Example 5.4.11.
You may recall this from pre-calculus the definition of an odd function, which is a function \(f\) such that \(f(x) = -f(-x)\text{.}\) Graphically, an odd function has rotational symmetry about the origin, in that rotating its graph \(180^{\circ}\) 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 \(180^{\circ}\) 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,
\begin{equation*}
\underbrace{\begin{bmatrix} 1 \amp 0 \\ 0 \amp -1 \end{bmatrix}}_{R_x} \underbrace{\begin{bmatrix} -1 \amp 0 \\ 0 \amp 1 \end{bmatrix}}_{R_y} = \underbrace{\begin{bmatrix} -1 \amp 0 \\ 0 \amp 1 \end{bmatrix}}_{R_{180^{\circ}}}
\end{equation*}
Subsection 5.4.7 Glide Reflection
Glide reflection is a reflection and a translation.
