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Applied Mathematics

Peter Staab

Section 3.4 Linear Transformations

This section discusses linear transformations. In short, such a transformation will map vectors to vectors in a linear way. The definition of a linear transformation (or linear map) is the following:
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Definition 3.4.1. Linear Transformation.

Let \(V\) and \(W\) be vector spaces. A linear transformation or linear map \(T\) from \(V\) to \(W\) is a function that assigns to each vector \(\boldsymbol{v} \in V\) a unique vector \(T\boldsymbol{v} \in W\) and that satisfies for each \(\boldsymbol{u}\) and \(\boldsymbol{v}\) in \(V\) and each scalar \(\alpha\text{,}\)
\begin{align} (\boldsymbol{u} + \boldsymbol{v}) \amp = T(\boldsymbol{u}) + T (\boldsymbol{v}), \tag{3.4.1}\\ T(\alpha \boldsymbol{v}) \amp = \alpha T (\boldsymbol{v}), \tag{3.4.2} \end{align}
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These are also called homomorphisms and the notation explaining that a map \(T\) goes from \(V\) to \(W\) is \(T: V \rightarrow W\text{.}\)
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Example 3.4.2. Reflection Map.

The reflection of any vector in \(\mathbb{R}^2\) across the horizontal axis is a linear map. Specifically this is given as
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\begin{equation*} T(\begin{bmatrix} x_1 \\ x_2 \end{bmatrix})= \begin{bmatrix} x_1 \\ -x_2 \end{bmatrix} \end{equation*}
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and geometrically you can see this as:
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Figure 3.4.3. Mapping a vector over the \(x\)-axis
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Show that this is a linear transformation.
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Solution.
Specifically, we need to show that \(T\) defined above satisfies (3.4.1) and (3.4.2). Let
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\begin{align*} \boldsymbol{x} \amp = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}, \amp \boldsymbol{y} \amp = \begin{bmatrix} y_1 \\ y_2 \end{bmatrix} \end{align*}
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and
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\(\alpha \in \mathbb{R}\text{,}\) then
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\begin{align*} T(\boldsymbol{x}+\boldsymbol{y}) \amp = T \biggl( \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}+ \begin{bmatrix} y_1 \\ y_2 \end{bmatrix} \biggr) = T \biggl( \begin{bmatrix} x_1 + y_1 \\ x_2 + y_2 \end{bmatrix} \biggr) \\ \amp = \begin{bmatrix} x_1 + y_1 \\ -(x_2+y_2) \end{bmatrix} = \begin{bmatrix} x_1 + y_1 \\ -x_2 -y_2 \end{bmatrix} \\ \amp = \begin{bmatrix} x_1 \\ -x_2 \end{bmatrix} + \begin{bmatrix} y_1 \\ -y_2 \end{bmatrix} = T(\boldsymbol{x}) + T(\boldsymbol{y}). \end{align*}
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so (3.4.1) is satisfied. Next,
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\begin{align*} T(\alpha \boldsymbol{x}) \amp = T \biggl( \alpha \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} \biggr) = T \biggl( \begin{bmatrix} \alpha x_1 \\ \alpha x_2 \end{bmatrix} \biggr) \\ \amp = \begin{bmatrix} \alpha x_1 \\ -\alpha x_2 \end{bmatrix} = \alpha \begin{bmatrix} x_1 \\ -x_2 \end{bmatrix} = \alpha T(\boldsymbol{x}). \end{align*}
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so (3.4.2) is satisfied.
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A very important linear map in \(\mathbb{R}^2\) is the rotational map that takes any vector in the plane and rotates it a given angle. The next example, derives this map.
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Example 3.4.4. Rotational Map in \(\mathbb{R}^2\).

Let
\begin{equation*} \boldsymbol{x} = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} \end{equation*}
be a vector in \(\mathbb{R}^2\text{.}\) Let the function \(T\) take the vector \(\boldsymbol{x}\) and rotate it by \(\theta\) radians in the counterclockwise direction. Call the new vector
\begin{equation*} \boldsymbol{y} =\begin{bmatrix} y_1 \\ y_2 \end{bmatrix} \end{equation*}
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Figure 3.4.5. Diagram of the rotational map
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Derive the formula for the rotational map.
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Solution.
Let \(r = ||\boldsymbol{x}||\) and since \(\boldsymbol{y}\) is the rotated version of \(\boldsymbol{x}\) it has the same length, therefore \(r=||\boldsymbol{y}||\text{.}\) The values \(x_1, x_2, y_1,y_2\) can be written in terms of \(\alpha\text{,}\) the angle that the vector \(\boldsymbol{x}\) makes with the positive horizontal axis, and \(\theta\) the angle between the vectors as follows.
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\begin{align*} x_1 \amp = r \cos \alpha, \amp x_2 \amp = r \sin \alpha, \\ y_1 \amp = r \cos (\alpha+\theta), \amp y_2 \amp= r \sin (\alpha + \theta) \\ \amp = r \cos \alpha \cos \theta - r \sin \theta \sin \alpha, \amp y_2 \amp = r \sin \theta \cos \alpha + r \sin \alpha \cos \theta \\ \amp = x_1 \cos \theta - x_2 \sin \theta, \amp \amp = x_1 \sin \theta + x_2 \cos \theta \end{align*}
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and note that these can be written:
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\begin{align*} \begin{bmatrix} y_1 \\ y_2 \end{bmatrix} \amp = \begin{bmatrix} \cos \theta \amp - \sin \theta \\ \sin \theta \amp \cos \theta \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} \end{align*}
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This is a linear transformation (as we will explain later) and is called a rotational transformation.
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Example 3.4.6. Linear Scale Map.

Consider the map \(S: \mathbb{R}^2 \rightarrow \mathbb{R}^2\) that scales any vector in the plane by a factor of \(k\) given by
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\begin{equation*} S(\boldsymbol{x}) = k \boldsymbol{x} \end{equation*}
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which can be visualized in the following diagram where \(k=2\text{:}\)
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Figure 3.4.7. Scaling Map
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where every vector under the map results in a new vector that is twice as long as the original. In general, the scale \(k\) will scale the vector by a factor of \(k\) and recall that if \(k \lt 0\text{,}\) then the direction changes. Show that this is a linear map.
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Solution.
Again, we show that (3.4.1) and (3.4.2) are satisfied. Let \(\boldsymbol{x}\) and \(\boldsymbol{y}\) be elements of \(\mathbb{R}^2\text{.}\)
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\begin{equation*} S(\boldsymbol{x}+\boldsymbol{y}) = k(\boldsymbol{x}+\boldsymbol{y}) = k\boldsymbol{x} + k \boldsymbol{y} = S(\boldsymbol{x}) + S(\boldsymbol{y}) \end{equation*}
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so (3.4.1) is satisfied and
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\begin{equation*} S(\alpha \boldsymbol{x}) = k (\alpha \boldsymbol{x} ) = \alpha (k \boldsymbol{x}) = \alpha S(\boldsymbol{x}) \end{equation*}
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so (3.4.1) is satisfied so \(S\) is a linear map.
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Proof.

This is a consequence of matrix operations.
\begin{align*} T(\boldsymbol{x}+\boldsymbol{y}) &= A(\boldsymbol{x}+\boldsymbol{y}) = A\boldsymbol{x} + A \boldsymbol{y} = T(\boldsymbol{x}) + T(\boldsymbol{y}) \\ T(\alpha \boldsymbol{x}) & = A (\alpha \boldsymbol{x}) = \alpha A\boldsymbol{x} = \alpha T(\boldsymbol{x}). \end{align*}
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Note that the rotational transformation that was defined in Example 3.4.4 is easily shown to be a linear transformation because from Theorem 3.4.8, any transformation shown as a matrix, is a linear transformation. The next theorem shows the counter direction to Theorem 3.4.8, that is that any linear transformation can be written as a matrix.
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We won’t prove this here, but instead will motivate this below. In short, if \(T\) is a linear transformation, then the matrix \(A_T\) corresponding to the linear transformation is called the transformation matrix.
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Subsection 3.4.1 Finding the Matrix Form of a Linear Transformation

The theorem above shows that any linear transformation, \(T: V \rightarrow W\) can be written in matrix form. This section explains how to find it. Let \(B_V=(\boldsymbol{v}_1,\boldsymbol{v}_2, \ldots, \boldsymbol{v}_m)\) be a basis of \(V\) and \(B_W=(\boldsymbol{w}_1,\boldsymbol{w}_2, \ldots, \boldsymbol{w}_n)\) be a basis of \(W\text{.}\) Any vector \(\boldsymbol{x}\) in \(V\) can be written
\begin{equation*} \boldsymbol{x} = c_1 \boldsymbol{v}_1 + c_2 \boldsymbol{v}_2 + \cdots + c_m \boldsymbol{v}_m = \sum_{j=1}^m c_j \boldsymbol{v}_j \end{equation*}
or in other words \(\text{Rep}_{B_V}(\boldsymbol{x}) = \boldsymbol{c}\text{.}\) Applying the map to \(\boldsymbol{x}\) is
\begin{equation*} T(\boldsymbol{x}) = T\biggl(\sum_{j=1}^m c_j \boldsymbol{v}_j \biggr) \notag \end{equation*}
and since it is a linear map
\begin{equation} T(\boldsymbol{x})= \sum_{j=1}^m c_j T(\boldsymbol{v}_j)\tag{3.4.3} \end{equation}
Next, we write the transformation in terms of the basis vectors of \(W\) or
\begin{equation} T(\boldsymbol{v}_j) = a_{1,j} \boldsymbol{w}_1 + a_{2,j} \boldsymbol{w}_2 + \cdots + a_{n,j} \boldsymbol{w}_n = \sum_{i=1}^n a_{i,j} \boldsymbol{w}_i\tag{3.4.4} \end{equation}
Substituting (3.4.4) into (3.4.3) results in
\begin{equation*} T(\boldsymbol{x}) = \sum_{j=1}^m c_j \sum_{i=1}^n a_{i,j} \boldsymbol{w}_i = \sum_{i=1}^n \sum_{j=1}^m a_{i,j} c_j \boldsymbol{w}_i \end{equation*}
and letting \(A_T\) be the \(m \times n\) matrix with entries \(a_{i,j}\) then
\begin{equation*} \text{Rep}_{B_W} (T(\boldsymbol{x})) = A_T\boldsymbol{c} \end{equation*}
or in other words, the matrix performs the map on the coefficients. Equation (3.4.4) also shows how the matrix \(A_T\) can be created from the linear map. That equation can also be thought of as a representation of the basis vectors or
\begin{equation*} \text{Rep}_{B_W} ( T(\boldsymbol{v}_j)) = \boldsymbol{a}_j \end{equation*}
where \(a_j\) is the \(j\)th column of \(A\text{.}\) The following summarizes how to find the matrix.
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Remark 3.4.10.

If \(A_T\) is the matrix representation of the map \(T\text{,}\) then the \(j\)th column of \(A_T\) is the vector \(\text{Rep}_{B_W}(T(\boldsymbol{v}_j))\text{,}\) the map applied to the \(j\)th basis vector of \(V\) written in terms of the basis of \(W\text{.}\)
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We now show many examples on how to apply this.
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Example 3.4.11.

The matrix representation of the reflection map from Example 3.4.2 given by
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\begin{equation*} T\biggl(\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} \biggr) = \begin{bmatrix} x_1 \\ -x_2 \end{bmatrix} \end{equation*}
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where the representation will be in terms of the standard basis vectors.
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Solution.
In this case, we need to determine how the standard basis vectors map under the reflection. Thus
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\begin{align*} \boldsymbol{v}_1 \amp = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \amp \boldsymbol{v}_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \end{align*}
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and
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\begin{align*} T(\boldsymbol{v}_1) \amp = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \amp T(\boldsymbol{v}_2) \amp = \begin{bmatrix} 0 \\ -1 \end{bmatrix} \end{align*}
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Because we are using the standard basis vectors, the representations of these vectors are themselves therefore,
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\begin{equation*} A_T = \begin{bmatrix} 1 \amp 0 \\ 0 \amp -1 \end{bmatrix} \end{equation*}
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and just to verify,
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\begin{align*} T\biggl(\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} \biggr) \amp = \begin{bmatrix} 1 \amp 0 \\ 0 \amp -1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} x_1 \\ - x_2 \end{bmatrix} \end{align*}
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This next example shows how to contruct the transformation matrix for the scale map.
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Example 3.4.12.

Find the matrix representation of the scale map in Example 3.4.6.
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Solution.
We need to map the basis vectors \([1\;\;0]^{\intercal}\) and \([0\;\;1]^{\intercal}\) to determine the columns of the matrix representation.
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\begin{align*} S \biggl(\begin{bmatrix} 1 \\ 0 \end{bmatrix} \biggr) \amp = \begin{bmatrix} k \\ 0 \end{bmatrix} \amp S\biggl( \begin{bmatrix} 0 \\ 1 \end{bmatrix} \biggr) \amp = \begin{bmatrix} 0 \\ k \end{bmatrix} \end{align*}
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so the matrix representation is
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\begin{equation*} A_S = \begin{bmatrix} k \amp 0 \\ 0 \amp k \end{bmatrix} \end{equation*}
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And the next example shows that a common matrix operation, the trace is a linear map.
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Example 3.4.13.

Let \(T: \mathcal{M}_{2 \times 2} \rightarrow \mathbb{R}\)
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\begin{equation*} \begin{bmatrix} a \amp b \\ c\amp d \end{bmatrix} \rightarrow a + d \end{equation*}
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which is the trace of a 2 by 2 matrix. Show that the trace is a linear map and find the matrix representation of the trace.
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Solution.
First, to prove that the trace is a linear map, we need to show that it satisfies (3.4.1) and (3.4.2). Let
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\begin{align*} A_1 \amp = \begin{bmatrix} a_1 \amp b_1 \\ c_1\amp d_1 \end{bmatrix} \amp A_2 \amp = \begin{bmatrix} a_2 \amp b_2 \\ c_2\amp d_2 \end{bmatrix} \end{align*}
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Then
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\begin{align*} T(A_1+A_2) \amp = T \left( \begin{bmatrix} a_1 \amp b_1 \\ c_1\amp d_1 \end{bmatrix} + \begin{bmatrix} a_2 \amp b_2 \\ c_2\amp d_2 \end{bmatrix} \right) = T \left( \begin{bmatrix} a_1 + a_2 \amp b_1 + b_2 \\ c_1 + c_2 \amp d_1 + d_2 \end{bmatrix} \right) \\ \amp = (a_1 + a_2) + (d_1 + d_2) \\ \amp = (a_1 + d_1) + (a_2 + d_2) = T(A_1) + T(A_2) \end{align*}
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Similarly,
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\begin{align*} T(\alpha A_1) \amp = T \biggl( \alpha \begin{bmatrix} a_1 \amp b_1 \\ c_1\amp d_1 \end{bmatrix}\biggr) = T\biggl(\begin{bmatrix} \alpha a_1 \amp \alpha b_1 \\ \alpha c_1\amp \alpha d_1 \end{bmatrix} \biggr) \\ \amp = \alpha a_1 + \alpha d_1 = \alpha (a_1 + d_1) = \alpha T(A_1) \end{align*}
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Next, we want to find the matrix representation of the trace. To do this, we need to determine how the map affects the basis of the vector space \(\mathcal{M}_{2 \times 2}\text{,}\) which is
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\begin{equation*} \left( \begin{bmatrix} 1 \amp 0 \\ 0 \amp 0 \end{bmatrix}, \begin{bmatrix} 0 \amp 1 \\ 0 \amp 0 \end{bmatrix}, \begin{bmatrix} 0 \amp 0 \\ 1 \amp 0 \end{bmatrix}, \begin{bmatrix} 0 \amp 0 \\ 0 \amp 1 \end{bmatrix} \right) \end{equation*}
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and since
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\begin{align*} \begin{bmatrix} 1 \amp 0 \\ 0 \amp 0 \end{bmatrix} \amp\mapsto 1, \amp \begin{bmatrix} 0 \amp 1 \\ 0 \amp 0 \end{bmatrix} \amp\mapsto 0, \\ \begin{bmatrix} 0 \amp 0 \\ 1 \amp 0 \end{bmatrix} \amp\mapsto 0, \amp \begin{bmatrix} 0 \amp 0 \\ 0 \amp 1 \end{bmatrix} \amp \mapsto 1, \end{align*}
The matrix representation is
\begin{equation*} A_T = \begin{bmatrix} 1 \amp 0 \amp 0 \amp 1 \end{bmatrix} \end{equation*}
Note that it may be surprising that the matrix representation is just a row vector, however, recall that the trace maps from a \(2 \times 2\) matrix that can be represented as a column vector of length 4 to the reals, so the matrix representation should be a \(1 \times 4\) matrix.
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To verify the above results, recall that from Example 3.2.24 that the representation of a matrix is the unfolded matrix or in the \(2 \times 2\) case, that
\begin{equation*} \text{Rep}_B \biggl( \begin{bmatrix} a \amp b \\ c \amp d \end{bmatrix}\biggr) = \begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix} \end{equation*}
where \(B\) is the natural basis of \(\mathcal{M}_{2 \times 2}\text{.}\) So the matrix trace can be written as
\begin{equation*} \begin{bmatrix} 1 \amp 0 \amp 0 \amp 1 \end{bmatrix}\text{Rep}_B (A) = \begin{bmatrix} 1 \amp 0 \amp 0 \amp 1 \end{bmatrix}\begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix} = a + d \end{equation*}
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