Skip to main content

Linear Algebra For Undergraduate Students

Yilan(Peter) Tan

Section 6.4 Properties of Linear Transformations

In this section, we study important properties of linear transformations that determine their behavior: whether they are one-to-one, onto, or invertible. These concepts help us understand when and how linear transformations can be "reversed" and how they can be combined.
🔗

Subsection 6.4.1 One-to-One and Onto Transformations

Definition 6.4.1. One-to-One and Onto.

Let \(T: V \rightarrow W\) be a linear transformation.
🔗
  1. \(T\) is called one-to-one (or injective) if different vectors in \(V\) are mapped to different vectors in \(W\text{.}\) That is,
    \begin{equation*} T(\mathbf{u}) = T(\mathbf{v}) \implies \mathbf{u} = \mathbf{v} \end{equation*}
    for all \(\mathbf{u}, \mathbf{v} \in V\text{.}\)
    🔗
    Equivalently, \(T\) is one-to-one if and only if \(T(\mathbf{v}) = \mathbf{0}\) implies \(\mathbf{v} = \mathbf{0}\text{.}\)
    🔗
    🔗
  2. \(T\) is called onto (or surjective) if every vector in \(W\) is the image of at least one vector in \(V\text{.}\) That is,
    \begin{equation*} \text{for every } \mathbf{w} \in W, \text{ there exists } \mathbf{v} \in V \text{ such that } T(\mathbf{v}) = \mathbf{w}. \end{equation*}
    🔗
    Equivalently, \(T\) is onto if and only if \(\operatorname{range}(T) = W\text{.}\)
    🔗
    🔗
🔗
🔗

Geometric Interpretation.

  • One-to-One: Think of \(T\) as a transformation that never collapses distinct points together. If two points are different before the transformation, they remain different after. Geometrically, \(T\) preserves distinctness — no "squashing" occurs.
    🔗
    Example: A rotation or reflection in \(\mathbb{R}^2\) is one-to-one because distinct points don’t overlap.
    🔗
    🔗
  • Onto: Think of \(T\) as a transformation that fills up the entire target space \(W\text{.}\) Every point in \(W\) is "hit" by the transformation. Geometrically, \(T\) covers all of \(W\) — no gaps remain.
    🔗
    Example: A projection from \(\mathbb{R}^3\) to \(\mathbb{R}^2\) can be onto because it covers the entire plane.
    🔗
    🔗
🔗
🔗
described in detail following the image
Illustrates the difference between one-to-one and onto transformations using mapping diagrams.
🔗
Figure 6.4.2. Visual Comparison: One-to-One vs Onto
🔗

Example 6.4.3. A Transformation Neither One-to-One Nor Onto.

Consider the linear transformation \(T:\mathbb{R}^{4}\rightarrow \mathbb{R}^{4}\) from Example 6.1.4. We showed that:
  • \(\displaystyle \operatorname{ker}(T) = \operatorname{span}\left\{\begin{pmatrix}-2\\2\\-1\\1\end{pmatrix}\right\} \neq \{\mathbf{0}\}\)
    🔗
    🔗
  • \(\operatorname{range}(T) = \operatorname{span}\left\{\begin{pmatrix}1\\0\\-1\\0\end{pmatrix}, \begin{pmatrix}1\\1\\0\\2\end{pmatrix}, \begin{pmatrix}0\\3\\1\\2\end{pmatrix}\right\}\text{,}\) which is a proper subspace of \(\mathbb{R}^4\)
    🔗
    🔗
🔗
Therefore:
  • \(T\) is not one-to-one because \(\operatorname{ker}(T) \neq \{\mathbf{0}\}\text{.}\) For instance, \(T\begin{pmatrix}-2\\2\\-1\\1\end{pmatrix} = \mathbf{0} = T(\mathbf{0})\text{,}\) but \(\begin{pmatrix}-2\\2\\-1\\1\end{pmatrix} \neq \mathbf{0}\text{.}\)
    🔗
    🔗
  • \(T\) is not onto because \(\operatorname{range}(T) \neq \mathbb{R}^4\text{.}\) In fact, \(\dim(\operatorname{range}(T)) = 3 < 4 = \dim(\mathbb{R}^4)\text{,}\) so there exist vectors in \(\mathbb{R}^4\) that are not in the range of \(T\text{.}\)
    🔗
    🔗
🔗
🔗

Proof.

(\(\Rightarrow\)) Assume \(T\) is one-to-one. We know \(\mathbf{0} \in \operatorname{ker}(T)\) because \(T(\mathbf{0}) = \mathbf{0}\text{.}\) Suppose \(\mathbf{v} \in \operatorname{ker}(T)\text{.}\) Then \(T(\mathbf{v}) = \mathbf{0} = T(\mathbf{0})\text{.}\) Since \(T\) is one-to-one, we must have \(\mathbf{v} = \mathbf{0}\text{.}\) Therefore, \(\operatorname{ker}(T) = \{\mathbf{0}\}\text{.}\)
🔗
(\(\Leftarrow\)) Assume \(\operatorname{ker}(T) = \{\mathbf{0}\}\text{.}\) Suppose \(T(\mathbf{u}) = T(\mathbf{v})\text{.}\) Then
\begin{equation*} T(\mathbf{u}) - T(\mathbf{v}) = \mathbf{0} \implies T(\mathbf{u} - \mathbf{v}) = \mathbf{0} \implies \mathbf{u} - \mathbf{v} \in \operatorname{ker}(T). \end{equation*}
Since \(\operatorname{ker}(T) = \{\mathbf{0}\}\text{,}\) we have \(\mathbf{u} - \mathbf{v} = \mathbf{0}\text{,}\) so \(\mathbf{u} = \mathbf{v}\text{.}\) Therefore, \(T\) is one-to-one.
🔗
🔗

Example 6.4.5. Checking One-to-One.

Determine whether the linear transformation \(T: \mathbb{R}^3 \rightarrow \mathbb{R}^3\) defined by
\begin{equation*} T\left(\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}\right) = \begin{bmatrix} x_1 + x_2 \\ x_2 + x_3 \\ x_1 + x_3 \end{bmatrix} \end{equation*}
is one-to-one.
🔗
Solution.
By Theorem 6.4.4, \(T\) is one-to-one if and only if \(\operatorname{ker}(T) = \{\mathbf{0}\}\text{.}\) We need to solve \(T(\mathbf{x}) = \mathbf{0}\text{:}\)
\begin{equation*} \begin{bmatrix} x_1 + x_2 \\ x_2 + x_3 \\ x_1 + x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \end{equation*}
This gives the system:
\begin{align*} x_1 + x_2 \amp= 0\\ x_2 + x_3 \amp= 0\\ x_1 + x_3 \amp= 0 \end{align*}
🔗
The augmented matrix for this system is:
\begin{equation*} \left[\begin{array}{ccc|c} 1 \amp 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp 1 \amp 0 \\ 1 \amp 0 \amp 1 \amp 0 \end{array}\right]\xrightarrow{\operatorname{RREF}}\left[\begin{array}{ccc|c} 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp 0 \end{array}\right] \end{equation*}
Therefore, the unique solution is \(x_1 = x_2 = x_3 = 0\text{.}\)
🔗
Since \(\operatorname{ker}(T) = \{\mathbf{0}\}\text{,}\) the transformation \(T\) is one-to-one.
🔗
🔗
🔗

Proof.

By definition, \(T\) is onto if and only if \(\operatorname{range}(T) = W\text{.}\) Since \(W\) is finite-dimensional, this occurs if and only if \(\dim(\operatorname{range}(T)) = \dim(W)\text{.}\)
🔗
🔗

Example 6.4.7. The Transformation from Example 6.4.5 is Also Onto.

Show that the linear transformation \(T: \mathbb{R}^3 \rightarrow \mathbb{R}^3\) from Example 6.4.5 is also onto.
🔗
Solution.
By Theorem 6.4.6, \(T\) is onto if and only if \(\dim(\operatorname{range}(T)) = \dim(\mathbb{R}^3) = 3\text{.}\) The standard matrix for \(T\) is
\begin{equation*} A = \begin{bmatrix} 1 \amp 1 \amp 0 \\ 0 \amp 1 \amp 1 \\ 1 \amp 0 \amp 1 \end{bmatrix}. \end{equation*}
🔗
In Example 6.4.5, we showed that \(\dim(\operatorname{ker}(T)) = 0\text{.}\) By the Rank-Nullity Theorem,
\begin{equation*} \dim(\operatorname{range}(T)) = \dim(\mathbb{R}^3) - \dim(\operatorname{ker}(T)) = 3 - 0 = 3. \end{equation*}
🔗
Since \(\dim(\operatorname{range}(T)) = 3 = \dim(\mathbb{R}^3)\text{,}\) the transformation \(T\) is onto.
🔗
🔗
🔗

Example 6.4.8. Checking Onto.

Determine whether the linear transformation \(T: \mathbb{R}^3 \rightarrow \mathbb{R}^2\) defined by
\begin{equation*} T\left(\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}\right) = \begin{bmatrix} x_1 + x_2 \\ x_2 + x_3 \end{bmatrix} \end{equation*}
is onto.
🔗
Solution.
By Theorem 6.4.6, \(T\) is onto if and only if \(\dim(\operatorname{range}(T)) = \dim(\mathbb{R}^2) = 2\text{.}\) The standard matrix for \(T\) is
\begin{equation*} A = \begin{bmatrix} 1 \amp 1 \amp 0 \\ 0 \amp 1 \amp 1 \end{bmatrix}. \end{equation*}
Since \(\operatorname{range}(T) = \operatorname{Col}(A)\text{,}\) we need to find the dimension of the column space of \(A\text{.}\)
🔗
Row reducing \(A\) to row echelon form:
\begin{equation*} \begin{bmatrix} 1 \amp 1 \amp 0 \\ 0 \amp 1 \amp 1 \end{bmatrix} \end{equation*}
The matrix is already in row echelon form with two pivot positions (in columns 1 and 2). Therefore, \(\dim(\operatorname{Col}(A)) = 2\text{.}\)
🔗
Since \(\dim(\operatorname{range}(T)) = 2 = \dim(\mathbb{R}^2)\text{,}\) the transformation \(T\) is onto.
🔗
🔗
🔗
🔗

Subsection 6.4.2 Invertible Linear Transformations

Definition 6.4.9. Invertible Transformation.

A linear transformation \(T: V \rightarrow W\) is called invertible if there exists a linear transformation \(S: W \rightarrow V\) such that
\begin{equation*} S \circ T = I_V \quad \text{and} \quad T \circ S = I_W, \end{equation*}
where \(I_V\) and \(I_W\) are the identity transformations on \(V\) and \(W\text{,}\) respectively.
🔗
The transformation \(S\) is called the inverse of \(T\text{,}\) denoted \(T^{-1}\text{.}\)
🔗
🔗

Proof.

(\(\Rightarrow\)) Assume \(T\) is invertible with inverse \(T^{-1}\text{.}\)
🔗
One-to-one: If \(T(\mathbf{u}) = T(\mathbf{v})\text{,}\) then applying \(T^{-1}\) to both sides gives
\begin{equation*} T^{-1}(T(\mathbf{u})) = T^{-1}(T(\mathbf{v})) \implies \mathbf{u} = \mathbf{v}. \end{equation*}
So \(T\) is one-to-one.
🔗
Onto: For any \(\mathbf{w} \in W\text{,}\) let \(\mathbf{v} = T^{-1}(\mathbf{w})\text{.}\) Then
\begin{equation*} T(\mathbf{v}) = T(T^{-1}(\mathbf{w})) = \mathbf{w}. \end{equation*}
So \(T\) is onto.
🔗
(\(\Leftarrow\)) Assume \(T\) is both one-to-one and onto. For each \(\mathbf{w} \in W\text{,}\) since \(T\) is onto, there exists at least one \(\mathbf{v} \in V\) such that \(T(\mathbf{v}) = \mathbf{w}\text{.}\) Since \(T\) is one-to-one, this \(\mathbf{v}\) is unique. Define \(T^{-1}(\mathbf{w}) = \mathbf{v}\text{.}\) One can verify that \(T^{-1}\) is linear and satisfies the inverse properties.
🔗
🔗

Proof.

By Theorem 6.4.10, \(T\) is invertible if and only if \(T\) is both one-to-one and onto.
🔗
(\(\Rightarrow\)) Assume \(T\) is invertible. Then \(T\) is one-to-one, so \(\operatorname{ker}(T) = \{\mathbf{0}\}\text{.}\) This means the only solution to \([T]_\alpha^\beta [\mathbf{x}]_\alpha = \mathbf{0}\) is \([\mathbf{x}]_\alpha = \mathbf{0}\text{.}\) Therefore, the columns of \([T]_\alpha^\beta\) are linearly independent. Since \([T]_\alpha^\beta\) is an \(n \times n\) matrix with \(n\) linearly independent columns, it is invertible.
🔗
(\(\Leftarrow\)) Assume \([T]_\alpha^\beta\) is invertible. Then the columns of \([T]_\alpha^\beta\) are linearly independent, which implies \(\operatorname{ker}(T) = \{\mathbf{0}\}\text{,}\) so \(T\) is one-to-one. By the Rank-Nullity Theorem,
\begin{equation*} \dim(\operatorname{range}(T)) = \dim(V) - \dim(\operatorname{ker}(T)) = n - 0 = n = \dim(W). \end{equation*}
By Theorem 6.4.6, \(T\) is onto. Since \(T\) is both one-to-one and onto, \(T\) is invertible.
🔗
🔗

Example 6.4.12.

Let \(T: \mathbb{R}^3 \rightarrow \mathbb{R}^3\) be a linear transformation such that
\begin{equation*} T\left(\begin{bmatrix} x \\ y \\ z \end{bmatrix}\right) = \begin{bmatrix}x-2y+z \\ 2x+2y-8z \\ 5x+y-5z\end{bmatrix}. \end{equation*}
🔗
Question: Is \(T\) invertible? If so, find \(T^{-1}\text{.}\)
🔗
Hint.
Use Theorem 6.4.11 to check if the standard matrix \(A\) is invertible. To find \(T^{-1}\text{,}\) compute the reduced row echelon form of the augmented matrix \([A \mid I]\text{,}\) where \(I\) is the \(3 \times 3\) identity matrix. If \(A\) is invertible, the RREF will be \([I \mid A^{-1}]\text{.}\)
🔗
🔗
Solution.
The standard matrix for \(T\) is
\begin{equation*} A = \begin{bmatrix} 1 \amp -2 \amp 1 \\ 2 \amp 2 \amp -8 \\ 5 \amp 1 \amp -5 \end{bmatrix}. \end{equation*}
By Theorem 6.4.11, \(T\) is invertible if and only if \(A\) is invertible.
🔗
We use SageMath to compute the reduced row echelon form of \([A \mid I]\text{:}\)
🔗
From the output, we see that the RREF is
\begin{equation*} \left[\begin{array}{ccc|ccc} 1 \amp 0 \amp 0 \amp -1 \amp -\frac{9}{2} \amp \frac{7}{2} \\ 0 \amp 1 \amp 0 \amp 15 \amp -5 \amp 5 \\ 0 \amp 0 \amp 1 \amp 4 \amp -\frac{11}{2} \amp \frac{3}{2} \end{array}\right]. \end{equation*}
Since the left side is the identity matrix, \(A\) is invertible. Therefore, \(T\) is invertible with inverse matrix
\begin{equation*} A^{-1} = \begin{bmatrix} -1 \amp -\frac{9}{2} \amp \frac{7}{2} \\ 15 \amp -5 \amp 5 \\ 4 \amp -\frac{11}{2} \amp \frac{3}{2} \end{bmatrix}. \end{equation*}
🔗
Thus, the inverse transformation is
\begin{equation*} T^{-1}\left(\begin{bmatrix} x \\ y \\ z \end{bmatrix}\right) = \begin{bmatrix} -x - \frac{9}{2}y + \frac{7}{2}z \\ 15x - 5y + 5z \\ 4x - \frac{11}{2}y + \frac{3}{2}z \end{bmatrix}. \end{equation*}
🔗
🔗
🔗

Checkpoint 6.4.13.

Let \(T: \mathbb{P}_2(x) \rightarrow \mathbb{P}_2(x)\) be a linear transformation such that
\begin{equation*} T\left(a+bx+cx^2\right) = (a-2b+c)+(2a+2b-8c)x+(5a+b-5c)x^2. \end{equation*}
🔗
Question: Is \(T\) invertible? If so, find \(T^{-1}(a+bx+cx^2)\text{.}\)
🔗
🔗

Example 6.4.14. Rotation is Invertible.

Consider the rotation transformation \(T: \mathbb{R}^2 \rightarrow \mathbb{R}^2\) that rotates vectors counterclockwise by \(\theta\) radians:
\begin{equation*} T\left(\begin{bmatrix} x \\ y \end{bmatrix}\right) = \begin{bmatrix} \cos\theta \amp -\sin\theta \\ \sin\theta \amp \cos\theta \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}. \end{equation*}
Show that \(T\) is invertible and find \(T^{-1}\text{.}\)
🔗
Solution.
The standard matrix is \(A = \begin{bmatrix} \cos\theta \amp -\sin\theta \\ \sin\theta \amp \cos\theta \end{bmatrix}\text{.}\) We compute
\begin{equation*} \det(A) = \cos^2\theta + \sin^2\theta = 1 \neq 0. \end{equation*}
Since \(\det(A) \neq 0\text{,}\) the matrix \(A\) is invertible, so \(T\) is invertible.
🔗
The inverse transformation rotates by \(-\theta\) (clockwise):
\begin{equation*} T^{-1}\left(\begin{bmatrix} x \\ y \end{bmatrix}\right) = \begin{bmatrix} \cos\theta \amp \sin\theta \\ -\sin\theta \amp \cos\theta \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}. \end{equation*}
🔗
Geometric interpretation: To undo a counterclockwise rotation by \(\theta\text{,}\) we rotate clockwise by \(\theta\text{.}\)
🔗
🔗
🔗

Example 6.4.15. Projection is Not Invertible.

Consider the projection \(T: \mathbb{R}^3 \rightarrow \mathbb{R}^3\) onto the \(xy\)-plane:
\begin{equation*} T\left(\begin{bmatrix} x \\ y \\ z \end{bmatrix}\right) = \begin{bmatrix} x \\ y \\ 0 \end{bmatrix}. \end{equation*}
Show that \(T\) is not invertible.
🔗
Solution.
To check if \(T\) is one-to-one, we find \(\operatorname{ker}(T)\text{:}\)
\begin{equation*} T\left(\begin{bmatrix} x \\ y \\ z \end{bmatrix}\right) = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \implies \begin{bmatrix} x \\ y \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \implies x = y = 0. \end{equation*}
So \(\operatorname{ker}(T) = \left\{\begin{bmatrix} 0 \\ 0 \\ z \end{bmatrix} : z \in \mathbb{R}\right\}\text{,}\) which is the \(z\)-axis.
🔗
Since \(\operatorname{ker}(T) \neq \{\mathbf{0}\}\text{,}\) the transformation \(T\) is not one-to-one, hence not invertible.
🔗
Geometric interpretation: All points along a vertical line (parallel to the \(z\)-axis) project to the same point in the \(xy\)-plane. Information about the \(z\)-coordinate is lost, so the transformation cannot be reversed.
🔗
🔗
🔗

Insight 6.4.16. The Power of Matrix Representation.

Throughout this section, we have seen that the matrix representation of a linear transformation \(T: \mathbb{R}^n \rightarrow \mathbb{R}^m\) plays a crucial role in determining its properties. Once we identify the standard matrix \(A\) of \(T\text{,}\) many fundamental questions about the transformation reduce to questions about the matrix:
🔗
  • \(T\) is one-to-one if and only if the columns of \(A\) are linearly independent, which we can verify by checking that \(\operatorname{rank}(A) = n\text{.}\)
    🔗
    🔗
  • \(T\) is onto if and only if the columns of \(A\) span \(\mathbb{R}^m\text{,}\) which means \(\operatorname{rank}(A) = m\text{.}\)
    🔗
    🔗
  • \(T\) is invertible if and only if \(A\) is an invertible matrix, which for square matrices means \(\det(A) \neq 0\text{.}\)
    🔗
    🔗
  • The kernel of \(T\) is the null space of \(A\text{,}\) found by solving the homogeneous system \(A\mathbf{x} = \mathbf{0}\text{.}\)
    🔗
    🔗
  • The range of \(T\) is the column space of \(A\text{,}\) which is spanned by the pivot columns of \(A\text{.}\)
    🔗
    🔗
🔗
This connection between transformations and matrices is profound: it transforms abstract questions about functions between vector spaces into concrete computational problems involving systems of linear equations, row reduction, and matrix algebra. The matrix is not just a convenient representation—it is the key that unlocks our ability to analyze and compute with linear transformations systematically.
🔗
🔗
🔗

Subsection 6.4.3 Composition of Linear Transformations

Definition 6.4.17. Composition of Transformations.

Let \(T: U \rightarrow V\) and \(S: V \rightarrow W\) be linear transformations. The composition of \(S\) and \(T\text{,}\) denoted \(S \circ T\text{,}\) is the transformation from \(U\) to \(W\) defined by
\begin{equation*} (S \circ T)(\mathbf{u}) = S(T(\mathbf{u})) \end{equation*}
for all \(\mathbf{u} \in U\text{.}\)
🔗
🔗

Proof.

We need to verify the two properties of linearity. Let \(\mathbf{u}_1, \mathbf{u}_2 \in U\) and \(c \in \mathbb{R}\text{.}\)
🔗
Additivity:
\begin{align*} (S \circ T)(\mathbf{u}_1 + \mathbf{u}_2) \amp= S(T(\mathbf{u}_1 + \mathbf{u}_2))\\ \amp= S(T(\mathbf{u}_1) + T(\mathbf{u}_2)) \quad \text{(since } T \text{ is linear)}\\ \amp= S(T(\mathbf{u}_1)) + S(T(\mathbf{u}_2)) \quad \text{(since } S \text{ is linear)}\\ \amp= (S \circ T)(\mathbf{u}_1) + (S \circ T)(\mathbf{u}_2) \end{align*}
🔗
Homogeneity:
\begin{align*} (S \circ T)(c\mathbf{u}_1) \amp= S(T(c\mathbf{u}_1))\\ \amp= S(cT(\mathbf{u}_1)) \quad \text{(since } T \text{ is linear)}\\ \amp= cS(T(\mathbf{u}_1)) \quad \text{(since } S \text{ is linear)}\\ \amp= c(S \circ T)(\mathbf{u}_1) \end{align*}
🔗
🔗

Proof.

For any \(\mathbf{x} \in \mathbb{R}^n\text{,}\)
\begin{align*} (S \circ T)(\mathbf{x}) \amp= S(T(\mathbf{x}))\\ \amp= S(A\mathbf{x})\\ \amp= B(A\mathbf{x})\\ \amp= (BA)\mathbf{x} \end{align*}
Therefore, the standard matrix of \(S \circ T\) is \(BA\text{.}\)
🔗
🔗

Example 6.4.20. Composition: Rotation followed by Scaling.

Let \(T: \mathbb{R}^2 \rightarrow \mathbb{R}^2\) be the rotation by \(90°\) counterclockwise, and let \(S: \mathbb{R}^2 \rightarrow \mathbb{R}^2\) be the scaling that doubles all coordinates. Find the standard matrix of \(S \circ T\text{.}\)
🔗
Solution.
The standard matrices are:
\begin{equation*} A = \begin{bmatrix} 0 \amp -1 \\ 1 \amp 0 \end{bmatrix} \quad \text{(rotation by } 90°\text{)} \end{equation*}
\begin{equation*} B = \begin{bmatrix} 2 \amp 0 \\ 0 \amp 2 \end{bmatrix} \quad \text{(scaling by 2)} \end{equation*}
🔗
By Theorem 6.4.19, the standard matrix of \(S \circ T\) is
\begin{equation*} BA = \begin{bmatrix} 2 \amp 0 \\ 0 \amp 2 \end{bmatrix} \begin{bmatrix} 0 \amp -1 \\ 1 \amp 0 \end{bmatrix} = \begin{bmatrix} 0 \amp -2 \\ 2 \amp 0 \end{bmatrix}. \end{equation*}
🔗
Interpretation: First rotate by \(90°\text{,}\) then scale by 2. For example, \((S \circ T)\left(\begin{bmatrix} 1 \\ 0 \end{bmatrix}\right) = \begin{bmatrix} 0 \\ 2 \end{bmatrix}\text{.}\)
🔗
🔗
🔗

Warning 6.4.21. Order Matters!

Composition of transformations is not commutative: in general, \(S \circ T \neq T \circ S\text{.}\) The order in which transformations are applied matters!
🔗
For example, in Example 6.4.20, if we compute \(T \circ S\) (scale first, then rotate), we get
\begin{equation*} AB = \begin{bmatrix} 0 \amp -1 \\ 1 \amp 0 \end{bmatrix} \begin{bmatrix} 2 \amp 0 \\ 0 \amp 2 \end{bmatrix} = \begin{bmatrix} 0 \amp -2 \\ 2 \amp 0 \end{bmatrix}, \end{equation*}
1 which happens to be the same. But this is not always the case!
🔗
🔗
🔗
🔗
PrevTopNext