Ch 1.8 - Introduction to Linear Transformations

Linear Dependence

• Definition of Linear Dependence Refresher: If a set of vectors ${V<em>1, V</em>2, V<em>3}$ is linearly dependent, it means there exist scalars $C</em>1$, $C<em>2$, and $C</em>3$, not all zero, such that $C<em>1V</em>1 + C<em>2V</em>2 + C<em>3V</em>3 = 0$.

• Extension to a Larger Set: If $C<em>1V</em>1 + C<em>2V</em>2 + C<em>3V</em>3 = 0$ holds, then adding $0 ext{ V}<em>4 = 0$ to both sides results in $C</em>1V<em>1 + C</em>2V<em>2 + C</em>3V<em>3 + 0V</em>4 = 0$. Since not all scalars (certainly $C<em>1, C</em>2, C<em>3$ and also $0$ for $V</em>4$) are zero, the set ${V<em>1, V</em>2, V<em>3, V</em>4}$ is also linearly dependent.

Introduction to Linear Transformations

Dynamic View of Matrix Multiplication

• Notation Equivalence: The difference between a matrix equation $Ax = b$ and its associated vector equation $x<em>1a</em>1 + ext{…} + x<em>na</em>n = b$ is primarily notational.

• Transformational Perspective: In various applications (e.g., computer graphics, signal processing), a matrix equation $Ax = b$ can be viewed dynamically. The matrix $A$ is considered an entity that "acts" on a vector $x$ through multiplication, resulting in a new vector called $Ax$.

  • Example: The equations $\begin{bmatrix} 1 &amp; 1 \ 0 &amp; 5 \ 1 &amp; 3 \end{bmatrix} \begin{bmatrix} 1 \ 3 \end{bmatrix} = \begin{bmatrix} 4 \ 15 \ 10 \end{bmatrix}$ and $\begin{bmatrix} 1 &amp; 1 \ 0 &amp; 5 \ 1 &amp; 3 \end{bmatrix} \begin{bmatrix} -3 \ 3 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \ 0 \end{bmatrix}$ show that multiplication by $A$ transforms $x = \begin{bmatrix} 1 \ 3 \end{bmatrix}$ into $b = \begin{bmatrix} 4 \ 15 \ 10 \end{bmatrix}$ and transforms $u = \begin{bmatrix} -3 \ 3 \end{bmatrix}$ into the zero vector.

• Solving $Ax = b$ from this Viewpoint: Finding all vectors $x$ in $R^n$ that are transformed into the vector $b$ in $R^m$ under the action of multiplication by $A$. For instance, in Figure 1, a vector from $R^3$ is transformed to a vector in $R^2$.

Definition of a Transformation (Function or Mapping)

• A transformation (or function or mapping) $T$ from $R^n$ to $R^m$ is a rule that assigns to each vector in $R^n$ a unique vector in $R^m$.

  • Domain: The set $R^n$ is called the domain of $T$.

  • Codomain: The set $R^m$ is called the codomain of $T$.

  • Notation: $T: R^n \rightarrow R^m$ indicates that $R^n$ is the domain and $R^m$ is the codomain.

  • Image: For a vector $x$ in $R^n$, the vector $T(x)$ in $R^m$ is called the image of $x$ (under the action of $T$).

  • Range: The set of all possible images ${T(x) \mid x \in R^n}$ is called the range of $T$. The range is a subset of the codomain. (See Figure 2 for an illustration of domain, codomain, and range).

• Importance of Dynamic View: This dynamic perspective of matrix-vector multiplication is crucial for understanding linear algebra concepts and building mathematical models of physical systems that evolve over time (e.g., in Sections 1.10, 4.8, and Chapter 5).

Matrix Transformations

• Definition: A matrix transformation is a mapping where for each $x$ in $R^n$, $T(x)$ is computed as $Ax$, where $A$ is an $m \times n$ matrix.

  • Simplified Notation: Sometimes denoted by $x \mapsto Ax$.

• Domain and Codomain of Matrix Transformations:

  • The domain of $T$ is $R^n$ when the matrix $A$ has $n$ columns.

  • The codomain of $T$ is $R^m$ when each column of $A$ has $m$ entries.

• Range of a Matrix Transformation: The range of $T$ is precisely the set of all linear combinations of the columns of $A$, because every image $T(x)$ is of the form $Ax$, which is by definition a linear combination of the columns of $A$.

Example 1: Image, Pre-image, Uniqueness, and Range Membership

Let $A = \begin{bmatrix} 1 &amp; -3 \ 3 &amp; 5 \ -1 &amp; 7 \end{bmatrix}$.
Define a transformation $T: R^2 \rightarrow R^3$ by $T(x) = Ax$, so that $T(\begin{bmatrix} x<em>1 \ x</em>2 \end{bmatrix}) = \begin{bmatrix} 1 &amp; -3 \ 3 &amp; 5 \ -1 &amp; 7 \end{bmatrix} \begin{bmatrix} x<em>1 \ x</em>2 \end{bmatrix} = \begin{bmatrix} x<em>1 - 3x</em>2 \ 3x<em>1 + 5x</em>2 \ -x<em>1 + 7x</em>2 \end{bmatrix}$.

a. Finding the image of $u = \begin{bmatrix} 3 \ 2 \end{bmatrix}$ under $T$ (i.e., $T(u)$):
$T(u) = A u = \begin{bmatrix} 1 &amp; -3 \ 3 &amp; 5 \ -1 &amp; 7 \end{bmatrix} \begin{bmatrix} 3 \ 2 \end{bmatrix} = \begin{bmatrix} (1)(3) + (-3)(2) \ (3)(3) + (5)(2) \ (-1)(3) + (7)(2) \end{bmatrix} = \begin{bmatrix} 3 - 6 \ 9 + 10 \ -3 + 14 \end{bmatrix} = \begin{bmatrix} -3 \ 19 \ 11 \end{bmatrix}$

b. Finding an $x$ in $R^2$ whose image under $T$ is $b = \begin{bmatrix} 3 \ 2 \ -5 \end{bmatrix}$ (i.e., solving $Ax = b$):
This requires solving the matrix equation $Ax = b$:
$\begin{bmatrix} 1 &amp; -3 \ 3 &amp; 5 \ -1 &amp; 7 \end{bmatrix} \begin{bmatrix} x<em>1 \ x</em>2 \end{bmatrix} = \begin{bmatrix} 3 \ 2 \ -5 \end{bmatrix}$
We form the augmented matrix and row reduce it:
$\begin{bmatrix} 1 &amp; -3 &amp; 3 \ 3 &amp; 5 &amp; 2 \ -1 &amp; 7 &amp; -5 \end{bmatrix} \sim \begin{bmatrix} 1 &amp; -3 &amp; 3 \ 0 &amp; 14 &amp; -7 \ 0 &amp; 4 &amp; -2 \end{bmatrix} \quad (R<em>2 - 3R</em>1 \rightarrow R<em>2, R</em>3 + R<em>1 \rightarrow R</em>3)$
$\sim \begin{bmatrix} 1 &amp; -3 &amp; 3 \ 0 &amp; 1 &amp; -.5 \ 0 &amp; 4 &amp; -2 \end{bmatrix} \quad (R<em>2 / 14 \rightarrow R</em>2)$
$\sim \begin{bmatrix} 1 &amp; -3 &amp; 3 \ 0 &amp; 1 &amp; -.5 \ 0 &amp; 0 &amp; 0 \end{bmatrix} \quad (R<em>3 - 4R</em>2 \rightarrow R<em>3)$ From the reduced echelon form: $x</em>2 = -.5$
$x<em>1 - 3x</em>2 = 3 \Rightarrow x<em>1 - 3(-.5) = 3 \Rightarrow x</em>1 + 1.5 = 3 \Rightarrow x_1 = 1.5$
Thus, the vector is $x = \begin{bmatrix} 1.5 \ -.5 \end{bmatrix}$. The image of this $x$ under $T$ is the given vector $b$.

c. Uniqueness of $x$: From the row reduction in part (b), the system $Ax=b$ has a unique solution (no free variables). Therefore, there is exactly one $x$ whose image under $T$ is $b$.

d. Determining if $c = \begin{bmatrix} 3 \ 2 \ -5 \end{bmatrix}$ (this seems to be a typo in the original text, as 'c' also represented the third column of the original augmented matrix in 'b'. Let's use the actual definition provided in 'd') $c = \begin{bmatrix} 3 \ 2 \ 8 \end{bmatrix}$ is in the range of $T$:
This asks if $c$ is the image of some $x$ in $R^2$, i.e., if $c = T(x)$ for some $x$. This means checking if the system $Ax=c$ is consistent. We row reduce the augmented matrix:
$\begin{bmatrix} 1 &amp; -3 &amp; 3 \ 3 &amp; 5 &amp; 2 \ -1 &amp; 7 &amp; 8 \end{bmatrix} \sim \begin{bmatrix} 1 &amp; -3 &amp; 3 \ 0 &amp; 14 &amp; -7 \ 0 &amp; 4 &amp; 11 \end{bmatrix} \quad (R<em>2 - 3R</em>1 \rightarrow R<em>2, R</em>3 + R<em>1 \rightarrow R</em>3)$
$\sim \begin{bmatrix} 1 &amp; -3 &amp; 3 \ 0 &amp; 1 &amp; -.5 \ 0 &amp; 4 &amp; 11 \end{bmatrix} \quad (R<em>2 / 14 \rightarrow R</em>2)$
$\sim \begin{bmatrix} 1 &amp; -3 &amp; 3 \ 0 &amp; 1 &amp; -.5 \ 0 &amp; 0 &amp; 13 \end{bmatrix} \quad (R<em>3 - 4R</em>2 \rightarrow R<em>3)$ The third row corresponds to the equation $0x</em>1 + 0x_2 = 13$, which simplifies to $0=13$. This is a contradiction, so the system is inconsistent. Therefore, $c$ is not in the range of $T$.

• Summary of Example 1: Example 1c is a uniqueness problem (is $b$ the image of a unique $x$?), and Example 1d is an existence problem (does there exist an $x$ whose image is $c$?).

Geometric Matrix Transformations

These examples illustrate the dynamic view of matrices as operators that transform vectors.

• Example 2: Projection Transformation

  • Let $A = \begin{bmatrix} 1 &amp; 0 &amp; 0 \ 0 &amp; 1 &amp; 0 \ 0 &amp; 0 &amp; 0 \end{bmatrix}$. The transformation $T(x) = Ax$ projects points in $R^3$ onto the $x<em>1x</em>2$-plane.

  • $T(\begin{bmatrix} x<em>1 \ x</em>2 \ x<em>3 \end{bmatrix}) = \begin{bmatrix} 1 &amp; 0 &amp; 0 \ 0 &amp; 1 &amp; 0 \ 0 &amp; 0 &amp; 0 \end{bmatrix} \begin{bmatrix} x</em>1 \ x<em>2 \ x</em>3 \end{bmatrix} = \begin{bmatrix} x<em>1 \ x</em>2 \ 0 \end{bmatrix}$.

• Example 3: Shear Transformation

  • Let $A = \begin{bmatrix} 1 &amp; 1 \ 0 &amp; 1 \end{bmatrix}$. The transformation $T: R^2 \rightarrow R^2$ defined by $T(x) = Ax$ is called a shear transformation.

  • It deforms a shape (e.g., a $2 \times 2$ square) into a sheared parallelogram (See Figure 4).

  • Key Idea: $T$ maps line segments onto line segments. By checking the images of the corners of the square, one can see the transformation.

    • Image of $\begin{bmatrix} 0 \ 2 \end{bmatrix}$ is $A \begin{bmatrix} 0 \ 2 \end{bmatrix} = \begin{bmatrix} 1 &amp; 1 \ 0 &amp; 1 \end{bmatrix} \begin{bmatrix} 0 \ 2 \end{bmatrix} = \begin{bmatrix} 2 \ 2 \end{bmatrix}$.

    • Image of $\begin{bmatrix} 2 \ 2 \end{bmatrix}$ is $A \begin{bmatrix} 2 \ 2 \end{bmatrix} = \begin{bmatrix} 1 &amp; 1 \ 0 &amp; 1 \end{bmatrix} \begin{bmatrix} 2 \ 2 \end{bmatrix} = \begin{bmatrix} 4 \ 2 \end{bmatrix}$.

  • Geometric Effect: The transformation deforms the square as if its top were pushed to the right while the base is held fixed. Shear transformations are observed in physics, geology, and crystallography.

Linear Transformations

Formal Definition

• Theorem 5 in Section 1.4 established properties for matrix transformations: $A(u+v) = Au+Av$ and $A(cu) = cAu$. These properties define the most important class of transformations in linear algebra.

• A transformation (or mapping) $T$ is linear if:

  • (i) Additivity: $T(u+v) = T(u) + T(v)$ for all vectors $u, v$ in the domain of $T$.

  • (ii) Homogeneity of Degree 1: $T(cu) = cT(u)$ for all scalars $c$ and all vectors $u$ in the domain of $T$.

Important Notes on Linear Transformations

• Matrix Transformations are Linear: Every matrix transformation ($T(x)=Ax$) is a linear transformation. (However, not all linear transformations are matrix transformations; this will be explored in Chapters 4 and 5).

• Preservation of Operations: Linear transformations preserve the operations of vector addition and scalar multiplication. Property (i) means that applying $T$ after adding vectors $u$ and $v$ is the same as applying $T$ to $u$ and $v$ separately and then adding their images.

Useful Facts derived from Linearity

If $T$ is a linear transformation, then:

1. Image of the Zero Vector: $T(0) = 0$

   • Proof: From condition (ii), $T(0) = T(0u) = 0T(u) = 0$ (where $0$ on the left is the scalar, and $0$ on the right is the zero vector in the codomain).

2. Linear Combination Property: $T(cu+dv) = cT(u) + dT(v)$

   • Proof: This property requires both (i) and (ii):
     $T(cu+dv) = T(cu) + T(dv) \quad \text{(by property (i))}$
     $= cT(u) + dT(v) \quad \text{(by property (ii))}$

   • Implication: If a transformation satisfies $T(cu+dv) = cT(u) + dT(v)$ for all $u, v$ and $c, d$, it must be linear. (Setting $c=d=1$ gives additivity, and setting $d=0$ gives scalar multiplication property).

• Superposition Principle (Generalization): Repeated application of the linear combination property leads to a useful generalization: $T(c<em>1v</em>1 + c<em>2v</em>2 + \text{…} + c<em>pv</em>p) = c<em>1T(v</em>1) + c<em>2T(v</em>2) + \text{…} + c<em>pT(v</em>p)$

  • Significance in Science/Engineering: This is widely known as the superposition principle in engineering and physics. If $v<em>1, \text{…}, v</em>p$ are inputs (signals) to a system, and $T(v<em>1), \text{…}, T(v</em>p)$ are the corresponding responses, then a system satisfies the superposition principle if its response to a linear combination of inputs is the same linear combination of the responses to the individual inputs. This concept is fundamental in many fields.

Examples of Linear Transformations

Example 4: Contraction and Dilation Transformation

• Given a scalar $r$, define $T: R^2 \rightarrow R^2$ by $T(x) = rx$.

  • This is a contraction when 0 < r < 1 (shrinks vectors).

  • This is a dilation when r > 1 (stretches vectors).

• Proof of Linearity (for $r=3$):
  Let $u, v$ be vectors in $R^2$ and $c, d$ be scalars.
  $T(cu+dv) = r(cu+dv) \quad \text{(by definition of } T)$
  $= c(ru) + d(rv) \quad \text{(by vector arithmetic)}$
  $= cT(u) + dT(v) \quad \text{(by definition of } T)$
  Since $T(cu+dv) = cT(u)+dT(v)$, the transformation $T$ is linear. (See Figure 5 for a dilation example).

Example 5: Rotation Transformation

• Define a linear transformation $T: R^2 \rightarrow R^2$ by $T(x) = \begin{bmatrix} 0 &amp; -1 \ 1 &amp; 0 \end{bmatrix}x$.

• Let $u = \begin{bmatrix} 4 \ 1 \end{bmatrix}$, $v = \begin{bmatrix} 1 \ 3 \end{bmatrix}$, and $u+v = \begin{bmatrix} 5 \ 4 \end{bmatrix}$.

  • $T(u) = \begin{bmatrix} 0 &amp; -1 \ 1 &amp; 0 \end{bmatrix} \begin{bmatrix} 4 \ 1 \end{bmatrix} = \begin{bmatrix} -1 \ 4 \end{bmatrix}$

  • $T(v) = \begin{bmatrix} 0 &amp; -1 \ 1 &amp; 0 \end{bmatrix} \begin{bmatrix} 1 \ 3 \end{bmatrix} = \begin{bmatrix} -3 \ 1 \end{bmatrix}$

  • $T(u+v) = \begin{bmatrix} 0 &amp; -1 \ 1 &amp; 0 \end{bmatrix} \begin{bmatrix} 5 \ 4 \end{bmatrix} = \begin{bmatrix} -4 \ 5 \end{bmatrix}$

• Verification of Additivity: Notice that $T(u)+T(v) = \begin{bmatrix} -1 \ 4 \end{bmatrix} + \begin{bmatrix} -3 \ 1 \end{bmatrix} = \begin{bmatrix} -4 \ 5 \end{bmatrix}$, which is indeed equal to $T(u+v)$.

• Geometric Interpretation: This transformation rotates vectors (e.g., $u, v, ext{ and } u+v$) counterclockwise about the origin through $90^\circ$ (See Figure 6). It transforms the entire parallelogram determined by $u$ and $v$ into the parallelogram determined by $T(u)$ and $T(v)$.

Example 6: Cost Transformation for Manufacturing Production

• Scenario: A company manufactures two products, B and C.

• Unit Cost Matrix $U$: From Section 1.3, we construct a