2.1 Identify linear transformations and properties

Linear transformations are the backbone of vector space operations. They preserve addition and scalar multiplication, making them crucial for understanding how vectors behave under different mappings. This topic dives into their properties, types, and representations.

Identifying linear transformations involves checking if they maintain vector addition and scalar multiplication. We'll explore how to spot these transformations, their key characteristics like injectivity and surjectivity, and how to represent them using matrices.

Linear Transformations and Properties

Definition and Key Properties

• A linear transformation is a function $T$ from a vector space $V$ to a vector space $W$ that preserves the operations of vector addition and scalar multiplication
• For any vectors $u$ and $v$ in $V$ and any scalar $c$, a linear transformation $T$ satisfies:
  • $T(u + v) = T(u) + T(v)$ (preserves vector addition)
  • $T(cu) = cT(u)$ (preserves scalar multiplication)
• The kernel (or null space) of a linear transformation $T$ is the set of all vectors $v$ in $V$ such that $T(v) = 0$
  • Example: For the linear transformation $T(x, y) = (x + y, x - y)$, the kernel is $\{(x, x) : x \in \mathbb{R}\}$
• The range (or image) of a linear transformation $T$ is the set of all vectors $w$ in $W$ such that $w = T(v)$ for some vector $v$ in $V$
  • Example: For the linear transformation $T(x, y) = (x + y, x - y)$, the range is $\mathbb{R}^2$

Injectivity, Surjectivity, and Bijectivity

• A linear transformation is injective (one-to-one) if and only if its kernel is $\{0\}$
  • Example: The linear transformation $T(x, y) = (x, y, 0)$ from $\mathbb{R}^2$ to $\mathbb{R}^3$ is injective
• A linear transformation is surjective (onto) if and only if its range is equal to the codomain $W$
  • Example: The linear transformation $T(x, y, z) = (x + y, z)$ from $\mathbb{R}^3$ to $\mathbb{R}^2$ is surjective
• A linear transformation is bijective (one-to-one and onto) if and only if it is both injective and surjective
  • Example: The linear transformation $T(x, y) = (2x, 3y)$ from $\mathbb{R}^2$ to $\mathbb{R}^2$ is bijective

Identifying Linear Transformations

Preserving Vector Addition and Scalar Multiplication

• To determine if a transformation $T$ is linear, check if it preserves vector addition and scalar multiplication
• For vector addition, verify that $T(u + v) = T(u) + T(v)$ for any vectors $u$ and $v$ in the domain
  • Example: For $T(x, y) = (x^2, y)$, $T((1, 1) + (2, 3)) \neq T(1, 1) + T(2, 3)$, so $T$ is not linear
• For scalar multiplication, verify that $T(cu) = cT(u)$ for any vector $u$ in the domain and any scalar $c$
  • Example: For $T(x, y) = (2x, 3y)$, $T(2(1, 1)) = 2T(1, 1)$, so $T$ preserves scalar multiplication

Conditions for Linearity

• If both properties hold for all vectors in the domain and all scalars, then the transformation is linear
• If either property fails for any vectors or scalars, then the transformation is not linear
  • Example: The transformation $T(x, y) = (xy, x + y)$ is not linear because $T((1, 1) + (2, 3)) \neq T(1, 1) + T(2, 3)$

Composition of Linear Transformations

Definition and Properties

• The composition of two linear transformations $S$ and $T$, denoted $S \circ T$, is defined by $(S \circ T)(v) = S(T(v))$ for any vector $v$ in the domain of $T$
• The composition of linear transformations is associative: $(R \circ S) \circ T = R \circ (S \circ T)$ for any linear transformations $R$, $S$, and $T$
  • Example: Let $R(x, y) = (x + y, x - y)$, $S(x, y) = (2x, 3y)$, and $T(x, y) = (x, 0)$. Then, $((R \circ S) \circ T)(x, y) = (R \circ S)(x, 0) = R(2x, 0) = (2x, 2x) = (R \circ (S \circ T))(x, y)$
• The composition of linear transformations is distributive over addition: $(S + T) \circ R = (S \circ R) + (T \circ R)$ and $R \circ (S + T) = (R \circ S) + (R \circ T)$ for any linear transformations $R$, $S$, and $T$
  • Example: Let $R(x, y) = (x, y)$, $S(x, y) = (x + y, 0)$, and $T(x, y) = (0, x - y)$. Then, $((S + T) \circ R)(x, y) = (S + T)(x, y) = (x + y, x - y) = (S \circ R)(x, y) + (T \circ R)(x, y)$

Kernel and Range of Compositions

• The kernel of the composition $S \circ T$ contains the kernel of $T$, and the range of $S \circ T$ is a subset of the range of $S$
  • Example: Let $S(x, y) = (x + y, 0)$ and $T(x, y) = (x, x)$. The kernel of $T$ is $\{(x, x) : x \in \mathbb{R}\}$, which is contained in the kernel of $S \circ T$. The range of $S \circ T$ is $\{(2x, 0) : x \in \mathbb{R}\}$, which is a subset of the range of $S$
• If $S$ and $T$ are both injective, then $S \circ T$ is injective. If $S$ and $T$ are both surjective, then $S \circ T$ is surjective
  • Example: Let $S(x, y) = (2x, y)$ and $T(x, y) = (x, 2y)$. Both $S$ and $T$ are injective, so $S \circ T$ is also injective. However, neither $S$ nor $T$ is surjective, so $S \circ T$ is not surjective

Matrix Representation of Linear Transformations

Calculating the Matrix Representation

• Choose ordered bases for the domain and codomain of the linear transformation $T$
• For each basis vector $v_i$ in the domain, calculate its image $T(v_i)$ under the transformation
• Express each $T(v_i)$ as a linear combination of the basis vectors in the codomain
  • Example: Let $T(x, y) = (x + y, x - y)$ and choose the standard basis $\{(1, 0), (0, 1)\}$ for both the domain and codomain. Then, $T(1, 0) = (1, 1)$ and $T(0, 1) = (1, -1)$
• The coefficients of these linear combinations form the columns of the matrix representation $A$ of the linear transformation $T$ with respect to the chosen bases
  • Example: For the linear transformation $T(x, y) = (x + y, x - y)$ with the standard basis, the matrix representation is $A = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$

Coordinate Vectors and Matrix Multiplication

• For any vector $x$ in the domain expressed as a coordinate vector $[x]_B$ with respect to the chosen basis $B$, the coordinate vector of its image $T(x)$ with respect to the chosen basis $C$ in the codomain is given by $A[x]_B = [T(x)]_C$
  • Example: For the linear transformation $T(x, y) = (x + y, x - y)$ with matrix representation $A = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$, if $x = (2, 3)$, then $[x]_B = \begin{pmatrix} 2 \\ 3 \end{pmatrix}$ and $[T(x)]_C = A[x]_B = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 5 \\ -1 \end{pmatrix}$