Linear Maps

Linear Map

Definition

A linear map from \(V\) to \(W\) is a function \(T: V \to W\) with the following properties:

• additivity: \(T(u + v) = T(u) + T(v)\) for all \(u, v \in V\)

• homogeneity: \(T(c \cdot v) = c \cdot T(v)\) for all \(v \in V\) and all scalars \(c\)

Notation

The set of all linear maps from \(V\) to \(W\) is denoted \( \mathcal{L}(V; W)\).

Linear maps and basis of domain

Theorem

Suppose \(v_1, \ldots, v_n\) is a basis of \(V\) and \(w_1, \ldots, w_n \in W\). Then there exists a unique linear map \(T: V \to W\) such that

\(T(v_i) = w_i\) for all \(i = 1, \ldots, n\).

Proof

Since \(\{v_1, \ldots, v_n\}\) is a basis for \(V\), every \(v \in V\) can be written uniquely as

\[ v = c_1 v_1 + c_2 v_2 + \cdots + c_n v_n, \]

where \(c_1, c_2, \ldots, c_n\) are scalars. Define \(T: V \to W\) by

\[ T(v) = c_1 w_1 + c_2 w_2 + \cdots + c_n w_n. \]

This is well-defined due to the uniqueness of the representation. We verify that \(T\) is linear:

• Let \(v, u \in V\), with \(v = \sum_{i=1}^n c_i v_i\) and \(u = \sum_{i=1}^n d_i v_i\). Then

  \[ v + u = \sum_{i=1}^n (c_i + d_i) v_i, \]

  so

  \[ T(v + u) = \sum_{i=1}^n (c_i + d_i) w_i = \sum_{i=1}^n c_i w_i + \sum_{i=1}^n d_i w_i = T(v) + T(u). \]

• For any scalar \(a\),

  \[ a v = \sum_{i=1}^n (a c_i) v_i, \] so \[ T(a v) = \sum_{i=1}^n (a c_i) w_i = a \sum_{i=1}^n c_i w_i = a T(v). \]

Thus, \(T\) is linear. Moreover, for each basis vector \(v_j\),

\[ v_j = 0 \cdot v_1 + \cdots + 1 \cdot v_j + \cdots + 0 \cdot v_n, \]

so

\[ T(v_j) = 0 \cdot w_1 + \cdots + 1 \cdot w_j + \cdots + 0 \cdot w_n = w_j. \]

Therefore, \(T\) satisfies \(T(v_i) = w_i\) for all \(i\).

Suppose \(T: V \to W\) is a linear map such that \(T(v_i) = w_i\) for all \(i\). For any \(v \in V\), write

\[ v = \sum_{i=1}^n c_i v_i. \]

Then by linearity of \(T\),

\[ T(v) = T\left( \sum_{i=1}^n c_i v_i \right) = \sum_{i=1}^n c_i T(v_i) = \sum_{i=1}^n c_i w_i. \]

This shows that \(T(v)\) is completely determined by the values \(T(v_i) = w_i\). Hence, \(T\) is unique.

$\square$

Addition and scalar multiplication on \( \mathcal{L}(V; W)\)

Definition

Let \(T_1, T_2 \in \mathcal{L}(V; W)\) and \(c \in \mathbb{F}\). We define addition and scalar multiplication on \( \mathcal{L}(V; W)\) as follows:

• Addition: \((T_1 + T_2)(v) = T_1(v) + T_2(v)\) for all \(v \in V\).

• Scalar multiplication: \((c \cdot T)(v) = c \cdot T(v)\) for all \(v \in V\).

Vector space \( \mathcal{L}(V; W)\)

Theorem

The set \( \mathcal{L}(V; W)\) is a vector space over the field \( \mathbb{F}\).

Proof

To show that \( \mathcal{L}(V; W)\) is a vector space, we need to verify the following properties:

• Closure under addition: Let \(T_1, T_2 \in \mathcal{L}(V; W)\). Then \((T_1 + T_2)(v) = T_1(v) + T_2(v)\) is a linear map, so \(T_1 + T_2 \in \mathcal{L}(V; W)\).

• Closure under scalar multiplication: Let \(T \in \mathcal{L}(V; W)\) and \(c \in \mathbb{F}\). Then \((c \cdot T)(v) = c \cdot T(v)\) is a linear map, so \(c \cdot T \in \mathcal{L}(V; W)\).

• Existence of zero vector: The zero map \(0: V \to W\) defined by \(0(v) = 0\) for all \(v \in V\) is in \( \mathcal{L}(V; W)\).

• Existence of additive inverses: For each \(T \in \mathcal{L}(V; W)\), the map \(-T\) defined by \((-T)(v) = -T(v)\) is in \( \mathcal{L}(V; W)\).

Since all vector space axioms are satisfied, we conclude that \( \mathcal{L}(V; W)\) is a vector space over the field \( \mathbb{F}\).

$\square$

Product of Linear Maps

Definition

Let \(T_1: V \to W\) and \(T_2: W \to U\) be linear maps. The product of \(T_1\) and \(T_2\), denoted \(T_2 \circ T_1\), is defined by

\[ (T_2 \circ T_1)(v) = T_2(T_1(v)) \]

for all \(v \in V\).

Algebraic properties of products of linear maps

• associativity

  $$ (T_3 \cdot T_2) \cdot T_1 = T_3 \cdot (T_2 \cdot T_1) $$

  Whenever \(T_1\), \(T_2\), and \(T_3\) are linear maps such that the products make sense.

• identity

  $$ T \cdot I = T \cdot I = T $$

  Whenever \(T \in \mathcal{L}(V; W)\).

• distributive properties

  $$ (S_1 + S_2) T = S_1 T + S_2 T \quad \text{and} \quad S (T_1 + T_2) = S T_1 + S T_2 $$

  Whenever \(T, T_1, T_2 \in \mathcal{L}(U; V)\) and \(S, S_1, S_2 \in \mathcal{L}(V; W)\).

Linear maps take 0 to 0

Theorem

Suppose $T$ is a linear map from $V$ to $W$. Then $T(0) = 0$.

Proof

Let $v \in V$. Then by linearity, we have:

\[ T(0) = T(0 \cdot v) = 0 \cdot T(v) = 0. \]

$\square$