Step 1: Understanding the Concept:
A mapping \( T: V \to W \) between two vector spaces over a field \( F \) is a linear transformation if it satisfies two conditions for all vectors \( \mathbf{u}, \mathbf{v} \in V \) and scalars \( c \in F \):
1. Additivity: \( T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v}) \)
2. Homogeneity: \( T(c\mathbf{u}) = cT(\mathbf{u}) \)
Step 2: Detailed Explanation:
Let us evaluate each of the given options:
- Option A: "Nullity($T$) is a subspace of $W$."
The null space (kernel) of $T$, defined as \( \text{Ker}(T) = \{ \mathbf{v} \in V \mid T(\mathbf{v}) = \mathbf{0}_W \} \), is a subspace of the domain $V$.
Nullity is the *dimension* of this null space (a scalar), so it cannot be a subspace, and the null space itself lies in $V$, not $W$. Thus, Option A is false.
- Option B: "Rank($T$) is a subspace of $V$."
The range (image) of $T$, defined as \( \text{Im}(T) = \{ T(\mathbf{v}) \mid \mathbf{v} \in V \} \), is a subspace of the codomain $W$.
Rank is the *dimension* of the range (a scalar), so Option B is false.
- Option C: "T(\(0_V\)) = \(0_W\)."
Let us prove this fundamental property of any linear transformation using homogeneity with scalar \( c = 0 \):
\[ T(\mathbf{0}_V) = T(0 \cdot \mathbf{v}) \]
Using the homogeneity property:
\[ T(\mathbf{0}_V) = 0 \cdot T(\mathbf{v}) \]
Since multiplying any vector in $W$ by the scalar $0$ yields the zero vector of $W$:
\[ T(\mathbf{0}_V) = \mathbf{0}_W \]
This proof shows that a linear transformation must map the zero vector of the domain to the zero vector of the codomain. Thus, Option C is always true.
Step 3: Final Answer:
The correct relationship is $T(0_V) = 0_W$.