Step 1: Understanding the Concept:
We need to analyze the given linear transformation T by finding the dimensions of its null space (kernel), \(N(T)\), and its range (image), \(R(T)\). The Rank-Nullity Theorem states that for a linear map \(T: V \to W\), \(\dim V = \dim N(T) + \dim R(T)\).
Step 2: Key Formula or Approach:
1. Find the Null Space (Kernel): The null space \(N(T)\) is the set of all vectors \(\vec{v}\) in the domain such that \(T(\vec{v}) = \vec{0}\). We will solve the equation \(T(a,b,c) = (0,0,0)\) to find a basis for \(N(T)\) and its dimension (the nullity).
2. Find the Range (Image): The range \(R(T)\) is the set of all possible output vectors. It is the span of the transformations of the basis vectors of the domain. The dimension of the range is the rank.
3. Use the Rank-Nullity Theorem: \(\dim(\mathbb{R}^3) = \dim N(T) + \dim R(T)\), which is \(3 = \text{nullity}(T) + \text{rank}(T)\).
Step 3: Detailed Explanation:
1. Finding the Null Space \(N(T)\):
We set \(T(a,b,c) = (0,0,0)\):
\[ (a-b, a-b, 0) = (0,0,0) \]
This gives the condition \(a-b = 0\), which means \(a=b\). The variable \(c\) can be any real number.
So, a vector in the null space has the form \((a, a, c)\).
We can write this vector as a linear combination:
\[ (a, a, c) = (a, a, 0) + (0, 0, c) = a(1, 1, 0) + c(0, 0, 1) \]
The null space is spanned by the vectors \(\{(1,1,0), (0,0,1)\}\). These two vectors are clearly linearly independent.
Therefore, they form a basis for \(N(T)\). The number of vectors in the basis is 2.
So, the dimension of the null space (nullity) is \(\dim N(T) = 2\). This matches option (A).
2. Finding the Range \(R(T)\):
An arbitrary vector in the range has the form \((a-b, a-b, 0)\).
Let \(k = a-b\). Then the vector is \((k, k, 0) = k(1,1,0)\).
This shows that the range is spanned by the single vector \((1,1,0)\).
A basis for the range is \(\{(1,1,0)\}\).
The dimension of the range (rank) is \(\dim R(T) = 1\).
3. Verifying with Rank-Nullity Theorem:
\[ \dim N(T) + \dim R(T) = 2 + 1 = 3 \]
This matches the dimension of the domain \(\mathbb{R}^3\), so our calculations are consistent.
Evaluating the options:
- (A) \(\dim N(T) = 2\): Correct.
- (B) \(\dim R(T) = 2\): Incorrect, it is 1.
- (C) \(R(T) = N(T)\): Incorrect. \(R(T) = \text{span}\{(1,1,0)\}\) and \(N(T) = \text{span}\{(1,1,0), (0,0,1)\}\). They are not equal.
- (D) \(\dim R(T) = 3\): Incorrect, it is 1.
Step 4: Final Answer:
The dimension of the null space of T is 2. This corresponds to option (A).