Question

Let \( f_1, f_2, f_3 \) be nonzero linear transformations from \( \mathbb{R}^4 \) to \( \mathbb{R} \) and \[ \ker(f_1) \subset \ker(f_2) \cap \ker(f_3). \] Let \( T : \mathbb{R}^4 \to \mathbb{R}^3 \) be the linear transformation defined by \[ T(v) = (f_1(v), f_2(v), f_3(v)) \quad \text{for all } v \in \mathbb{R}^4. \] Then, the nullity of \( T \) is equal to:

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

The nullity of a linear transformation is the dimension of its kernel. Analyzing the relationships between the kernels of the transformations allows us to determine the nullity.

Answer 1

The Correct Option is C

Step 1: Analyze the nullity of \( T \).
The nullity of \( T \) is the dimension of the kernel of \( T \). Since \( T(v) = (f_1(v), f_2(v), f_3(v)) \), the kernel of \( T \) consists of vectors \( v \in \mathbb{R}^4 \) such that \( f_1(v) = f_2(v) = f_3(v) = 0 \). Step 2: Use the given information.
We know that \( \ker(f_1) \subset \ker(f_2) \cap \ker(f_3) \). This implies that \( \ker(f_1) \) is the most restrictive kernel among the three. The kernel of \( T \) is formed by the vectors in \( \ker(f_1) \), which has a dimension of 3 (because the nullity of a linear map from \( \mathbb{R}^4 \) to \( \mathbb{R} \) is 3). Final Answer: \[ \boxed{3} \]