Question

Let \( S \) and \( T \) be linear transformations from a finite dimensional vector space \( V \) to itself such that \( S(T(v)) = 0 \) for all \( v \in V \). Then

• A. \( \text{rank}(T) \geq \text{nullity}(S) \)
• B. \( \text{rank}(S) \geq \text{nullity}(T) \)
• C. \( \text{rank}(T) \leq \text{nullity}(S) \)
• D. \( \text{rank}(S) \leq \text{nullity}(T) \)

Hint:

When dealing with compositions of linear transformations, use the rank-nullity theorem and understand the inclusion of images and kernels.

Answer 1

The Correct Option is C, D

Step 1: Understand the condition \( S(T(v)) = 0 \).
The condition \( S(T(v)) = 0 \) for all \( v \in V \) implies that the image of \( T \) is contained within the kernel of \( S \), i.e., \( \text{Im}(T) \subseteq \ker(S) \).
Step 2: Relate rank and nullity.
From the rank-nullity theorem, we know that: \[ \text{rank}(T) + \text{nullity}(T) = \dim V \quad \text{and} \quad \text{rank}(S) + \text{nullity}(S) = \dim V. \] Since \( \text{Im}(T) \subseteq \ker(S) \), we have \( \text{rank}(T) \leq \text{nullity}(S) \).
Step 3: Conclusion.
Thus, the correct answer is \( \boxed{(C)} \).