Question

Let \(V\) be a finite-dimensional vector space and \(T: V \to V\) be a linear transformation. Let \(\mathcal{R}(T)\) denote the range of \(T\) and \(\mathcal{N}(T)\) denote the null space of \(T\). If \(\operatorname{rank}(T) = \operatorname{rank}(T^2)\), then which of the following is/are necessarily true?

• A. \(\mathcal{N}(T) = \mathcal{N}(T^2).\)
• B. \(\mathcal{R}(T) = \mathcal{R}(T^2).\)
• C. \(\mathcal{N}(T) \cap \mathcal{R}(T) = \{0\}.\)
• D. \(\mathcal{N}(T) = \{0\}.\)

Hint:

If \(\operatorname{rank}(T) = \operatorname{rank}(T^2)\), then \(T\) is said to be a *semisimple operator*, and its null space and range remain stable under further applications.

Answer 1

The Correct Option is A, B, C

Step 1: Apply rank-nullity theorem.
Since \(\operatorname{rank}(T) = \operatorname{rank}(T^2)\), \[ \dim(\mathcal{N}(T)) = \dim(V) - \operatorname{rank}(T) = \dim(V) - \operatorname{rank}(T^2) = \dim(\mathcal{N}(T^2)). \]
Step 2: Relation between null spaces.
We know \(\mathcal{N}(T) \subseteq \mathcal{N}(T^2)\). Since their dimensions are equal, they must be equal as sets: \[ \mathcal{N}(T) = \mathcal{N}(T^2). \]
Step 3: Relation between ranges.
Also, \( \mathcal{R}(T^2) \subseteq \mathcal{R}(T)\). Since they have the same dimension, we get \[ \mathcal{R}(T^2) = \mathcal{R}(T). \]
Step 4: Conclusion.
Thus, both (A) and (B) are necessarily true.