Question

Let V(F) be a finite dimensional vector space and T: V \(\to\) V be a linear transformation. Let R(T) denote the range of T and N(T) denote the null space of T. If rank(T) = rank(T\textsuperscript{2}), then which of the following are correct?
A. N(T) = R(T)
B. N(T) = N(T\textsuperscript{2})
C. N(T) \(\cap\) R(T) = \{0\
D. R(T) = R(T\textsuperscript{2})}

• A. A, B and D only
• B. A, B and C only
• C. A, B, C and D
• D. B, C and D only

Hint:

The condition \(\text{rank}(T) = \text{rank}(T^2)\) on a finite-dimensional space is equivalent to having \(V = N(T) \oplus R(T)\), a direct sum decomposition. This means their intersection is \{0\} (C is correct) and every vector in V can be uniquely written as a sum of a vector in N(T) and a vector in R(T). This is a powerful result to remember.

Answer 1

The Correct Option is D

Step 1: Problem Overview:
This question examines the relationship between the range (image) and null space (kernel) of a linear transformation \(T\) and its square \(T^2\), given that their ranks are equal. This relates to fundamental properties of linear maps on finite-dimensional vector spaces, including the Rank-Nullity Theorem.

Step 2: Key Concepts:
Rank-Nullity Theorem: For a linear map \(T: V \to V\) on a finite-dimensional vector space \(V\), \(\text{dim}(V) = \text{rank}(T) + \text{nullity}(T)\), where \(\text{rank}(T) = \text{dim}(R(T))\) and \(\text{nullity}(T) = \text{dim}(N(T))\).
Subspace Properties: For any linear operator \(T\), we have \(R(T^2) \subseteq R(T)\) and \(N(T) \subseteq N(T^2)\). If two nested subspaces have the same dimension, they are equal.

Step 3: Detailed Analysis:
Let \(\text{dim}(V) = n\). We are given \(\text{rank}(T) = \text{rank}(T^2)\).
Statement D: \(R(T) = R(T\textsuperscript{2})\)
\(R(T^2) = T(R(T))\), so \(R(T^2)\) is a subspace of \(R(T)\).
We are given \(\text{dim}(R(T)) = \text{rank}(T) = \text{rank}(T^2) = \text{dim}(R(T^2))\).
Since \(R(T^2)\) is a subspace of \(R(T)\) and they have the same dimension, they are equal. Thus, D is correct.
Statement B: \(N(T) = N(T\textsuperscript{2})\)
Using the Rank-Nullity Theorem: \(\text{nullity}(T) = n - \text{rank}(T)\) \(\text{nullity}(T^2) = n - \text{rank}(T^2)\) Since \(\text{rank}(T) = \text{rank}(T^2)\), then \(\text{nullity}(T) = \text{nullity}(T^2)\).
If \(x \in N(T)\), then \(T(x) = 0\). Applying \(T\) again, \(T(T(x)) = T(0) = 0\), so \(T^2(x) = 0\). This means \(x \in N(T^2)\). Therefore, \(N(T) \subseteq N(T^2)\).
Since \(N(T)\) is a subspace of \(N(T^2)\) and \(\text{dim}(N(T)) = \text{dim}(N(T^2))\), they must be equal. Thus, B is correct.
Statement C: \(N(T) \cap R(T) = \{0\}\)
Let \(v \in N(T) \cap R(T)\).
Since \(v \in R(T)\), there exists \(u \in V\) such that \(v = T(u)\).
Since \(v \in N(T)\), \(T(v) = 0\).
Substituting \(v=T(u)\) gives \(T(T(u)) = 0\), so \(T^2(u) = 0\), and \(u \in N(T^2)\).
From statement B, \(N(T^2) = N(T)\). Thus, \(u \in N(T)\).
If \(u \in N(T)\), then \(T(u) = 0\).
But \(v = T(u)\), so \(v = 0\).
Therefore, the intersection contains only the zero vector. Thus, C is correct.
Statement A: \(N(T) = R(T)\)
This is not generally true. Consider the identity transformation \(T=I\) on a non-zero vector space \(V\). Then \(T^2 = I\), so \(\text{rank}(T) = \text{rank}(T^2) = \text{dim}(V)\). Here, \(N(T) = \{0\}\) and \(R(T) = V\). Clearly \(N(T) eq R(T)\). Thus, A is incorrect.

Step 4: Conclusion:
The correct statements are B, C, and D, corresponding to option (D).