Chapter 6: Problem 37
If \(V\) is a finite-dimensional vector space and \(T: V \rightarrow V\) is a linear transformation such that rank(T) = \(\operatorname{rank}\left(T^{2}\right),\) prove that range \((T) \cap \operatorname{ker}(T)=\\{0\\} .[\) Hint: \(T^{2}\) denotes \(T \circ T .\) Use the Rank Theorem to help show that the kernels of \(T\) and \(T^{2}\) are the same.
Short Answer
Step by step solution
Understanding the Problem
Apply Rank-Nullity Theorem
Express Nullity of T and T² with Rank
Relate Kernels of T and T²
Showing Intersection is Zero Vector
Conclusion
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Rank-Nullity Theorem
\[ \text{dim}(V) = \text{rank}(T) + \text{nullity}(T) \]
Here, \( \text{rank}(T) \) refers to the dimension of the range of \( T \), while \( \text{nullity}(T) \) refers to the dimension of the kernel of \( T \).
- **Rank (\(T\))**: The rank of \( T \) is the number of linearly independent columns in the matrix representation of \( T \), or simply the dimension of the image of \( T \).
- **Nullity (\(T\))**: The nullity measures the "failure" of \( T \) to be injective; it is the dimension of the kernel of \( T \).
Using the Rank-Nullity Theorem in the given exercise, we see why \( \text{rank}(T) = \text{rank}(T^{2}) \) implies \( \text{nullity}(T) = \text{nullity}(T^{2}) \). Since the total dimension \( \text{dim}(V) \) is composed of both rank and nullity, keeping rank the same for both transformation \( T \) and its square \( T^2 \) necessitates the nullities to be identical.
Vector Space
Some key characteristics of a vector space include:
- **Closure**: The sum of two vectors in the space is also a vector within the same space, and scalar multiplication of a vector results in another vector in the space.
- **Zero Vector**: There exists a zero vector \( \mathbf{0} \) in the space such that adding it to any other vector \( v \) yields \( v \).
- **Inverse Elements**: For each vector \( v \), there exists an inverse \( -v \) such that \( v + (-v) = \mathbf{0} \).
In the context of the exercise, we're dealing with a finite-dimensional vector space \( V \), which means \( V \) has a finite basis, and all linear transformations including \( T \) involve vectors from this space.
Kernel and Range
- **Kernel**: The kernel of a linear transformation \( T : V \rightarrow W \) is the set of all vectors in \( V \) that \( T \) maps to the zero vector in \( W \). Formally, it's denoted as:
\[ \text{ker}(T) = \{ v \in V \mid T(v) = 0 \} \]
The kernel is crucial for determining the nullity of \( T \) and understanding how \( T \) might not be injective (one-to-one).
- **Range**: The range (or image) of \( T \) is the set of all vectors in \( W \) that can be written as \( T(v) \) for some vector \( v \) in \( V \). The range gives us the rank of \( T \), showing which vectors are "covered" by the transformation.
\[ \text{range}(T) = \{ T(v) \mid v \in V \} \]
In this exercise, proving that the intersection \( \text{range}(T) \cap \text{ker}(T) = \{0\} \) helps to determine the characteristics of non-singular transformations, highlighting that \( T \) has no overlap (except at the zero vector) between the space it "annihilates" and the space it actually "covers" with transformations.