Chapter 6: Problem 70
Proof Let \(T: V \rightarrow W\) be a linear transformation, and let \(U\) be a subspace of \(W\). Prove that the set \(T^{-1}(U)=\\{\mathbf{v} \in V: T(\mathbf{v}) \in U\\}\) is a subspace of \(V .\) What is \(T^{-1}(U)\) when \(U=\\{\mathbf{0}\\} ?\)
Short Answer
Expert verified
The set \(T^{-1}(U)\) is a subspace of \(V\). When \(U = \{\mathbf{0}\}\), \(T^{-1}(\{\mathbf{0}\})\) is the kernel (or nullspace) of \(T\).
Step by step solution
01
Recognize the structure of a subspace
A subspace of a vector space is itself a vector space that adheres to the properties of vector space: closure under addition and scalar multiplication, the existence of a zero element and additive inverses.
02
Prove closure under addition for the inverse image
We need to show that if \(\mathbf{v_1}\) and \(\mathbf{v_2}\) are in \(T^{-1}(U)\), then \( \mathbf{v_1} + \mathbf{v_2}\) is also in \(T^{-1}(U)\). Because \(\mathbf{v_1}\) and \(\mathbf{v_2}\) are in \(T^{-1}(U)\), so \(T(\mathbf{v_1})\) and \(T(\mathbf{v_2})\) are in \(U\). Since \(U\) is a subspace of \(W\), it is closed under addition, so \(T(\mathbf{v_1}) + T(\mathbf{v_2}) = T(\mathbf{v_1} + \mathbf{v_2})\) is in \(U\). Therefore, \( \mathbf{v_1} + \mathbf{v_2}\) is in \(T^{-1}(U)\).
03
Prove closure under scalar multiplication for the inverse image
We need to show that if \(\mathbf{v}\) is in \(T^{-1}(U)\) and \(c\) is any scalar, then \(c\mathbf{v}\) is also in \(T^{-1}(U)\). Because \(\mathbf{v}\) is in \(T^{-1}(U)\), so \(T(\mathbf{v})\) is in \(U\). Since \(U\) is a subspace of \(W\), it is closed under scalar multiplication, so \(cT(\mathbf{v}) = T(c\mathbf{v})\) is in \(U\). Therefore, \(c\mathbf{v}\) is in \(T^{-1}(U)\).
04
Show the existence of the zero element and additive inverses
The zero vector of \(V\) is in \(T^{-1}(U)\) because \(T(\mathbf{0})\) is in \(U\), as \(U\) as a subspace must contain the zero vector. For any vector \(\mathbf{v}\) in \(T^{-1}(U)\), its additive inverse \(-\mathbf{v}\) is also in \(T^{-1}(U)\), because the image under \(T\) of \(-\mathbf{v}\) is the additive inverse of the image of \(\mathbf{v}\), which is in \(U\). Therefore, \( -\mathbf{v}\) is in \(T^{-1}(U)\).
05
Determine \(T^{-1}(U)\) when \(U=\{\mathbf{0}\}\)
If \(U = \{\mathbf{0}\}\), then \(T^{-1}(U)\) would be the set of all vectors \(\mathbf{v}\) in \(V\) that \(T\) maps to the zero vector. In other words, \(T^{-1}(U)\) is the kernel (nullspace) of the transformation \(T\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Subspace
A subspace is essentially a smaller, contained vector space, which sits inside a larger vector space. It must adhere to certain rules that preserve its structure as a vector space.
To qualify as a subspace, a set must satisfy three crucial properties:
To qualify as a subspace, a set must satisfy three crucial properties:
- Closure Under Addition: If you take any two vectors in the subspace and add them together, the resulting vector should still be in the subspace.
- Closure Under Scalar Multiplication: If you multiply any vector in the subspace by a scalar (a real number), the result should remain in the subspace.
- Zero Vector: The subspace must contain the zero vector, the origin of the vector space, ensuring that it includes a neutral element for addition.
Exploring the Inverse Image
The concept of an inverse image in linear algebra pertains to mapping elements from one space back to their pre-mapped counterparts in another. When dealing with a linear transformation \( T: V \rightarrow W \), finding the inverse image \( T^{-1}(U) \) involves identifying those vectors in \( V \) that map into a specific subspace \( U \) of \( W \).
The inverse image \( T^{-1}(U) \) has to be a subspace of \( V \).
Using the rules of subspaces:
The inverse image \( T^{-1}(U) \) has to be a subspace of \( V \).
Using the rules of subspaces:
- When two vectors from \( T^{-1}(U) \) are added, their image via \( T \) is also in \( U \) because \( U \) is a subspace.
- Similarly, multiplying a vector from \( T^{-1}(U) \) by a scalar keeps the transformed result within \( U \), ensuring closure under scalar multiplication.
- The zero vector in \( V \) must map to the zero vector in \( U \), and this establishes the foundation of the zero element in \( T^{-1}(U) \).
Decoding Kernel (Nullspace)
The kernel, often referred to as the nullspace, is a foundational concept in linear algebra focused on understanding the structure of linear transformations.
Mathematically, the kernel of a transformation \( T: V \rightarrow W \) consists of vectors in \( V \) that map to the zero vector in \( W \).
You can think of the kernel as capturing how much of \( V \) collapses into \( W \) at the zero vector.
This can be visualized as follows:
Mathematically, the kernel of a transformation \( T: V \rightarrow W \) consists of vectors in \( V \) that map to the zero vector in \( W \).
You can think of the kernel as capturing how much of \( V \) collapses into \( W \) at the zero vector.
This can be visualized as follows:
- When you run the transformation, any vector that ends up at the zero vector in \( W \) is part of the kernel.
- The kernel \( T^{-1}(\{\mathbf{0}\}) \) inherently forms a subspace of \( V \). It satisfies closure under addition and scalar multiplication simply because zero maps are inherent in vector space calculations.
- It’s critical to note that when the kernel comprises only the zero vector, the transformation \( T \) is injective (one-to-one).