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Chapter 9: Problem 22

Let \(V\) and \(W\) be vector spaces. If \(X\) is a subset of \(V\), define \(X^{0}=\\{T\) in \(\mathbf{L}(V, W) \mid T(\mathbf{v})=0\) for all \(\mathbf{v}\) in \(X\\}\) a. Show that \(X^{0}\) is a subspace of \(\mathbf{L}(V, W)\). b. If \(X \subseteq X_{1}\), show that \(X_{1}^{0} \subseteq X^{0}\). c. If \(U\) and \(U_{1}\) are subspaces of \(V\), show that \(\left(U+U_{1}\right)^{0}=U^{0} \cap U_{1}^{0}\)

Short Answer

Expert verified
(a) Yes, (b) Yes, (c) Yes.

Step by step solution

01

Understanding the Definition

To solve these problems, we need to understand that for any set \(X\), the set \(X^0\) consists of linear transformations \(T\) in \(\mathbf{L}(V, W)\) such that \(T(\mathbf{v}) = 0\) for every vector \(\mathbf{v}\) in \(X\). This is akin to considering functions or transformations that 'annihilate' every vector in \(X\).
02

Prove (a) Subspace Criterion for \(X^0\)

To prove \(X^0\) is a subspace, we must show it contains the zero transformation, is closed under addition, and scalar multiplication. 1. **Zero Transformation**: The zero transformation, which maps every vector \(\mathbf{v} \in V\) to the zero vector in \(W\), is in \(X^0\) since \(0(\mathbf{v}) = 0 \forall \mathbf{v} \in X\).2. **Closure under Addition**: If \(T_1, T_2 \in X^0\), then for any \(\mathbf{v} \in X\), \((T_1 + T_2)(\mathbf{v}) = T_1(\mathbf{v}) + T_2(\mathbf{v}) = 0 + 0 = 0\). Thus, \(T_1 + T_2 \in X^0\).3. **Closure under Scalar Multiplication**: If \(T \in X^0\) and \(c\) is a scalar, then for any \(\mathbf{v} \in X\), \((cT)(\mathbf{v}) = c \cdot T(\mathbf{v}) = c \cdot 0 = 0\). Thus, \(cT \in X^0\). Thus, \(X^0\) is a subspace.
03

Prove (b) Subset Inclusion Property

Assume \(X \subseteq X_1\). We need to show that \(X_1^0 \subseteq X^0\). Let \(T \in X_1^0\); this means \(T(\mathbf{v}) = 0\) for all \(\mathbf{v} \in X_1\). Since every \(\mathbf{v} \in X\) is also in \(X_1\) (because \(X \subseteq X_1\)), it follows that \(T(\mathbf{v}) = 0\) for all \(\mathbf{v} \in X\). Hence, \(T \in X^0\). Therefore, \(X_1^0 \subseteq X^0\).
04

Prove (c) for Subspaces \(U\) and \(U_1\)

We need to show that \((U+U_1)^0 = U^0 \cap U_1^0\). 1. **\((U+U_1)^0 \subseteq U^0 \cap U_1^0\)**: Let \(T \in (U+U_1)^0\). For any \(\mathbf{v} \in U\), since \(\mathbf{v} \in U+U_1\), we have \(T(\mathbf{v}) = 0\), implying \(T \in U^0\). Similarly, for any \(\mathbf{w} \in U_1\), \(T(\mathbf{w}) = 0\). Thus, \(T \in U_1^0\). Therefore, \(T \in U^0 \cap U_1^0\).2. **\(U^0 \cap U_1^0 \subseteq (U+U_1)^0\)**: Let \(T \in U^0 \cap U_1^0\). For any \(\mathbf{x} \in U+U_1\), \(\mathbf{x} = \mathbf{u} + \mathbf{u_1}\), where \(\mathbf{u} \in U\) and \(\mathbf{u_1} \in U_1\). Then, \(T(\mathbf{x}) = T(\mathbf{u} + \mathbf{u_1}) = T(\mathbf{u}) + T(\mathbf{u_1}) = 0 + 0 = 0\). Hence, \(T \in (U+U_1)^0\).Both inclusions hold, so \((U+U_1)^0 = U^0 \cap U_1^0\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Subspace
Understanding what makes a subset a subspace is crucial for many areas of linear algebra. A subspace is a subset of a vector space that is itself a vector space. To prove a set is a subspace, three conditions must be met:
  • It contains the zero vector.
  • It is closed under vector addition.
  • It is closed under scalar multiplication.
When applying these rules to our exercise, we focus on the set \(X^0\), which includes all linear transformations that send every vector in \(X\) to zero. This set inherently contains the zero transformation, confirming the first criterion. When adding two transformations from \(X^0\), their combined effect also sends vectors in \(X\) to zero, thus satisfying closure under addition. Similarly, multiplying a transformation by a scalar keeps the transformation in \(X^0\), fulfilling the last condition. By meeting these requirements, \(X^0\) is confirmed to be a subspace of \(\mathbf{L}(V, W)\).
Linear Transformations
Linear transformations are fundamental to understanding vector spaces. They are mappings between vector spaces that preserve vector addition and scalar multiplication. Given vector spaces \(V\) and \(W\), a transformation \(T: V \to W\) is linear if for any vectors \(\mathbf{u}, \mathbf{v} \in V\) and any scalar \(c\), the following properties hold:
  • Additivity: \(T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v})\)
  • Homogeneity: \(T(c\mathbf{u}) = cT(\mathbf{u})\)
In the context of \(X^0\), a transformation is linear and sends each vector in the subset \(X\) of \(V\) to the zero vector in \(W\). This reflects the idea of an 'annihilating' transformation, reinforcing how transformations interact within the structure of a vector space. Being linear ensures these transformations fit the framework of subspaces perfectly.
Closure Properties
Closure properties are fundamental for proving whether a structure within a vector space is a valid subspace. In our exercise, closure under both addition and scalar multiplication are tested. For a subset to claim the status of a subspace, it must remain intact under these operations.Closure under addition ensures that the proper sum of any two elements from the set still belongs to the set. For \(X^0\), if two transformations annihilate every vector in \(X\), then their sum also achieves this, showing closure under addition.

Closure under scalar multiplication insists that multiplying any element by a scalar results in an element still within the set. It affirms that any scalar multiple of a transformation in \(X^0\) remains within \(X^0\) due to the linear nature of transformations. By satisfying these closure properties, \(X^0\) robustly qualifies as a subspace. Hence, closure properties are critical in the verification process of a subspace's legitimacy.

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Most popular questions from this chapter

Let \(T: V \rightarrow V\) be a linear operator where \(\operatorname{dim} V=n\). Show that \(V\) has a basis of eigenvectors if and only if \(V\) has a basis \(B\) such that \(M_{B}(T)\) is diagonal.

Let \(T: V \rightarrow V\) be a linear operator where \(V\) is finite dimensional. If \(U \subseteq V\) is a subspace, let \(\bar{U}=\left\\{\mathbf{u}_{0}+T\left(\mathbf{u}_{1}\right)+T^{2}\left(\mathbf{u}_{2}\right)+\cdots+T^{k}\left(\mathbf{u}_{k}\right) \mid \mathbf{u}_{i}\right.\) in \(U, k \geq\) 0\\}. Show that \(\bar{U}\) is the smallest \(T\) -invariant subspace containing \(U\) (that is, it is \(T\) -invariant, contains \(U\), and is contained in every such subspace).

Let \(V\) be a vector space. If \(f: V \rightarrow \mathbb{R}\) is a linear transformation and \(\mathrm{z}\) is a vector in \(V,\) define \(T_{f, \mathbf{z}}: V \rightarrow V\) by \(T_{f, \mathbf{z}}(\mathbf{v})=f(\mathbf{v}) \mathbf{z}\) for all \(\mathbf{v}\) in \(V\). Assume that \(f \neq 0\) and \(\mathbf{z} \neq \mathbf{0}\). a. Show that \(T_{f, z}\) is a linear operator of rank 1 . b. If \(f \neq 0\), show that \(T_{f, \mathrm{z}}\) is an idempotent if and only if \(f(\mathbf{z})=1\). (Recall that \(T: V \rightarrow V\) is called an idempotent if \(T^{2}=T\).) c. Show that every idempotent \(T: V \rightarrow V\) of rank 1 has the form \(T=T_{f, \mathbf{z}}\) for some \(f: V \rightarrow \mathbb{R}\) and some \(\mathbf{z}\) in \(V\) with \(f(\mathbf{z})=1\). [Hint: Write \(\operatorname{im} T=\mathbb{R} \mathbf{z}\) and show that \(T(\mathbf{z})=\mathbf{z}\). Then use \(\mathrm{Ex}\) - ercise \(9.3 .23 .]\)

Let \(S\) and \(T\) be linear operators on \(V\) and assume that \(S T=T S\). a. Show that \(\mathrm{im} S\) and ker \(S\) are \(T\) -invariant. b. If \(U\) is \(T\) -invariant, show that \(S(U)\) is \(T\) -invariant.

Find \(P_{D \leftarrow B}\) if \(B=\left\\{\mathbf{b}_{1}, \mathbf{b}_{2}, \mathbf{b}_{3}, \mathbf{b}_{4}\right\\}\) and \(D=\left\\{\mathbf{b}_{2}, \mathbf{b}_{3}, \mathbf{b}_{1}, \mathbf{b}_{4}\right\\} .\) Change matrices arising when the bases differ only in the order of the vectors are called permutation matrices.
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