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Let \(P: V \rightarrow \mathbb{R}\) and \(Q: V \rightarrow \mathbb{R}\) be linear transformations, where \(V\) is a vector space. Define \(T: V \rightarrow \mathbb{R}^{2}\) by \(T(\mathbf{v})=(P(\mathbf{v}), Q(\mathbf{v}))\) a. Show that \(T\) is a linear transformation. b. Show that \(\operatorname{ker} T=\operatorname{ker} P \cap \operatorname{ker} Q,\) the set of vec- tors in both \(\operatorname{ker} P\) and \(\operatorname{ker} Q\).

Short Answer

Expert verified
T is a linear transformation; \( \operatorname{ker} T = \operatorname{ker} P \cap \operatorname{ker} Q \).

Step by step solution

01

Define Linear Transformation

A transformation \( T: V \rightarrow \mathbb{R}^2 \) is linear if it satisfies two properties: 1) Additivity: \( T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v}) \) for any vectors \( \mathbf{u}, \mathbf{v} \in V \), and 2) Scalar Multiplication: \( T(c\mathbf{v}) = cT(\mathbf{v}) \) for any scalar \( c \) and vector \( \mathbf{v} \in V \).
02

Check Additivity for T

To show additivity, consider two vectors \( \mathbf{u}, \mathbf{v} \in V \). Then, \( T(\mathbf{u} + \mathbf{v}) = (P(\mathbf{u} + \mathbf{v}), Q(\mathbf{u} + \mathbf{v})) \). Since \( P \) and \( Q \) are linear, \( P(\mathbf{u} + \mathbf{v}) = P(\mathbf{u}) + P(\mathbf{v}) \) and \( Q(\mathbf{u} + \mathbf{v}) = Q(\mathbf{u}) + Q(\mathbf{v}) \). Thus, \( T(\mathbf{u} + \mathbf{v}) = (P(\mathbf{u}) + P(\mathbf{v}), Q(\mathbf{u}) + Q(\mathbf{v})) = T(\mathbf{u}) + T(\mathbf{v}) \), confirming additivity.
03

Check Scalar Multiplication for T

For scalar multiplication, consider \( c \in \mathbb{R} \) and vector \( \mathbf{v} \in V \). Then, \( T(c \mathbf{v}) = (P(c \mathbf{v}), Q(c \mathbf{v})) \). Since \( P \) and \( Q \) are linear, \( P(c \mathbf{v}) = cP(\mathbf{v}) \) and \( Q(c \mathbf{v}) = cQ(\mathbf{v}) \). Thus, \( T(c \mathbf{v}) = (cP(\mathbf{v}), cQ(\mathbf{v})) = c(P(\mathbf{v}), Q(\mathbf{v})) = cT(\mathbf{v}) \), confirming scalar multiplication.
04

Conclusion of Linearity

Since both additivity and scalar multiplication have been verified, \( T \) is a linear transformation.
05

Define Kernel of T

The kernel of \( T \), denoted \( \operatorname{ker} T \), is the set of all vectors \( \mathbf{v} \in V \) such that \( T(\mathbf{v}) = \mathbf{0}_{\mathbb{R}^2} = (0,0) \).
06

Express Conditions for Ker T

For \( T(\mathbf{v}) = (P(\mathbf{v}), Q(\mathbf{v})) = (0,0) \), it must be that \( P(\mathbf{v}) = 0 \) and \( Q(\mathbf{v}) = 0 \).
07

Compare Ker T, Ker P, and Ker Q

Since \( \operatorname{ker} P \) is the set of vectors \( \mathbf{v} \) such that \( P(\mathbf{v}) = 0 \) and \( \operatorname{ker} Q \) is the set where \( Q(\mathbf{v}) = 0 \), the vectors that satisfy both conditions are in \( \operatorname{ker} T \). Thus, \( \operatorname{ker} T = \operatorname{ker} P \cap \operatorname{ker} Q \).
08

Conclusion of Kernel Intersection

The kernel of \( T \), \( \operatorname{ker} T \), is indeed the intersection of \( \operatorname{ker} P \) and \( \operatorname{ker} Q \), meaning it consists of all vectors that lie in both \( \operatorname{ker} P \) and \( \operatorname{ker} Q \).

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Key Concepts

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

Vector Spaces
A vector space is a fundamental concept in linear algebra, encompassing a set of vectors along with operations of vector addition and scalar multiplication. Each vector space is equipped with some basic rules or axioms that all its operations must satisfy, ensuring consistency and predictability. This structure forms the basis for many advanced mathematical concepts and is crucial for understanding linear transformations and their properties.

Here are the primary characteristics of vector spaces:
  • Closure of Addition and Scalar Multiplication. The sum of any two vectors in the space, or the product of a vector by a scalar, must also be within the space.
  • Existence of a Zero Vector. There exists a special vector, the zero vector, such that adding it to any other vector leaves that vector unchanged.
  • Inverses. For every vector, there exists an opposite vector, so that their sum is the zero vector.
  • Distributive and Associative Properties. These ensure that combining vectors and scalars behaves as expected, similar to basic arithmetic operations.
Understanding the framework of vector spaces is pivotal for working with linear transformations, such as those in the problem where transformations map vectors from one space to another, requiring adherence to these foundational properties.
Kernel of a Transformation
The kernel of a linear transformation is an essential concept that describes the "solution set" for the transformation mapping to the zero vector. Specifically, for a transformation \( T: V \rightarrow W \), the kernel, often denoted as \( \text{ker} \, T \), comprises all vectors \( \mathbf{v} \in V \) for which \( T(\mathbf{v}) \) is the zero vector in \( W \).

To further grasp the kernel's role, consider the following:
  • Subspace. The kernel itself forms a subspace within the original vector space \( V \). This means it possesses all the properties of a vector space, including closure under addition and scalar multiplication.
  • Dimension Insight. The dimension of the kernel offers insight into the rank-nullity theorem, which relates dimensions of a vector space, its image (or range), and its kernel.
  • Analysis of Dependency. A non-zero kernel indicates linear dependence within the transformation, where certain input vectors do not uniquely map to outputs.
The original exercise emphasizes this concept by showing that the kernel of a combined transformation \( T \) is the intersection of the kernels of the individual transformations \( P \) and \( Q \). This occurs because only vectors transforming to the zero vector under both \( P \) and \( Q \) will map to the zero vector under \( T \).
Additivity and Scalar Multiplication
Additivity and scalar multiplication are key properties that define linear transformations. For a transformation to be linear, it must satisfy specific criteria related to these two concepts, which are akin to essential rules of conduct for transformations in linear algebra.

Here's what you need to know about these two properties:
  • Additivity: This property implies that a transformation preserves vector addition. If \( T \) is linear, then for any vectors \( \mathbf{u} \) and \( \mathbf{v} \), \( T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v}) \). This rule ensures that the transformation respects the algebraic structure of vector spaces.
  • Scalar Multiplication: A linear transformation must respect scalar multiplication. This means for any vector \( \mathbf{v} \) and scalar \( c \), \( T(c\mathbf{v}) = cT(\mathbf{v}) \). This property guarantees consistency in scaling vectors before and after transformation.
These two properties were used in the given exercise to verify the linearity of transformation \( T \). Since all steps confirming both additivity and scalar multiplication were rightly followed, \( T \) was indeed shown to be linear, mapping vectors from \( V \) to \( \mathbb{R}^2 \) while maintaining the core principles of linear transformations.

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Most popular questions from this chapter

Let \(S\) and \(T\) be linear transformations \(V \rightarrow W,\) where \(\operatorname{dim} V=n\) and \(\operatorname{dim} W=m\) a. Show that \(\operatorname{ker} S=\operatorname{ker} T\) if and only if \(T=R S\) for some isomorphism \(R: W \rightarrow W\). [Hint: Let \(\left\\{\mathbf{e}_{1}, \ldots, \mathbf{e}_{r}, \ldots, \mathbf{e}_{n}\right\\}\) be a basis of \(V\) such that \(\left\\{\mathbf{e}_{r+1}, \ldots, \mathbf{e}_{n}\right\\}\) is a basis of \(\operatorname{ker} S=\operatorname{ker} T\). Use Theorem 7.2 .5 to extend \(\left\\{S\left(\mathbf{e}_{1}\right), \ldots, S\left(\mathbf{e}_{r}\right)\right\\}\) and \(\left\\{T\left(\mathbf{e}_{1}\right), \ldots, T\left(\mathbf{e}_{r}\right)\right\\}\) to bases of \(\left.W .\right]\) b. Show that \(\operatorname{im} S=\operatorname{im} T\) if and only if \(T=S R\) for some isomorphism \(R: V \rightarrow V\). [Hint: Show that \(\operatorname{dim}(\operatorname{ker} S)=\operatorname{dim}(\) ker \(T)\) and choose bases \(\left\\{\mathbf{e}_{1}, \ldots, \mathbf{e}_{r}, \ldots, \mathbf{e}_{n}\right\\}\) and \(\left\\{\mathbf{f}_{1}, \ldots, \mathbf{f}_{r}, \ldots, \mathbf{f}_{n}\right\\}\) of \(V\) where \(\left\\{\mathbf{e}_{r+1}, \ldots, \mathbf{e}_{n}\right\\}\) and \(\left\\{\mathbf{f}_{r+1}, \ldots, \mathbf{f}_{n}\right\\}\) are bases of ker \(S\) and ker \(T,\) respectively. If \(1 \leq i \leq r,\) show that \(S\left(\mathbf{e}_{i}\right)=T\left(\mathbf{g}_{i}\right)\) for some \(\mathbf{g}_{i}\) in \(V,\) and prove that \(\left\\{\mathbf{g}_{1}, \ldots, \mathbf{g}_{r}, \mathbf{f}_{r+1}, \ldots, \mathbf{f}_{n}\right\\}\) is a basis of \(\left.V .\right]\)

\(\begin{array}{lll}\text { Exercise } & 7.3 .24 & \text { Let } V \text { consist of all sequences }\end{array}\) \(\left[x_{0}, x_{1}, x_{2}, \ldots\right)\) of numbers, and define vector operations $$ \begin{aligned} \left[x_{o}, x_{1}, \ldots\right)+\left[y_{0}, y_{1}, \ldots\right) &=\left[x_{0}+y_{0}, x_{1}+y_{1}, \ldots\right) \\ r\left[x_{0}, x_{1}, \ldots\right) &=\left[r x_{0}, r x_{1}, \ldots\right) \end{aligned} $$ a. Show that \(V\) is a vector space of infinite dimension. b. Define \(T: V \rightarrow V\) and \(S: V \rightarrow V\) by \(T\left[x_{0}, x_{1}, \ldots\right)=\left[x_{1}, x_{2}, \ldots\right)\) and \(S\left[x_{0}, x_{1}, \ldots\right)=\left[0, x_{0}, x_{1}, \ldots\right) .\) Show that \(T S=1_{V},\) so \(T S\) is one-to-one and onto, but that \(T\) is not one-to-one and \(S\) is not onto.

If \(T: \mathbf{M}_{n n} \rightarrow \mathbb{R}\) is any linear transformation satisfying \(T(A B)=T(B A)\) for all \(A\) and \(B\) in \(\mathbf{M}_{n n}\) show that there exists a number \(k\) such that \(T(A)=k \operatorname{tr} A\) for all A. (See Lemma 5.5.1.) [Hint: Let \(E_{i j}\) denote the \(n \times n\) matrix with 1 in the \((i, j)\) position and zeros elsewhere. Show that \(E_{i k} E_{l j}=\left\\{\begin{array}{cl}0 & \text { if } k \neq l \\\ E_{i j} & \text { if } k=l\end{array}\right.\). Use this to show that \(T\left(E_{i j}\right)=0\) if \(i \neq j\) and \(T\left(E_{11}\right)=T\left(E_{22}\right)=\cdots=T\left(E_{n n}\right) .\) Put \(k=T\left(E_{11}\right)\) and use the fact that \(\left\\{E_{i j} \mid 1 \leq i, j \leq n\right\\}\) is a basis of \(\mathbf{M}_{n n}\).

Let \(T: V \rightarrow \mathbb{R}\) be a nonzero linear transformation, where \(\operatorname{dim} V=n\). Show that there is a basis \(\left\\{\mathbf{e}_{1}, \ldots, \mathbf{e}_{n}\right\\}\) of \(V\) so that \(T\left(r_{1} \mathbf{e}_{1}+r_{2} \mathbf{e}_{2}+\cdots+r_{n} \mathbf{e}_{n}\right)=r_{1}\)

Let \(f \neq 0\) be a fixed polynomial of degree \(m \geq 1\). If \(p\) is any polynomial, recall that \((p \circ f)(x)=p[f(x)] .\) Define \(T_{f}: P_{n} \rightarrow P_{n+m}\) by \(T_{f}(p)=p \circ f\) a. Show that \(T_{f}\) is linear. b. Show that \(T_{f}\) is one-to-one.

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