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If \(U\) and \(V\) are vector spaces, define the Cartesian product of \(U\) and \(V\) to be \(U \times V=\\{(\mathbf{u}, \mathbf{v}): \mathbf{u} \text { is in } U \text { and } \mathbf{v} \text { is in } V\\}\) Prove that \(U \times V\) is a vector space.

Short Answer

Expert verified
The Cartesian product \(U \times V\) is a vector space.

Step by step solution

01

Define the Elements

The Cartesian product of the vector spaces \(U\) and \(V\) is defined as \(U \times V = \{ (\mathbf{u}, \mathbf{v}): \mathbf{u} \text{ is in } U \text{ and } \mathbf{v} \text{ is in } V \}\). An element of \(U \times V\) is a tuple \((\mathbf{u}, \mathbf{v})\).
02

Verify Vector Space Axioms

To show that \(U \times V\) is a vector space, we must verify that it satisfies the vector space axioms: closure under addition and scalar multiplication, existence of a zero vector, existence of additive inverses, commutative and associative properties of addition, and distributive properties.
03

Closure Under Addition

Consider two elements \((\mathbf{u_1}, \mathbf{v_1})\) and \((\mathbf{u_2}, \mathbf{v_2})\) in \(U \times V\). Their sum is \((\mathbf{u_1} + \mathbf{u_2}, \mathbf{v_1} + \mathbf{v_2})\). Since \(U\) and \(V\) are vector spaces, \(\mathbf{u_1} + \mathbf{u_2}\) is in \(U\) and \(\mathbf{v_1} + \mathbf{v_2}\) is in \(V\), so their sum is in \(U \times V\).
04

Closure Under Scalar Multiplication

For an element \((\mathbf{u}, \mathbf{v})\) in \(U \times V\) and a scalar \(c\), consider \(c \cdot (\mathbf{u}, \mathbf{v}) = (c \cdot \mathbf{u}, c \cdot \mathbf{v})\). As \(U\) and \(V\) are vector spaces, \(c \cdot \mathbf{u}\) is in \(U\) and \(c \cdot \mathbf{v}\) is in \(V\), so the result is in \(U \times V\).
05

Existence of Zero Vector

The zero vector in \(U \times V\) is \((\mathbf{0}_U, \mathbf{0}_V)\), where \(\mathbf{0}_U\) is the zero vector in \(U\) and \(\mathbf{0}_V\) is the zero vector in \(V\). This is, indeed, an element of \(U \times V\).
06

Existence of Additive Inverses

For each element \((\mathbf{u}, \mathbf{v})\) in \(U \times V\), the additive inverse is \((-\mathbf{u}, -\mathbf{v})\), since \((-\mathbf{u}, -\mathbf{v}) + (\mathbf{u}, \mathbf{v}) = (\mathbf{0}_U, \mathbf{0}_V)\), which is the zero vector in \(U \times V\).
07

Verify Commutative and Associative Properties

Addition in \(U \times V\) is commutative: \((\mathbf{u_1}, \mathbf{v_1}) + (\mathbf{u_2}, \mathbf{v_2}) = (\mathbf{u_2}, \mathbf{v_2}) + (\mathbf{u_1}, \mathbf{v_1})\), and associative: \(((\mathbf{u_1}, \mathbf{v_1}) + (\mathbf{u_2}, \mathbf{v_2})) + (\mathbf{u_3}, \mathbf{v_3}) = (\mathbf{u_1}, \mathbf{v_1}) + ((\mathbf{u_2}, \mathbf{v_2}) + (\mathbf{u_3}, \mathbf{v_3}))\).
08

Verify Distributive Properties

The operations are distributive over vector addition and scalar addition: \(c \cdot ((\mathbf{u_1}, \mathbf{v_1}) + (\mathbf{u_2}, \mathbf{v_2})) = c \cdot (\mathbf{u_1} + \mathbf{u_2}, \mathbf{v_1} + \mathbf{v_2})\) and \((c + d) \cdot (\mathbf{u}, \mathbf{v}) = c \cdot (\mathbf{u}, \mathbf{v}) + d \cdot (\mathbf{u}, \mathbf{v})\).
09

Final Conclusion

Since \(U \times V\) satisfies all the axioms necessary for vector spaces as shown, we can conclude that \(U \times V\) is indeed a vector space.

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Key Concepts

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

Cartesian product
The Cartesian product is a mathematical operation that returns a set of all possible ordered pairs or tuples created by taking an element from each of multiple sets. For vector spaces, the Cartesian product of two vector spaces \( U \) and \( V \) is denoted as \( U \times V \). This is defined by the set of all possible pairs \((\mathbf{u}, \mathbf{v})\), where \(\mathbf{u}\) is an element in \( U \) and \(\mathbf{v}\) is in \( V \).
In practical terms, by understanding the Cartesian product, you're recognizing how elements from separate vector spaces can be combined to form ordered pairs, allowing us to investigate more complex structures. Whether it's combining coordinates or configurations, the Cartesian product grants a foundational tool for broadening vector space applications.
closure under addition
Closure under addition is a fundamental property a vector space must satisfy. It implies that when you add any two vectors in a vector space, the result will also be a vector within the same space. For \( U \times V \), this means that if you consider two elements like \( (\mathbf{u_1}, \mathbf{v_1}) \) and \( (\mathbf{u_2}, \mathbf{v_2}) \), their sum \((\mathbf{u_1} + \mathbf{u_2}, \mathbf{v_1} + \mathbf{v_2})\) will also belong to \( U \times V \).

This characteristic ensures that the vector space is stable under the operation of addition, and you can freely add vectors without leaving the space, maintaining consistency in operations.
scalar multiplication
Scalar multiplication involves multiplying a vector by a scalar (a real number), and it is crucial for vector spaces. This operation must keep the vectors in the same vector space, known as closure under scalar multiplication. For the Cartesian product \( U \times V \), given a vector \((\mathbf{u}, \mathbf{v})\) and a scalar \(c\), the product is \((c \cdot \mathbf{u}, c \cdot \mathbf{v})\).

Such an operation scales both components of the tuple by the scalar. This concept expands the flexibility of vectors by allowing them to maintain linear properties when enlarged or shrunk, keeping operations predictable and standardized.
zero vector
The zero vector is the vector equivalent of the number zero; for any vector space, there exists a special vector, known as the zero vector, which acts like an additive identity. In \( U \times V \), the zero vector is represented as \((\mathbf{0}_U, \mathbf{0}_V)\), where \(\mathbf{0}_U\) and \(\mathbf{0}_V\) are zero vectors in \( U \) and \( V \) respectively.

This vector has the unique property where adding it to any vector \((\mathbf{u}, \mathbf{v})\) results in \((\mathbf{u}, \mathbf{v})\) itself, embodying the idea of doing 'nothing' in the sense of vector addition. The presence of a zero vector is critical as it confirms the existence of an additive identity, ensuring every vector space has a centerpiece for operation symmetry.
additive inverse
In a vector space, every vector \((\mathbf{u}, \mathbf{v})\) must have an additive inverse, which when added to the vector results in the zero vector. The additive inverse of \((\mathbf{u}, \mathbf{v})\) is \((-\mathbf{u}, -\mathbf{v})\). When these vectors are added together, \((\mathbf{u}, \mathbf{v}) + (-\mathbf{u}, -\mathbf{v})\ = (\mathbf{0}_U, \mathbf{0}_V)\), which is the zero vector in the space \( U \times V \).

Understanding the concept of an additive inverse is crucial as it ensures balance within the vector space, allowing vectors and their inverses to neutralize each other, which is critical for solving equations and understanding the structure of vector spaces.

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Most popular questions from this chapter

A pendulum consists of a mass, called a bob, that is affixed to the end of a string of length \(L\) (see Figure 6.26 . When the bob is moved from its rest position and released, it swings back and forth. The time it takes the pendulum to swing from its farthest right position to its farthest left position and back to its next farthest right position is called the period of the pendulum. (Figure can't copy) Let \(\theta=\theta(t)\) be the angle of the pendulum from the vertical. It can be shown that if there is no resistance, then when \(\theta\) is small it satisfies the differential equation \\[ \theta^{\prime \prime}+\frac{g}{L} \theta=0 \\] where \(g\) is the constant of acceleration due to gravity, approximately \(9.7 \mathrm{m} / \mathrm{s}^{2} .\) Suppose that \(L=1 \mathrm{m}\) and that the pendulum is at rest (i.e., \(\theta=0\) ) at time \(t=0\) second. The bob is then drawn to the right at an angle of \(\theta_{1}\) radians and released. (a) Find the period of the pendulum. (b) Does the period depend on the angle \(\theta_{1}\) at which the pendulum is released? This question was posed and answered by Galileo in \(1638 .\) [Galileo Galilei \((1564-1642)\) studied medicine as a student at the University of Pisa, but his real interest was always mathematics. In \(1592,\) Galileo was appointed professor of mathematics at the University of Padua in Venice, where he taught primarily geometry and astronomy. He was the first to use a telescope to look at the stars and planets, and in so doing, he produced experimental data in support of the Copernican view that the planets revolve around the sun and not the earth. For this, Galileo was summoned before the Inquisition, placed under house arrest, and forbidden to publish his results. While under house arrest, he was able to write up his research on falling objects and pendulums. His notes were smuggled out of Italy and published as Discourses on Two New Sciences in \(1638 .\)

Find a formula for the number of invertible matrices \(\operatorname{in} M_{n n}\left(\mathbb{Z}_{p}\right) .\) [Hint: This is the same as determining the number of different bases for \(\mathbb{Z}_{p^{*}}^{n}\) (Why?) Count the number of ways to construct a basis for \(\mathbb{Z}_{p}^{n}\), one vector at a time.

A linear transformation \(T: V \rightarrow V\) is given. If possible, find a basis \(\mathcal{C}\) for \(V\) such that the matrix \([T]_{c}\) of \(T\) with respect to \(\mathcal{C}\) is diagonal. $$\begin{array}{l} T: \mathscr{P}_{1} \rightarrow \mathscr{P}_{1} \text { defined by } T(a+b x)=(4 a+2 b)+ \\ (a+3 b) x \end{array}$$

Which of the codes in Exercises are linear codes? $$C=\left\\{\left[\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right],\left[\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right],\left[\begin{array}{l} 0 \\ 1 \\ 1 \end{array}\right],\left[\begin{array}{l} 0 \\ 0 \\ 1 \end{array}\right]\right\\}$$

T: U \(\rightarrow\) Vand \(S: V \rightarrow W\) are linear transformations and \(\mathcal{B}, \mathcal{C},\) and \(\mathcal{D}\) are bases for \(U, V,\) and \(W\) respectively. Compute \([S \circ T]_{D \leftarrow B}\) in two ways: \((a) b y\) finding \(S \circ\) T directly and then computing its matrix and (b) by finding the matrices of S and T separately and using Theorem 6.27. $$\begin{array}{l} T: \mathscr{P}_{1} \rightarrow \mathbb{R}^{2} \text { defined by } T(p(x))=\left[\begin{array}{l} p(0) \\ p(1) \end{array}\right], S: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2} \\ \text { defined by } S\left[\begin{array}{l} a \\ b \end{array}\right]=\left[\begin{array}{l} a-2 b \\ 2 a-b \end{array}\right], \mathcal{B}=\\{1, x\\} \\ \mathcal{C}=\mathcal{D}=\left\\{\mathbf{e}_{1}, \mathbf{e}_{2}\right\\} \end{array}$$

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