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Chapter 1: Problem 44

Let \(\mathrm{V}_{1}=\\{\mathrm{t}(1,1): \mathrm{t} \in \mathrm{R}\\}\) and \(\mathrm{V}_{2}=\left\\{\left(\mathrm{x}_{1}, \mathrm{x}_{2}\right) \in \mathrm{R}^{2}:(1,1)\left(\mathrm{x}_{1} \mathrm{x}_{2}\right)=0\right\\}\) be two subsets of \(\mathrm{R}^{2}\) Show that \(\mathrm{R}^{2}\) is the direct sum of \(\mathrm{V}_{1}\) and \(\mathrm{V}_{2}: \mathrm{R}^{2}=\mathrm{V}_{1} \oplus \mathrm{V}_{2}\).

Short Answer

Expert verified
The given subsets of 鈩澛 are V鈧 = {t(1,1): t 鈭 鈩潁, representing all scalar multiples of the vector (1,1), and V鈧 = {(x鈧,x鈧) 鈭 鈩澛: (1,1)(x鈧亁鈧) = 0}, representing all vectors orthogonal to (1,1). To show that 鈩澛 is the direct sum of V鈧 and V鈧, we need to prove two conditions: 1. V鈧 + V鈧 = 鈩澛, which means every vector in 鈩澛 can be written as the sum of a vector in V鈧 and a vector in V鈧. By solving a system of equations and finding a unique representation for a vector (a,b) 鈭 鈩澛, we show that V鈧 + V鈧 = 鈩澛. 2. V鈧 鈭 V鈧 = {0}, which means the only vector common to V鈧 and V鈧 is the zero vector. By supposing there exists a non-zero vector in V鈧 鈭 V鈧 and reaching a contradiction, we show that V鈧 鈭 V鈧 = {0}. Hence, 鈩澛 is the direct sum of V鈧 and V鈧: 鈩澛 = V鈧 鈯 V鈧.

Step by step solution

01

1. Understand the given subsets

First, let us analyze the given subsets. V鈧 = {t(1,1): t 鈭 鈩潁 represents all scalar multiples of the vector (1,1). V鈧 = {(x鈧,x鈧) 鈭 鈩澛: (1,1)(x鈧亁鈧) = 0} represents all vectors orthogonal to (1,1), as their dot product is zero.
02

2. Show V鈧 + V鈧 = 鈩澛

We need to show that any vector in 鈩澛 can be written as the sum of a vector in V鈧 and a vector in V鈧. Let (a,b) be a vector in 鈩澛. We want to find real numbers t, x鈧, and x鈧 such that: (a,b) = t(1,1) + (x鈧,x鈧) with the condition that (1,1)(x鈧亁鈧) = 0 (since (x鈧,x鈧) 鈭 V鈧). Expanding the equation, we get: (a,b) = (t+x鈧, t+x鈧) Now, we can solve the system of equations: - t + x鈧 = a - t + x鈧 = b (1,1)(x鈧,x鈧) = x鈧 + x鈧 = 0 Using the condition x鈧 + x鈧 = 0, we have x鈧 = -x鈧. Now, we can express x鈧 in terms of a and b, and substitute it into the equation for x鈧: - a - b = x鈧 - x鈧 = x鈧 = -(a+b) - x鈧 = a+b Therefore, there exists a unique representation for (a,b) with t = (a+b)/2, x鈧 = a+b, and x鈧 = -(a+b) as elements in V鈧 and V鈧, respectively.
03

3. Show V鈧 鈭 V鈧 = {0}

Now, we need to prove that the only vector common to V鈧 and V鈧 is the zero vector. Suppose there exists a non-zero vector v = (v鈧, v鈧) 鈭 V鈧 鈭 V鈧. Then, v is a scalar multiple of (1,1) (from V鈧) and orthogonal to (1,1) (from V鈧). So, v = t(1,1) for some t 鈮 0, and (1,1)(v鈧,v鈧) = 0. Expanding and substituting v: (1,1)(t, t) = t + t = 2t = 0 However, this implies t= 0, which contradicts the assumption that t 鈮 0. Therefore, no such non-zero vector exists in V鈧 鈭 V鈧 and V鈧 鈭 V鈧 = {0}. Since we have shown that V鈧 + V鈧 = 鈩澛 and V鈧 鈭 V鈧 = {0}, we conclude that 鈩澛 is the direct sum of V鈧 and V鈧: 鈩澛 = V鈧 鈯 V鈧.

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Key Concepts

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

Vector Spaces
Vector spaces are foundational in linear algebra and consist of a set of vectors that can be added together and multiplied by scalars. The concept is essential for understanding more complex structures like direct sums.

In our exercise, we have two subsets of \(\mathbb{R}^2\):
  • \(V_1 = \{ t(1,1) : t \in \mathbb{R} \}\), which includes all vectors that are scalar multiples of \((1,1)\). This is a one-dimensional subspace, meaning it stretches along a single line in the plane.
  • \(V_2 = \{ (x_1, x_2) \in \mathbb{R}^2 : (1,1) \cdot (x_1, x_2) = 0 \}\) contains vectors orthogonal to \((1,1)\). These are all the vectors whose dot product with \((1,1)\) equals zero, forming another one-dimensional subspace.
Understanding these subsets as vector spaces helps in analyzing how they can be combined to form the entire space \(\mathbb{R}^2\). This combination is what we explore next with the direct sum concept.
Orthogonal Vectors
Orthogonal vectors have a dot product of zero, signifying they are perpendicular to each other in a geometric sense. This property is integral to understanding the relationship between subspaces.

In the exercise, \(V_2\) is made up of vectors perpendicular to \((1,1)\). Let's break down why this is important:
  • Orthogonality ensures that vectors in \(V_2\) do not align with those in \(V_1\), which includes vectors like \(t(1,1)\).
  • Since the vectors are orthogonal, \(x_1 + x_2 = 0\), ensuring no contribution along the direction of \((1,1)\).
This orthogonal nature is critical when showing \(\mathbb{R}^2\) as a direct sum of the two spaces, as it allows them to complement each other fully without overlapping.
Intersection of Subspaces
The intersection of subspaces deals with the common elements shared by two subspaces. To consider the vector spaces \(V_1\) and \(V_2\) as a direct sum, they should only share the zero vector in common.

Here's why that's vital:
  • If \(V_1 \cap V_2 = \{0\}\), it means \(V_1\) and \(V_2\) have no other vectors in common.
  • Our exercise verifies this by demonstrating any vector being both in \(V_1\) and \(V_2\) leads to a contradiction unless it's the zero vector.
This result is pivotal because it ensures that any vector in \(\mathbb{R}^2\) can uniquely be expressed as a sum of vectors from \(V_1\) and \(V_2\), achieving the condition for a direct sum. By understanding this, we can see how \(\mathbb{R}^2 = V_1 \oplus V_2\).

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Most popular questions from this chapter

Give an example of a pair of non-collinear vectors in \(R^{2}\). Then, show that the point \(\left(\mathrm{x}_{1}, \mathrm{x}_{2}\right)=(8,7)\) can be expressed as a linear combination of the non-collinear vectors.

Show that \(\mathrm{V}\), the set of all functions from a set \(\mathrm{S} \neq \boldsymbol{c}\) to the field \(R\), is a vector space under the following operations: if \(\mathrm{f}(\mathrm{s})\) and \(\mathrm{g}(\mathrm{s}) \in \mathrm{V}\), then \((\mathrm{f}+\mathrm{g})(\mathrm{s})=\mathrm{f}(\mathrm{s})+\mathrm{g}(\mathrm{s}) .\) If \(\mathrm{c}\) is a scalar from \(R\), then \((c f)(s)=c f(s)\).

Are the following sets subspaces? i) \(\mathrm{W} \subset \mathrm{R}^{3}\) consists of all vectors of the form \((\mathrm{a}, \mathrm{b}, 1)\) where \(\mathrm{a}\) and \(\mathrm{b}\) are any real numbers. ii) \(W \subset R^{4}\) contains all vectors of the form \((a, b, a-b,\), \(a+2 b\) ) for \(a, b \in R\)

Let \(\mathrm{T}\) C \(\mathrm{R}^{3}\) be \(\\{(1,0,0),(0,1,1)\\}\). What is the span of \(\mathrm{T}\), \(\mathrm{Sp}(\mathrm{T}) ?\)

Find the components of the vector \(\mathrm{v}=\mathrm{P}_{1} \mathrm{P}_{2}\) with initial point \(\mathrm{P}_{1}(2,-1,4)\) and terminal point \(\mathrm{P}_{2}(7,5,-8)\).
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