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Chapter 4: Problem 31

Let \(\mathbf{u}\) and \(\mathbf{v}\) be vectors in a vector space \(V,\) and let \(H\) be any subspace of \(V\) that contains both \(\mathbf{u}\) and \(\mathbf{v} .\) Explain why \(H\) also contains Span \(\\{\mathbf{u}, \mathbf{v}\\} .\) This shows that \(\operatorname{Span}\\{\mathbf{u}, \mathbf{v}\\}\) is the smallest subspace of \(V\) that contains both \(\mathbf{u}\) and \(\mathbf{v} .\)

Short Answer

Expert verified
The subspace \(H\) contains \(\text{Span}\{\mathbf{u}, \mathbf{v}\}\) because it includes all linear combinations of \(\mathbf{u}\) and \(\mathbf{v}\), proving it's the smallest subspace containing them.

Step by step solution

01

Understand the Given Information

We have two vectors \(\mathbf{u}\) and \(\mathbf{v}\) in a vector space \(V\), and a subspace \(H\) of \(V\) that contains both \(\mathbf{u}\) and \(\mathbf{v}\). We need to show why \(H\) also contains \(\text{Span}\{\mathbf{u}, \mathbf{v}\}\).
02

Define the Span of Vectors

The span of vectors \(\mathbf{u}\) and \(\mathbf{v}\), denoted \(\text{Span}\{\mathbf{u}, \mathbf{v}\}\), is the set of all linear combinations of the vectors \(\mathbf{u}\) and \(\mathbf{v}\). This means any vector \(\mathbf{w}\) in the span is expressed as \(\mathbf{w} = a\mathbf{u} + b\mathbf{v}\), where \(a\) and \(b\) are scalars.
03

Utilize Subspace Properties

Since \(H\) is a subspace and contains \(\mathbf{u}\) and \(\mathbf{v}\), it must, by the properties of subspaces, also contain any linear combination of \(\mathbf{u}\) and \(\mathbf{v}\). This means \(H\) contains \(a\mathbf{u} + b\mathbf{v}\) for any scalars \(a\) and \(b\).
04

Conclude with Subspace Inclusion

Since any vector in \(\text{Span}\{\mathbf{u}, \mathbf{v}\}\) can be expressed as a linear combination of \(\mathbf{u}\) and \(\mathbf{v}\), and \(H\) contains all such linear combinations, it follows that \(\text{Span}\{\mathbf{u}, \mathbf{v}\}\) is a subset of \(H\).
05

Prove Minimality of the Span

Since \(\text{Span}\{\mathbf{u}, \mathbf{v}\}\) includes all linear combinations of \(\mathbf{u}\) and \(\mathbf{v}\) and is contained in any subspace that contains \(\mathbf{u}\) and \(\mathbf{v}\), it is the smallest subspace containing both vectors.

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Key Concepts

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

Vector Space
A vector space is a fundamental concept in linear algebra. It is a collection of objects, called vectors, that can be added together and multiplied by scalars, which are commonly real numbers. The most familiar example of a vector space is the set of all ordered pairs of real numbers
  • Vectors: Elements like \(\mathbf{u}\) and \(\mathbf{v}\) are vectors within this space.
  • Addition and Scalar Multiplication: In a vector space \(V\), for any vectors \(\mathbf{x}, \mathbf{y}\) in \(V\) and scalars \(a, b\), the vector \(a\mathbf{x} + b\mathbf{y}\) is also in \(V\).
Vector spaces are defined by a couple of key properties:
  • Closure: If you add two vectors or multiply a vector by a scalar, you remain within the vector space.
  • Zero Vector: There is a zero vector among elements in the vector space, which acts like the number zero in addition.
Understanding the principles of a vector space lays the groundwork for more complex concepts such as subspaces and span.
Subspace
A subspace is a set of vectors that forms a vector space itself within a larger vector space. If you imagine vector space \(V\) as a huge space with lots of vectors, a subspace \(H\) is like a smaller section of \(V\) that also fulfills all the conditions of a vector space.
  • Containing Vectors: In any subspace \(H\), if it contains vectors \(\mathbf{u}\) and \(\mathbf{v}\), it must also contain any vector derived as a linear combination of \(\mathbf{u}\) and \(\mathbf{v}\), such as \(a\mathbf{u} + b\mathbf{v}\).
  • Properties: A subspace must contain the zero vector from \(V\) and is closed under both vector addition and scalar multiplication.
A subspace should "inherit" the properties of its parent vector space. This inheritance ensures any operation performed within a subspace is consistent with those in the larger space.
Span
The span of a set of vectors is the collection of all possible vectors you can make using linear combinations of those vectors. It's like clay; from a few basic vectors (\(\mathbf{u}\) and \(\mathbf{v}\)), you can mold infinitely many shapes (vectors).
  • Definition: The span of vectors \(\{\mathbf{u}, \mathbf{v}\}\), denoted as \(\operatorname{Span}\{\mathbf{u}, \mathbf{v}\}\), includes every vector \(\mathbf{w}\) which can be written as \(a\mathbf{u} + b\mathbf{v}\) where \(a, b\) are scalars.
  • Smallest Subspace: The concept of span is crucial because it represents the smallest subspace that contains \(\mathbf{u}\) and \(\mathbf{v}\).
By constructing spans, we can define entire spaces using a minimal starting set of vectors.
Linear Combinations
Linear combinations are a way of creating new vectors from existing ones. When you take vectors, let's say \(\mathbf{u}\) and \(\mathbf{v}\), and combine them using scalars \(a\) and \(b\), you're forming a new vector \(a\mathbf{u} + b\mathbf{v}\).
  • Combining Elements: You take any vectors and multiply them by numbers, then add the results; this is the simplest way to mix and match vectors.
  • Creating New Vectors: Every new vector formed is still part of the same vector space or subspace, ensuring consistency in the structure.
Understanding how linear combinations work is vital because they are the building blocks for spans and tell us how vectors relate and interact within a space.

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Most popular questions from this chapter

Let \(\mathcal{B}=\left\\{1, \cos t, \cos ^{2} t, \ldots, \cos ^{6} t\right\\}\) and \(\mathcal{C}=\\{1, \cos t\) \(\cos 2 t, \ldots, \cos 6 t \\} .\) Assume the following trigonometric identities (see Exercise 37 in Section 4.1 ). \(\cos 2 t=-1+2 \cos ^{2} t\) \(\cos 3 t=-3 \cos t+4 \cos ^{3} t\) \(\cos 4 t=1-8 \cos ^{2} t+8 \cos ^{4} t\) \(\cos 5 t=5 \cos t-20 \cos ^{3} t+16 \cos ^{5} t\) \(\cos 6 t=-1+18 \cos ^{2} t-48 \cos ^{4} t+32 \cos ^{6} t\) Let \(H\) be the subspace of functions spanned by the functions in \(\mathcal{B} .\) Then \(\mathcal{B}\) is a basis for \(H,\) by Exercise 38 in Section \(4.3 .\) a. Write the \(\mathcal{B}\) -coordinate vectors of the vectors in \(\mathcal{C},\) and use them to show that \(\mathcal{C}\) is a linearly independent set in \(H .\) b. Explain why \(\mathcal{C}\) is a basis for \(H\)

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