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Chapter 1: Problem 8

Prove that the intersection of any collection of subspaces of \(V\) is a subspace of \(V\)

Short Answer

Expert verified
The intersection of a collection of subspaces \(W_i\) of \(V\) can be defined as the set \(W = \bigcap_{i = 1}^n W_i\). We can prove that the intersection \(W\) is a subspace of \(V\) by showing that it satisfies the subspace criteria: containing the zero vector, closure under vector addition, and closure under scalar multiplication. Since each subspace \(W_i\) contains the zero vector and is closed under both vector addition and scalar multiplication, the intersection \(W\) must also satisfy these criteria, proving that the intersection is indeed a subspace of \(V\).

Step by step solution

01

Define the intersection of subspaces

Let \(W_i\) be a collection of subspaces of V, where \(i = 1, 2, 3, ..., n\). Define the intersection of these subspaces as the set \(W = \bigcap_{i = 1}^n W_i\). By definition, a vector \(v\) is contained in the intersection W if and only if \(v\) is contained in every subspace \(W_i\).
02

Prove the zero vector is in the intersection

To prove that the intersection W contains the zero vector, we can show that the zero vector is present in each of the subspaces \(W_i\). Since each \(W_i\) is a subspace, they all contain the zero vector. Therefore, the intersection of all subspaces W must also contain the zero vector (0).
03

Prove closure under vector addition

Consider any two vectors \(u\) and \(v\) in the intersection W. Since \(u\) and \(v\) are in W, they are also in each of the subspaces \(W_i\). Since each \(W_i\) is a subspace and is closed under vector addition, the sum \(u + v\) must also be in each \(W_i\). Therefore, the sum \(u + v\) is in the intersection W, and W is closed under vector addition.
04

Prove closure under scalar multiplication

Lastly, consider a vector \(u\) in the intersection W and a scalar constant \(c\). Since \(u\) is in W, it is also in each of the subspaces \(W_i\). Because each \(W_i\) is a subspace and is closed under scalar multiplication, the scalar product \(cu\) must also be in each \(W_i\). Thus, the scalar product \(cu\) is in the intersection W, and W is closed under scalar multiplication.
05

Conclusion

Since the intersection W satisfies all three subspace criteria (contains the zero vector, is closed under vector addition, and closed under scalar multiplication), it must also be a subspace of the vector space V. Therefore, we have proved that the intersection of any collection of subspaces of \(V\) is a subspace of \(V\).

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Key Concepts

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

Linear Algebra
At its core, linear algebra is the branch of mathematics concerning linear equations and their representations through vectors and matrices. It is essential in various fields such as physics, engineering, computer science, and more.

In tackling problems within linear algebra, understanding the nature of vector spaces and their subspaces becomes crucial. The exercise regarding the intersection of subspaces exemplifies a practical application of linear algebra principles. It is through the lens of linear algebra that one explores the properties and structures which govern the interaction of these mathematical objects.
Vector Space
A vector space, also known as a linear space, is a fundamental concept in linear algebra. It consists of a collection of vectors that can be scaled and added together while adhering to specific rules.

In a vector space, two operations are defined: vector addition and scalar multiplication. The criteria for a vector space ensure that these operations do not lead outside the space, which means vectors remain within the space regardless of how they are combined using these operations. When we analyze the intersection of subspaces, we are particularly interested in seeing whether the intersection retains these properties of a vector space.
Subspace Criteria
A subspace is quite literally a 'space within a space'. More formally, it's a subset of a vector space that is, itself, a vector space under the same operations of addition and scalar multiplication.

To determine if a subset is a subspace, there are several criteria that it must meet. These are known as the subspace criteria, which include containing the zero vector, closure under vector addition, and closure under scalar multiplication. In the context of our exercise, these criteria are utilized to prove that the intersection of subspaces is indeed a subspace.
Closure Under Vector Addition
Closure under vector addition is one of the critical attributes a set must have to be considered a subspace. This means that if you take any two vectors in the set and add them together, the resulting vector must also be in the set.

Our exercise's Step 3 efficiently demonstrates this property for the intersection of subspaces. It shows that the addition of any two vectors, both belonging to the intersection, yields another vector that still lies within this intersection, satisfying the criteria for vector addition closure.
Closure Under Scalar Multiplication
The other pivotal subspace characteristic is closure under scalar multiplication. This property stipulates that if a vector is part of the set and we multiply it by any scalar, the result remains within the set.

In Step 4, the exercise clarifies that any vector from the intersection of subspaces, when multiplied by a scalar, produces a vector that doesn't stray from the intersection. This property's verification is instrumental in establishing the intersection as a subspace, in accordance with the underlying principle of scalar multiplication closure.

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Most popular questions from this chapter

For each of the following subsets of \(\mathbf{F}^{3}\), determine whether it is a subspace of \(\mathbf{F}^{3}:\) (a) \(\quad\left\\{\left(x_{1}, x_{2}, x_{3}\right) \in \mathbf{F}^{3}: x_{1}+2 x_{2}+3 x_{3}=0\right\\}\) (b) \(\quad\left\\{\left(x_{1}, x_{2}, x_{3}\right) \in \mathbf{F}^{3}: x_{1}+2 x_{2}+3 x_{3}=4\right\\}\) (c) \(\quad\left\\{\left(x_{1}, x_{2}, x_{3}\right) \in \mathbf{F}^{3}: x_{1} x_{2} x_{3}=0\right\\}\) (d) \(\quad\left\\{\left(x_{1}, x_{2}, x_{3}\right) \in \mathbf{F}^{3}: x_{1}=5 x_{3}\right\\}\)

Prove or give a counterexample: if \(U_{1}, U_{2}, W\) are subspaces of \(V\) such that \\[ U_{1}+W=U_{2}+W \\] \(\operatorname{then} U_{1}=U_{2}\)

Suppose \(a\) and \(b\) are real numbers, not both 0. Find real numbers \(c\) and \(d\) such that \\[ 1 /(a+b i)=c+d i \\]

Is the operation of addition on the subspaces of \(V\) commutative? Associative? (In other words, if \(U_{1}, U_{2}, U_{3}\) are subspaces of \(V\), is \\[ \left.U_{1}+U_{2}=U_{2}+U_{1} ? \text { Is }\left(U_{1}+U_{2}\right)+U_{3}=U_{1}+\left(U_{2}+U_{3}\right) ?\right) \\]

Does the operation of addition on the subspaces of \(V\) have an additive identity? Which subspaces have additive inverses?
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