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Chapter 1: Problem 12

Show that a subset \(W\) of a vector space \(V\) is a subspace of \(V\) if and only if \(\operatorname{span}(\mathrm{W})=\mathrm{W}\).

Short Answer

Expert verified
To prove that a subset \(W\) of a vector space \(V\) is a subspace if and only if \(\operatorname{span}(W) = W\), we need to show two things: (1) If \(W\) is a subspace, then \(\operatorname{span}(W) = W\). (2) If \(\operatorname{span}(W) = W\), then \(W\) is a subspace. For part (1), we showed that since \(W\) is a subspace, it follows the properties of having the zero vector, being closed under addition, and closed under scalar multiplication. Hence, \(\operatorname{span}(W) \subseteq W\) and \(W \subseteq \operatorname{span}(W)\), which implies \(\operatorname{span}(W) = W\). For part (2), we proved that if \(\operatorname{span}(W) = W\), then \(W\) contains the zero vector and is closed under addition and scalar multiplication, satisfying the properties of a subspace. Thus, \(W\) is a subspace of \(V\). In conclusion, subset \(W\) of a vector space \(V\) is a subspace if and only if \(\operatorname{span}(W) = W\).

Step by step solution

01

Part 1: If W is a subspace, then span(W) = W

Assume W is a subspace of V. We need to show that span(W) = W, i.e., the set of all possible linear combinations of the vectors in W is equal to W. Since W is a subspace, it follows the three properties mentioned above. Therefore, any linear combination of vectors in W will also be in W, which means span(W) 鈯� W. Now, we need to show that W 鈯� span(W). Since vectors in W are linear combinations of themselves (with scalar coefficients 1), every vector in W is inside the span(W). Therefore, W 鈯� span(W). Since span(W) 鈯� W and W 鈯� span(W), we can conclude that span(W) = W.
02

Part 2: If span(W) = W, then W is a subspace

Now assume that span(W) = W. We need to show that W follows the three properties of a subspace. a) The zero vector is in W: Since the span includes all possible linear combinations, multiplying any vector from W by the scalar 0 will result in the zero vector. Hence, the zero vector is in W. b) W is closed under vector addition: If two vectors u and v are in W, then they are also in span(W) since span(W) = W. Any linear combination of u and v is also in the span, thus the sum of u and v (u+v) is also in W. This means that W is closed under vector addition. c) W is closed under scalar multiplication: If a vector u is in W, then it is also in the span(W). Multiplying u by a scalar c will result in a linear combination cu, which is also in the span. Hence, cu is in W, and W is closed under scalar multiplication. Since W satisfies all three properties of a subspace, we can conclude that W is a subspace of V. Therefore, we have shown that if W is a subspace, then span(W) = W and if span(W) = W, then W is a subspace. This concludes the proof.

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Key Concepts

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

Span of a Set
The concept of the 'span of a set' is fundamental in understanding vector spaces. Imagine you have a box of building blocks; using these blocks, you can create various structures. Similarly, in vector spaces, the span of a set of vectors is like the collection of all different 'structures' that can be made by combining those vectors.

The mathematical way to describe this is to consider the span of a set of vectors as all possible linear combinations that can be formed with those vectors. Put simply, if you have a set of vectors, say \( \{v_1, v_2, ..., v_k\} \), the span of this set is every vector that can be expressed as \( c_1v_1 + c_2v_2 + ... + c_kv_k \), where \( c_1, c_2, ..., c_k \) are scalars. Therefore, when the question asks to show that \(\operatorname{span}(\mathrm{W})=\mathrm{W}\), it's asking us to prove that every vector in \(\mathrm{W}\) can be described as a linear combination of vectors within \(\mathrm{W}\) itself, and vice versa.
Linear Combinations
Now let's dive into the heart of the span: linear combinations. Assume you have a recipe that dictates how to mix specific ingredients to create a dish. In the world of vector spaces, a linear combination is just like that recipe, but for vectors and scalars.

To craft a linear combination, you take several vectors and 'mix' them together using scalars. The formal definition is: and a linear combination of vectors \(v_1, v_2, ..., v_n\) with corresponding scalars \(c_1, c_2, ..., c_n\) is the vector \(c_1v_1 + c_2v_2 + ... + c_nv_n\). It's the core method by which we construct the span of a set. The set of all such linear combinations constitutes the span, which can potentially represent an infinite set of vectors, depending on the original vectors chosen.
Properties of a Subspace
Let's think of a club with strict entry rules; a subspace is like that club within the broader vector space. For a subset \(W\) to be considered a subspace of a vector space \(V\), it must satisfy three main properties:

  • It must contain the zero vector,
  • It must be closed under vector addition, meaning that adding any two vectors in \(W\) gives a result that is still in \(W\),
  • And it must be closed under scalar multiplication, so if you multiply any vector in \(W\) by a scalar, the result is again a member of \(W\).
If \(W\) meets all these criteria, then it's officially a subspace, kind of like having an exclusive membership to the 'Subspace Club'!
Closure Under Vector Addition
Continuing our club analogy, closure under vector addition is one of the 'entry rules.' It specifically states that if you take any two members (vectors) of the club and combine them (add them together), the result must also be a member of the club (subspace).

Mathematically, for all vectors \(u\) and \(v\) in \(W\), the sum \(u+v\) must also be in \(W\). When showing that \(\operatorname{span}(\mathrm{W})=\mathrm{W}\), this property is crucial. It ensures that when vectors are added together, they don't produce a result that somehow escapes the set \(W\), thereby preserving the integrity of the subspace.
Closure Under Scalar Multiplication
The second 'entry rule' for our 'Subspace Club' is closure under scalar multiplication. This rule guarantees that when a scalar acts like a 'multiplier' or a 'stretcher' of a vector within the subspace, the result doesn't stretch or shrink so much that it ends up outside of the subspace.

Formally, for a vector \(u\) in \(W\) and a scalar \(c\), the product \(cu\) must remain in \(W\). That means you can adjust the magnitude of a vector as much as you want; as long as you're using scalars and the original vector is in \(W\), the new, magnified vector will still be a part of \(W\). In terms of proving \(\operatorname{span}(\mathrm{W})=\mathrm{W}\), this aspect demonstrates that scaling vectors doesn't lead to leaving the set \(W\).

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