Question

If \(U\) and \(W\) are subspaces of a vector space \(V\), let \(U \cup W=\\{\mathbf{v} \mid \mathbf{v}\) is in \(U\) or \(\mathbf{v}\) is in \(W\\}\). Show that \(U \cup W\) is a subspace if and only if \(U \subseteq W\) or \(W \subset U\).

Short answer

\(U \cup W\) is a subspace if and only if \(U \subseteq W\) or \(W \subseteq U\).

Step-by-step solution

Verify the Basics

Recall that for a set to be a subspace, it must be non-empty, closed under addition, and closed under scalar multiplication.

Assume Condition \\(U \cup W\\) is a Subspace

Suppose that \(U \cup W\) is a subspace. Since subspaces must include the zero vector, 0 must be in \(U \cup W\), ensuring \(0 \in U\) or \(0 \in W\).

Check Closure Under Addition

If \(U \cup W\) is a subspace, then it must be closed under addition. Take \(\mathbf{u} \in U\) and \(\mathbf{w} \in W\), both in \(U \cup W\). Then \(\mathbf{u} + \mathbf{w}\) must also be in \(U \cup W\).

Confirm Subset Condition from Addition

The closure under addition means if \(\mathbf{u} \in U\) and \(\mathbf{w} \in W\), then \(\mathbf{u} + \mathbf{w} \in U\) or \(\mathbf{u} + \mathbf{w} \in W\). Therefore, \(U \subseteq W\) or \(W \subseteq U\), otherwise the initial assumption fails.

Assume Subset Condition \\(U \subseteq W\\) or \\(W \subseteq U\\)

If \(U \subseteq W\), then \(U \cup W = W\), which is a subspace. Similarly, if \(W \subseteq U\), then \(U \cup W = U\), which is also a subspace.

Conclusion

Hence, if \(U \cup W\) is a subspace, then \(U \subseteq W\) or \(W \subseteq U\), and vice versa. This establishes the necessary and sufficient condition.

Key concepts

Subspaces

A subspace is a key concept in linear algebra and is essentially a "space within a space." It's a subset of a vector space that is itself a vector space under the same operations of vector addition and scalar multiplication. To qualify as a subspace, a set must satisfy three essential conditions:

• It must be non-empty, typically containing the zero vector since the zero vector is in every vector space.
• It must be closed under vector addition, meaning if you add any two vectors in the subspace, the resulting vector must also be in the subspace.
• It must be closed under scalar multiplication, meaning if you multiply any vector in the subspace by a scalar (a real number), the resulting vector must also be in the subspace.

Understanding these properties helps in identifying and working with subspaces in vector spaces.

Set Theory in Mathematics

Set theory is the mathematical foundation that deals with the collection of objects or elements. In the context of subspaces, set operations help us understand how two subspaces combine or interact within a larger vector space. Here, we explore the union of sets.
The union of two sets, noted as \(U \cup W\), includes every element that belongs to either set \(U\) or set \(W\). It essentially combines the elements of both sets into one larger set. However, this union is not always a subspace.
In the specific case of being a union of subspaces, \(U \cup W\) forms a subspace only if one subspace is contained within the other, ensuring closure properties are maintained. Thus, using set theory helps in formulating and understanding the conditions necessary for \(U \cup W\) to be a subspace.

Closure Properties

Closure properties are essential in ensuring a union of subspaces forms a valid subspace itself. Closure under addition and scalar multiplication are both pivotal to determining the subspace status.

• Closure under Addition: This means that if two vectors are in the subspace, then their sum must also be in the same subspace. For \(U \cup W\) to be closed under addition, if you have a vector from \(U\) and another from \(W\), their sum must lie within either \(U\) or \(W\).
• Closure under Scalar Multiplication: This property states that multiplying any vector in the subspace by a scalar should still yield a vector within the subspace. This is straightforward when considering the condition \(U \subseteq W\) or \(W \subseteq U\), as all vectors abide by the operation constraints of one subspace.

Closure properties thus ensure that the set can operate independently within the vector space without stepping outside the defined boundaries of the subspace.