Question

We often write vectors in \(\mathbb{R}^{n}\) as rows. Let \(U\) and \(W\) be subspaces of \(\mathbb{R}^{n}\). Define their intersection \(U \cap W\) and their sum \(U+W\) as follows: $$ U \cap W=\left\\{\mathbf{x} \in \mathbb{R}^{n} \mid \mathbf{x} \text { belongs to both } U \text { and } W\right\\} $$ \(U+W=\left\\{\mathbf{x} \in \mathbb{R}^{n} \mid \mathbf{x}\right.\) is a sum of a vector in \(U\) and a vector in \(W\\}\). a. Show that \(U \cap W\) is a subspace of \(\mathbb{R}^{n}\). b. Show that \(U+W\) is a subspace of \(\mathbb{R}^{n}\).

Short answer

Both \(U \cap W\) and \(U + W\) are subspaces of \(\mathbb{R}^n\).

Step-by-step solution

Understanding Intersection Criteria

For part (a), we need to show that the intersection of two subspaces, \(U \cap W\), is itself a subspace of \(\mathbb{R}^n\). A subset is a subspace if it satisfies three conditions: it contains the zero vector, is closed under vector addition, and is closed under scalar multiplication. Let's verify each of these for the intersection.

Intersection Contains Zero Vector

Since both \(U\) and \(W\) are subspaces, they both contain the zero vector \(\mathbf{0}\). Therefore, \(\mathbf{0} \in U\) and \(\mathbf{0} \in W\), which implies \(\mathbf{0} \in U \cap W\). Thus, the intersection contains the zero vector.

Intersection Closed Under Addition

Take any two vectors \(\mathbf{a}\) and \(\mathbf{b}\) in \(U \cap W\). By definition of intersection, both vectors are in \(U\) and \(W\). Since subspaces are closed under vector addition, \(\mathbf{a} + \mathbf{b} \in U\) and \(\mathbf{a} + \mathbf{b} \in W\). Therefore, \(\mathbf{a} + \mathbf{b} \in U \cap W\). The intersection is closed under addition.

Intersection Closed Under Scalar Multiplication

Take any vector \(\mathbf{a} \in U \cap W\) and any scalar \(c\). Since \(\mathbf{a}\) belongs to both \(U\) and \(W\), and both are subspaces, \(c\mathbf{a} \in U\) and \(c\mathbf{a} \in W\). Therefore, \(c\mathbf{a} \in U \cap W\). The intersection is closed under scalar multiplication.

Intersection Conclusion

Since \(U \cap W\) includes the zero vector and is closed under both addition and scalar multiplication, it is a subspace of \(\mathbb{R}^n\).

Understanding Sum Criteria

For part (b), we need to show that the sum of two subspaces, \(U + W\), is a subspace of \(\mathbb{R}^n\). Like the intersection, the sum must satisfy the same three subspace criteria.

Sum Contains Zero Vector

The zero vector in \(\mathbb{R}^n\) can be written as the sum of the zero vectors in \(U\) and \(W\) (i.e., \(\mathbf{0}_U + \mathbf{0}_W = \mathbf{0}\)). Thus, \(\mathbf{0} \in U + W\).

Sum Closed Under Addition

Take \(\mathbf{a} = \mathbf{u}_1 + \mathbf{w}_1\) and \(\mathbf{b} = \mathbf{u}_2 + \mathbf{w}_2\) with \(\mathbf{u}_1, \mathbf{u}_2 \in U\) and \(\mathbf{w}_1, \mathbf{w}_2 \in W\). Their sum, \(\mathbf{a} + \mathbf{b} = (\mathbf{u}_1 + \mathbf{w}_1) + (\mathbf{u}_2 + \mathbf{w}_2) = (\mathbf{u}_1 + \mathbf{u}_2) + (\mathbf{w}_1 + \mathbf{w}_2)\), also belongs to \(U + W\) because \(\mathbf{u}_1 + \mathbf{u}_2 \in U\) and \(\mathbf{w}_1 + \mathbf{w}_2 \in W\). Thus, \(U + W\) is closed under addition.

Sum Closed Under Scalar Multiplication

For \(\mathbf{a} = \mathbf{u} + \mathbf{w} \) in \(U + W\) and scalar \(c\), \(c\mathbf{a} = c(\mathbf{u} + \mathbf{w}) = c\mathbf{u} + c\mathbf{w}\). Since \(c\mathbf{u} \in U\) and \(c\mathbf{w} \in W\), \(c\mathbf{a} \in U + W\). Hence, \(U + W\) is closed under scalar multiplication.

Sum Conclusion

Since \(U + W\) includes the zero vector and is closed under both addition and scalar multiplication, it is a subspace of \(\mathbb{R}^n\).

Key concepts

Subspaces

In linear algebra, a subspace is a collection of vectors that exists within a larger vector space, like \( \mathbb{R}^{n} \). To qualify as a subspace, a set must meet three critical conditions: it must contain the zero vector, be closed under vector addition, and be closed under scalar multiplication.
These requirements ensure that any linear combination of vectors from the subspace also remains within the subspace itself.
For instance, think of a plane (which is a subspace) within a three-dimensional space. Every vector on that plane satisfies those three conditions.
Understanding these properties allows us to examine more complex vector constructions like intersections and sums of subspaces.

Intersection of Subspaces

The intersection of subspaces refers to the set of vectors common to two subspaces. Formally, if \( U \) and \( W \) are subspaces of \( \mathbb{R}^n \), their intersection \( U \cap W \) is defined as the set of vectors that lie in both \( U \) and \( W \).

To determine if this intersection forms a subspace, we apply the same three subspace criteria:

• Contains the Zero Vector: Since both \( U \) and \( W \) are subspaces, they each contain the zero vector. Therefore, their intersection must also include the zero vector.
• Closed Under Addition: If two vectors belong to the intersection, then their sum remains in both \( U \) and \( W \), making the sum part of the intersection.
• Closed Under Scalar Multiplication: Multiplying any vector from the intersection by a scalar results in a vector still within both subspaces.

The intersection is a more "restrictive" subspace, containing only vectors common to both original subspaces.

Sum of Subspaces

The sum of subspaces involves combining vectors from two subspaces to form another subspace. Suppose \( U \) and \( W \) are subspaces of \( \mathbb{R}^n \). Their sum, \( U+W \), consists of all possible sums of a vector from \( U \) and a vector from \( W \).

To show that \( U + W \) satisfies the subspace criteria, consider the following:

• Contains the Zero Vector: The zero vector can be seen as the sum of the zero vector from \( U \) and \( W \), so it must be included in \( U + W \).
• Closed Under Addition: For any two vectors in \( U + W \), their sum can be constructed as the addition of vectors from \( U \) and \( W \), ensuring the resultant vector remains within \( U + W \).
• Closed Under Scalar Multiplication: Scaling any vector in \( U + W \) involves scaling the individual components from \( U \) and \( W \), resulting in a vector still in \( U + W \).

The sum tends to be larger and less restrictive than the intersection, encapsulating the full span created by both subspaces.