Problem 31 - Differential Equations and Linear Algebra

Question

Let \(S_{1}\) and \(S_{2}\) be subspaces of a vector space \(V .\) Let $$ \begin{array}{l} S_{1} \cup S_{2}=\left\\{\mathbf{v} \in V: \mathbf{v} \in S_{1} \text { or } \mathbf{v} \in S_{2}\right\\} \\ S_{1} \cap S_{2}=\left\\{\mathbf{v} \in V: \mathbf{v} \in S_{1} \text { and } \mathbf{v} \in S_{2}\right\\} \end{array} $$ and let $$S_{1}+S_{2}=\left\\{\mathbf{v} \in V: \mathbf{v}=\mathbf{x}+\mathbf{y} \text { for some } \mathbf{x} \in S_{1} \text { and } \mathbf{y} \in S_{2}\right\\}$$ (a) Show that, in general, \(S_{1} \cup S_{2}\) is not a subspace of \(V\) (b) Show that \(S_{1} \cap S_{2}\) is a subspace of \(V\). (c) Show that \(S_{1}+S_{2}\) is a subspace of \(V\).

Step-by-Step Solution

Answer

In summary, the union, intersection, and sum of subspaces \(S_1\) and \(S_2\) of vector space \(V\) have the following properties: (a) \(S_{1} \cup S_{2}\) is not a subspace of \(V\) in general, as it may fail to be closed under vector addition. (b) \(S_{1} \cap S_{2}\) is a subspace of \(V\), as it contains the zero vector and is closed under vector addition and scalar multiplication. (c) \(S_{1}+S_{2}\) is a subspace of \(V\), as it contains the zero vector and is closed under vector addition and scalar multiplication.

Step 1: (a) Verify that \(S_{1} \cup S_{2}\) is not a subspace of \(V\) #

To show that \(S_{1} \cup S_{2}\) is not a subspace, we can provide a counterexample. Let \(\mathbf{v}_1 \in S_1\) and \(\mathbf{v}_2 \in S_2\) such that \(\mathbf{v}_1 \notin S_2\) and \(\mathbf{v}_2 \notin S_1\). Then, \(\mathbf{v}_1, \mathbf{v}_2 \in S_{1} \cup S_{2}\). Let's see if the resultant set is closed under vector addition. Let \(\mathbf{v}_1+\mathbf{v}_2=\mathbf{v}\). We have no guarantee that \(\mathbf{v} \in S_1\) or \(\mathbf{v} \in S_2\), hence it might not be true that \(\mathbf{v} \in S_1 \cup S_2\). As a consequence, the set \(S_1 \cup S_2\) could fail to be closed under vector addition and thus may not be a subspace of V. This is true for the union of two planes in \(\mathbb{R}^3\).

Step 2: (b) Verify that \(S_{1} \cap S_{2}\) is a subspace of \(V\) #

Let \(S = S_{1} \cap S_{2}\). To show that S is a subspace of V, we need to verify that S contains the zero vector and is closed under vector addition and scalar multiplication. 1. Since S₁ and S₂ are subspaces of V, they both contain the zero vector. Thus, the zero vector is also in their intersection, S. 2. Let \(\mathbf{v}_1, \mathbf{v}_2 \in S\). Then, \(\mathbf{v}_1, \mathbf{v}_2 \in S_1\) and \(\mathbf{v}_1, \mathbf{v}_2 \in S_2\). Since S₁ and S₂ are subspaces, their sums and scalar multiples also belong to S₁ and S₂. Therefore, \((\mathbf{v}_1+\mathbf{v}_2) \in S_1 \cap S_2\) and \(c\mathbf{v}_1 \in S_1 \cap S_2\), where c is any scalar. Since S contains the zero vector and is closed under vector addition and scalar multiplication, \(S_{1} \cap S_{2}\) is a subspace of V.

Step 3: (c) Verify that \(S_{1}+S_{2}\) is a subspace of \(V\) #

Let \(S = S_{1}+S_{2}\). To show that S is a subspace of V, we need to verify that S contains the zero vector and is closed under vector addition and scalar multiplication. 1. Since S₁ and S₂ are subspaces of V, they both contain the zero vector. Thus, zero vector can also be represented as a sum of these zero vectors: \(0 = 0 + 0\), and therefore, \(0 \in S_1 + S_2\). 2. Let \(\mathbf{v}_1, \mathbf{v}_2 \in S\). Then, \(\mathbf{v}_1 = \mathbf{x}_1 + \mathbf{y}_1\) for some \(\mathbf{x}_1 \in S_1\) and \(\mathbf{y}_1 \in S_2\), and \(\mathbf{v}_2 = \mathbf{x}_2 + \mathbf{y}_2\) for some \(\mathbf{x}_2 \in S_1\) and \(\mathbf{y}_2 \in S_2\). Now, we have to check if the sum \(\mathbf{v}_1 + \mathbf{v}_2\) and scalar multiple \(c\mathbf{v}_1\) belong to S: - \((\mathbf{v}_1+\mathbf{v}_2) = (\mathbf{x}_1 + \mathbf{y}_1) + (\mathbf{x}_2 + \mathbf{y}_2) = (\mathbf{x}_1 + \mathbf{x}_2) + (\mathbf{y}_1 + \mathbf{y}_2)\). Since S₁ and S₂ are subspaces of V, \((\mathbf{x}_1 + \mathbf{x}_2) \in S_1\) and \((\mathbf{y}_1 + \mathbf{y}_2) \in S_2\), so their sum belongs to S. - \(c\mathbf{v}_1 = c(\mathbf{x}_1 + \mathbf{y}_1) = c\mathbf{x}_1 + c\mathbf{y}_1\). Since S₁ and S₂ are subspaces of V, \(c\mathbf{x}_1 \in S_1\) and \(c\mathbf{y}_1 \in S_2\), so scalar multiple belongs to S. Since S contains the zero vector and is closed under vector addition and scalar multiplication, \(S_{1} + S_{2}\) is a subspace of V.

Key Concepts

Union of Subspaces

In vector spaces, when we talk about the union of two subspaces, we think of combining all the elements from both subspaces. Let’s say you have two subspaces, \( S_1 \) and \( S_2 \). The union \( S_1 \cup S_2 \) includes any vector that belongs to either \( S_1 \) or \( S_2 \) or both.
However, there is a catch! For this union to be a subspace, all subspace properties must hold, such as closure under addition and scalar multiplication. Unfortunately, in practice, unions of subspaces often don't satisfy these properties.

For instance, consider two planes in three-dimensional space (\( \mathbb{R}^3 \)). If you take any vector from the first plane and another from the second, their sum might not lie in either plane, failing the closure under addition requirement. Thus, \( S_1 \cup S_2 \) typically is not a subspace.

Intersection of Subspaces

The intersection of two subspaces is more predictable than the union. When you intersect \( S_1 \) and \( S_2 \), you only include vectors common to both subspaces. In other words, a vector must belong to both subspaces to be in their intersection, denoted as \( S_1 \cap S_2 \).

Why is this intersection always a subspace? Because it naturally inherits the properties needed. It contains the zero vector since both subspaces contain zero, and zero is common to any subspace. It’s closed under addition and scalar multiplication because if two vectors are in both subspaces, their sum or any scalar multiple is too!

• Contains the zero vector
• Closed under vector addition
• Closed under scalar multiplication

Thus, \( S_1 \cap S_2 \) will always be a subspace of \( V \).

Sum of Subspaces

The sum of subspaces \( S_1 \) and \( S_2 \), represented as \( S_1 + S_2 \), essentially takes all possible sums of vectors from \( S_1 \) and \( S_2 \). This means every vector in \( S_1 + S_2 \) can be expressed as \( \mathbf{v} = \mathbf{x} + \mathbf{y} \) where \( \mathbf{x} \in S_1 \) and \( \mathbf{y} \in S_2 \).

The sum of subspaces is always a subspace. Here’s why:

• Contains the zero vector: The zero vector results from adding zero vectors from \( S_1 \) and \( S_2 \).
• Closed under addition: If you take two vectors in \( S_1 + S_2 \), say \( \mathbf{v}_1 + \mathbf{v}_2 \), they are sums \( (\mathbf{x}_1 + \mathbf{y}_1) + (\mathbf{x}_2 + \mathbf{y}_2) \), which is \((\mathbf{x}_1 + \mathbf{x}_2) + (\mathbf{y}_1 + \mathbf{y}_2) \). Both new sums belong to \( S_1 \) and \( S_2 \) respectively.
• Closed under scalar multiplication: A vector can be scaled while remaining in \( S_1 + S_2 \) since scaling respects subspace properties.

Therefore, layouts of \( S_1 + S_2 \) form a subspace of \( V \).

Subspace Verification

Verifying if a set forms a subspace involves checking three fundamental criteria.

**Zero Vector Inclusion:** Every subspace must have the zero vector. This ensures that every subspace is non-empty and forms the base of vector operations.

**Closure under Addition:** When you add any two vectors, they should stay within the set. For instance, if \( \mathbf{u} \) and \( \mathbf{v} \) are both in the subspace, their sum \( \mathbf{u} + \mathbf{v} \) should also be in that subspace.

**Closure under Scalar Multiplication:** Multiplying vectors by a scalar should not take them outside the set. If \( \mathbf{u} \) is in the subspace, then any scalar multiple \( c\mathbf{u} \) should also remain within the subspace.

• Check for these properties on any subset to determine if it’s a subspace.
• If all conditions are true, you have a subspace.

This systematic check ensures that the subset behaves correctly under the rules of vector addition and scalar multiplication.