Question

Let \(W_1\) and \(W_2\) are two subspaces of a vector space V over the field F, then :

• A. \(W_1 + W_2\) not subspace of V
• B. \(W_1 \cup W_2\) is always subspace of V
• C. \(W_1 \cap W_2\) is always subspace of V
• D. \(W_1\) and \(W_2\) are always orthogonal

Hint:

Keep these fundamental rules in mind:
- Intersection (\(\cap\)) of subspaces is always a subspace.
- Sum (\(+\)) of subspaces is always a subspace.
- Union (\(\cup\)) of subspaces is only sometimes a subspace (when one is subset of the other).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
A subset \(W\) of a vector space \(V\) over a field \(F\) is called a subspace if it is itself a vector space under the operations of vector addition and scalar multiplication defined on \(V\).
This requires \(W\) to contain the zero vector, be closed under addition, and be closed under scalar multiplication.

Step 2: Detailed Explanation:
Let us analyze the standard theorems regarding operations on two subspaces \(W_1\) and \(W_2\):
1. Intersection of Subspaces: The intersection of any two subspaces \(W_1 \cap W_2\) is always a subspace of \(V\).
Let us prove this:
- Since \(W_1\) and \(W_2\) are subspaces, they both contain the zero vector \(0\). Thus, \(0 \in W_1 \cap W_2\).
- Let \(u, v \in W_1 \cap W_2\) and \(\alpha, \beta \in F\).
- Since \(u, v \in W_1\) and \(W_1\) is a subspace, \(\alpha u + \beta v \in W_1\) (by closure).
- Similarly, since \(u, v \in W_2\) and \(W_2\) is a subspace, \(\alpha u + \beta v \in W_2\).
- Since the linear combination lies in both \(W_1\) and \(W_2\), it must lie in their intersection: \(\alpha u + \beta v \in W_1 \cap W_2\).
Therefore, \(W_1 \cap W_2\) is closed under linear combinations and is always a subspace. This makes Option (C) correct.
Let us examine why the other options are incorrect:
- The sum of two subspaces \(W_1 + W_2 = \{w_1 + w_2 \mid w_1 \in W_1, w_2 \in W_2\}\) is always a subspace of \(V\). Option (A) claims it is not, which is false.
- The union \(W_1 \cup W_2\) is not always a subspace. It is a subspace if and only if one of the subspaces is contained in the other (\(W_1 \subseteq W_2\) or \(W_2 \subseteq W_1\)). Thus, Option (B) is false.
- Two subspaces do not have to be orthogonal to each other. Thus, Option (D) is false.

Step 3: Final Answer:
The correct option is (C).