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:

N/A

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
This question asks us to identify the correct statement about the operations on two subspaces \(W_1\) and \(W_2\) of a vector space \(V\) over a field \(F\).
Key Formula or Approach:
A subset \(W \subseteq V\) is a subspace if and only if:
1. \(\mathbf{0} \in W\) (The zero vector is in \(W\)).
2. If \(\mathbf{x}, \mathbf{y} \in W\), then \(\mathbf{x} + \mathbf{y} \in W\) (Closure under addition).
3. If \(\mathbf{x} \in W\) and \(c \in F\), then \(c\mathbf{x} \in W\) (Closure under scalar multiplication).

Step 2: Detailed Explanation:

• Let us evaluate each option systematically:
- Option (A) states "\(W_1 + W_2\) not subspace of \(V\)". This is incorrect.
The sum of two subspaces \(W_1 + W_2 = \{\mathbf{u} + \mathbf{v} \mid \mathbf{u} \in W_1, \mathbf{v} \in W_2\}\) is always a subspace of \(V\).

• Evaluate Option (B): "\(W_1 \cup W_2\) is always subspace of \(V\)". This is incorrect.
The union of two subspaces is a subspace if and only if one is contained in the other (i.e., \(W_1 \subseteq W_2\) or \(W_2 \subseteq W_1\)).
For example, in \(\mathbb{R}^2\), the x-axis and y-axis are subspaces, but their union is not a subspace (it is not closed under addition, e.g., \((1,0) + (0,1) = (1,1)\), which is not on either axis).

• Evaluate Option (C): "\(W_1 \cap W_2\) is always subspace of \(V\)". This is correct.
Let us prove that \(W_1 \cap W_2\) is always a subspace:
- Since \(W_1\) and \(W_2\) are subspaces, they both contain the zero vector \(\mathbf{0}\). Thus, \(\mathbf{0} \in W_1 \cap W_2\).
- If \(\mathbf{x}, \mathbf{y} \in W_1 \cap W_2\), then \(\mathbf{x}, \mathbf{y} \in W_1\) and \(\mathbf{x}, \mathbf{y} \in W_2\).
Since \(W_1\) and \(W_2\) are closed under addition and scalar multiplication:
\(\mathbf{x} + \mathbf{y} \in W_1\) and \(\mathbf{x} + \mathbf{y} \in W_2 \implies \mathbf{x} + \mathbf{y} \in W_1 \cap W_2\).
- For any scalar \(c \in F\), \(c\mathbf{x} \in W_1\) and \(c\mathbf{x} \in W_2 \implies c\mathbf{x} \in W_1 \cap W_2\).
Hence, the intersection is always a subspace.

• Option (D) states "\(W_1\) and \(W_2\) are always orthogonal", which is generally false. Subspaces do not have to be orthogonal.

Step 3: Final Answer:
The correct statement is \(W_1 \cap W_2\) is always a subspace of \(V\).