Question

Let \(\mathbb{R}^n = V\) is a vector space over the field of real numbers, then which of the following is a subspace ?

• A. \(W = \{(a_1, a_2, \dots, a_n) \mid \sum_{i=1}^{n} a_i = 0\}\)
• B. \(W = \{(a_1, a_2, \dots, a_n) \mid a_1 a_n > 0\}\)
• C. \(W = \{(a_1, a_2, \dots, a_n) \mid \frac{a_1}{a_2} = \lambda ; \lambda \in \mathbb{R}\}\)
• D. \(W = \{(a_1, a_2, \dots, a_n) \mid a_i \ge 0 \,\, \forall i = 1, 2, 3, \dots, n\}\)

Hint:

Any subset defined by a system of homogeneous linear equations (e.g., \(\sum a_i = 0\) or \(a_1 - 2a_2 = 0\)) is always a subspace.
Subsets defined by inequalities (\(a_i \ge 0\)) or non-linear relations (\(a_1 a_n > 0\)) are never subspaces because they fail scalar multiplication by negative numbers or basic closure properties.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We need to determine which of the given subsets of \(\mathbb{R}^n\) forms a vector subspace over the field of real numbers \(\mathbb{R}\).
Key Formula or Approach:
A subset \(W\) of a vector space \(V\) is a subspace if and only if it satisfies the three subspace criteria:
1. Zero vector: The zero vector \(\mathbf{0} \in W\).
2. Closure under addition: If \(\mathbf{x}, \mathbf{y} \in W\), then \(\mathbf{x} + \mathbf{y} \in W\).
3. Closure under scalar multiplication: If \(\mathbf{x} \in W\) and \(c \in \mathbb{R}\), then \(c\mathbf{x} \in W\).

Step 2: Detailed Explanation:

• Let us evaluate Option (A): \(W = \{(a_1, a_2, \dots, a_n) \mid \sum_{i=1}^{n} a_i = 0\}\).
- Zero vector: For \(\mathbf{0} = (0, 0, \dots, 0)\), we have \(\sum_{i=1}^{n} 0 = 0\), so \(\mathbf{0} \in W\).
- Closure under addition: Let \(\mathbf{x} = (x_1, \dots, x_n) \in W\) and \(\mathbf{y} = (y_1, \dots, y_n) \in W\). This means \(\sum x_i = 0\) and \(\sum y_i = 0\). For \(\mathbf{x} + \mathbf{y} = (x_1+y_1, \dots, x_n+y_n)\), the sum of components is: \[ \sum_{i=1}^{n} (x_i + y_i) = \sum_{i=1}^{n} x_i + \sum_{i=1}^{n} y_i = 0 + 0 = 0 \] So, \(\mathbf{x} + \mathbf{y} \in W\).
- Closure under scalar multiplication: For any scalar \(c \in \mathbb{R}\) and \(\mathbf{x} \in W\): \[ \sum_{i=1}^{n} (c x_i) = c \sum_{i=1}^{n} x_i = c(0) = 0 \] So, \(c\mathbf{x} \in W\).
Thus, Option (A) is a valid subspace.

• Let us see why other options are NOT subspaces:
- Option (B): For the zero vector, \(0 \cdot 0 = 0\), which is not \(> 0\). Thus, \(\mathbf{0} \notin W\). It is not a subspace.
- Option (C): If \(a_2 = 0\), the division \(\frac{a_1}{a_2}\) is undefined, so the zero vector \((0, 0, \dots, 0)\) cannot be easily validated without division by zero issues. Furthermore, it is not closed under addition.
- Option (D): Let \(\mathbf{x} = (1, 1, \dots, 1) \in W\). If we multiply by scalar \(c = -1\), we get \(c\mathbf{x} = (-1, -1, \dots, -1)\). Since the components are negative, \(c\mathbf{x} \notin W\). Thus, it fails closure under scalar multiplication.

Step 3: Final Answer:
The subset \(W = \{(a_1, a_2, \dots, a_n) \mid \sum_{i=1}^{n} a_i = 0\}\) is a subspace of \(\mathbb{R}^n\).