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 of \(\mathbb{R}^n\) defined by a system of homogeneous linear equations (such as \(\sum a_i = 0\) or \(2a_1 - 3a_2 = 0\)) is always a subspace.
Subsets defined by inequalities (\(a_i \ge 0\)), non-linear terms (\(a_1 a_n > 0\)), or divisions by variables are almost never subspaces.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
For a subset \(W\) of a vector space \(V\) to be a subspace, it must satisfy three criteria:
1. The zero vector must belong to \(W\) (\(\mathbf{0} \in W\)).
2. It must be closed under vector addition (if \(\mathbf{u}, \mathbf{v} \in W\), then \(\mathbf{u} + \mathbf{v} \in W\)).
3. It must be closed under scalar multiplication (if \(\mathbf{u} \in W\) and \(c \in \mathbb{R}\), then \(c \mathbf{u} \in W\)).

Step 2: Detailed Explanation:
Let us evaluate each option systematically:
- Option (A): \(W = \{\mathbf{a} \in \mathbb{R}^n \mid \sum_{i=1}^{n} a_i = 0\}\)
1. Zero vector: \(\mathbf{0} = (0, 0, \dots, 0)\). Since \(\sum_{i=1}^n 0 = 0\), \(\mathbf{0} \in W\).
2. Addition: Let \(\mathbf{u} = (u_1, \dots, u_n) \in W\) and \(\mathbf{v} = (v_1, \dots, v_n) \in W\). This means \(\sum u_i = 0\) and \(\sum v_i = 0\).
The sum is \(\mathbf{u} + \mathbf{v} = (u_1 + v_1, \dots, u_n + v_n)\).
Since \(\sum (u_i + v_i) = \sum u_i + \sum v_i = 0 + 0 = 0\), \(\mathbf{u} + \mathbf{v} \in W\).
3. Scalar multiplication: Let \(c \in \mathbb{R}\). The scalar product is \(c \mathbf{u} = (c u_1, \dots, c u_n)\).
Since \(\sum (c u_i) = c \sum u_i = c(0) = 0\), \(c \mathbf{u} \in W\).
Thus, Option (A) is a subspace (it is a hyperplane through the origin).
Let us see why the other options fail:
- Option (B): \(W = \{\mathbf{a} \in \mathbb{R}^n \mid a_1 a_n > 0\}\).
The zero vector \(\mathbf{0} = (0, 0, \dots, 0)\) has \(a_1 a_n = 0 \times 0 = 0\), which is not strictly greater than 0. So \(\mathbf{0} \notin W\). Thus, it is not a subspace.
- Option (C): \(W = \{\mathbf{a} \in \mathbb{R}^n \mid \frac{a_1}{a_2} = \lambda\}\).
This set is not well-defined if \(a_2 = 0\) (which is true for the zero vector). Thus, the zero vector cannot satisfy this condition. It is not a subspace.
- Option (D): \(W = \{\mathbf{a} \in \mathbb{R}^n \mid a_i \ge 0 \,\, \forall i\}\).
Let \(\mathbf{u} = (1, 1, \dots, 1) \in W\). If we multiply by a negative scalar \(c = -1\):
\(c \mathbf{u} = (-1, -1, \dots, -1)\). Here, the elements are negative, so \(c \mathbf{u} \notin W\). It fails closure under scalar multiplication.
Therefore, only Option (A) is a subspace.

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