Question

Which of the following is true?

• A. The set of all vectors \(\binom{x}{y}\) in \(\mathbb{R}^2\) such that \(x + y = 2\) is a subspace of \(\mathbb{R}^2\)
• B. The set of all vectors \(\binom{x}{y}\) in \(\mathbb{R}^2\) such that \(x + y = 0\) is a subspace of \(\mathbb{R}^2\)
• C. The union of first and third quadrants in \(\mathbb{R}^2\) is a subspace of \(\mathbb{R}^2\)
• D. The union of \(x\)-axis and \(y\)-axis in \(\mathbb{R}^2\) is a subspace of \(\mathbb{R}^2\)

Hint:

Exam Tip:
To test if a set is a subspace:

• Check if the zero vector is in the set.
• Check closure under addition.
• Check closure under scalar multiplication.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
A subspace of \(\mathbb{R}^2\) must satisfy three conditions:
• Contains the zero vector \((0, 0)\).
• Closed under vector addition.
• Closed under scalar multiplication.

Step 2: Analyzing Each Option:

• Option (A): \(x + y = 2\).
The zero vector \((0, 0)\) gives \(0 + 0 = 0 \neq 2\). So, the zero vector is not in the set.
Not a subspace.
• Option (B): \(x + y = 0\).
Zero vector: \(0 + 0 = 0\), so it contains \((0, 0)\).
Let \(u = (x_1, y_1)\) and \(v = (x_2, y_2)\) be in the set, so \(x_1 + y_1 = 0\) and \(x_2 + y_2 = 0\).
Addition: \(u + v = (x_1 + x_2, y_1 + y_2)\).
\((x_1 + x_2) + (y_1 + y_2) = (x_1 + y_1) + (x_2 + y_2) = 0 + 0 = 0\). So, closed under addition.
Scalar multiplication: For any scalar \(c\), \(c u = (c x_1, c y_1)\).
\(c x_1 + c y_1 = c(x_1 + y_1) = c \cdot 0 = 0\). So, closed under scalar multiplication.
This is a subspace.
• Option (C): Union of first and third quadrants.
This set includes vectors with \(x \geq 0, y \geq 0\) or \(x \leq 0, y \leq 0\).
Not closed under addition: \((1, 0)\) and \((0, 1)\) are in the set (first quadrant), but \((1, 0) + (0, 1) = (1, 1)\) is in the first quadrant, so that's fine.
But \((1, 0)\) and \((-1, 0)\) are both in the set (first and third quadrants respectively), but \((1, 0) + (-1, 0) = (0, 0)\) is in the set.
However, \((1, 1)\) and \((-1, -1)\) are in the set, but \((1, 1) + (-1, -1) = (0, 0)\) is in the set.
The issue is with scalar multiplication: \((-1) \cdot (1, 1) = (-1, -1)\), which is in the set.
Actually, the union of first and third quadrants is not closed under addition.
Example: \((1, 2)\) is in the first quadrant, \((-1, -2)\) is in the third quadrant, but their sum is \((0, 0)\), which is in the set (since it's on the boundary).
Another example: \((1, 2)\) and \((-2, -1)\) are both in the set, but their sum is \((-1, 1)\), which is in the second quadrant and not in the set.
So, not closed under addition.
Not a subspace.
• Option (D): Union of x-axis and y-axis.
This set includes all vectors with \(y = 0\) or \(x = 0\).
Not closed under addition: \((1, 0)\) and \((0, 1)\) are both in the set, but their sum is \((1, 1)\), which is not on either axis.
Not a subspace.

Step 3: Final Answer:
Only option (B) is true. Therefore, option (B) is correct.