Question

Which of the following is a subspace of the real vector space \(\R^3\) ?

• A. {(x, y, z) ∈ \(\R^3\) : (y + z)² + (2x - 3y)² = 0}
• B. {(x, y, z) ∈ \(\R^3\) : y ∈ Q}
• C. {(x, y, z) ∈ \(\R^3\) : yz = 0}
• D. {(x, y, z) ∈ \(\R^3\) : x + 2y - 3z + 1 = 0}

Answer 1

The Correct Option is A

To determine which set is a subspace of the real vector space \(\mathbb{R}^3\), we need to consider the properties of a subspace. A subset \(W\) of a vector space \(V\) is a subspace if it satisfies the following three conditions:

1. Contains the zero vector: The zero vector of \(V\) is in \(W\).
2. Closed under addition: If \(u, v \in W\), then \(u + v \in W\).
3. Closed under scalar multiplication: If \(v \in W\) and \(c\) is a scalar, then \(cv \in W\).

Let's examine each option:

Option 1: \(\{(x, y, z) \in \mathbb{R}^3 : (y + z)^2 + (2x - 3y)^2 = 0\}\)

This set includes all vectors \((x, y, z) \) for which \((y + z)^2 + (2x - 3y)^2 = 0\).

• To satisfy this equation, both \((y + z)^2 = 0\) and \((2x - 3y)^2 = 0\) must hold, which implies \(y + z = 0\) and \(2x - 3y = 0\).
• The solutions are \(z = -y\) and \(x = \frac{3}{2}y\).
• For all such equations, setting \(y = 0\) results in \(x = 0\) and \(z = 0\), so \((0, 0, 0) \in W\).
• Under vector addition and scalar multiplication, sums and scaled vectors of \((x, y, z)\) still satisfy these equations.

This indicates the set is indeed a subspace of \(\mathbb{R}^3\).

Option 2: \(\{(x, y, z) \in \mathbb{R}^3 : y \in \mathbb{Q}\}\)

This set contains vectors where \(y\) is a rational number.

• The zero vector \((0, 0, 0)\) is included since 0 is a rational number.
• However, vector addition and scalar multiplication can produce irrational results for \(y\), which would not remain in the set, violating the subspace criteria.

Hence, this set is not a subspace of \(\mathbb{R}^3\).

Option 3: \(\{(x, y, z) \in \mathbb{R}^3 : yz = 0\}\)

This set specifies that either \(y = 0\) or \(z = 0\).

• The zero vector is included since \(0\cdot0 = 0\).
• However, the sum of two vectors \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2)\), where \(y_1z_1 = 0\) and \(y_2z_2 = 0\), may not satisfy the condition unless both have the same zero coordinate, thereby not satisfying closed under addition.

This set is not a subspace of \(\mathbb{R}^3\).

Option 4: \(\{(x, y, z) \in \mathbb{R}^3 : x + 2y - 3z + 1 = 0\}\)

This describes a plane in \(\mathbb{R}^3\) not passing through the origin.

• The zero vector \((0, 0, 0)\) is not included since it results in \(1 \neq 0\).
• Non-zero constant prevents closure under addition and scalar multiplication.

This set is not a subspace of \(\mathbb{R}^3\).

Conclusion

Thus, the set \(\{(x, y, z) \in \mathbb{R}^3 : (y + z)^2 + (2x - 3y)^2 = 0\}\) satisfies all the criteria for being a subspace of \(\mathbb{R}^3\).