Question

Which of the following are subspaces of vector space \(\mathbb{R}^3\):
A. \( \{(x,y,z) : x+y=0\ \)
B. \( \{(x,y,z) : x-y=0\} \)
C. \( \{(x,y,z) : x+y=1\} \)
D. \( \{(x,y,z) : x-y=1\} \)}

• A. A and C only
• B. A, B and C only
• C. A and B only
• D. A and D only

Hint:

A very quick first check for a subspace is the zero vector test. If the set doesn't contain the zero vector of the parent space, it cannot be a subspace. For sets defined by homogeneous linear equations (like \(ax+by+cz=0\)), they are always subspaces. For non-homogeneous equations (like \(ax+by+cz=k\) where \(k \neq 0\)), they are never subspaces.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
A subset \(S\) of a vector space \(V\) is a subspace if it satisfies three conditions: 1. The zero vector of \(V\) is in \(S\). 2. \(S\) is closed under vector addition: If \(\mathbf{u}, \mathbf{v} \in S\), then \(\mathbf{u}+\mathbf{v} \in S\). 3. \(S\) is closed under scalar multiplication: If \(\mathbf{u} \in S\) and \(c\) is a scalar, then \(c\mathbf{u} \in S\).

Step 2: Detailed Explanation:
Let's check each set against these conditions. The zero vector in \(\mathbb{R}^3\) is \((0,0,0)\).
A. \( S_A = \{(x,y,z) : x+y=0\ \):} 1. Zero Vector: Is \((0,0,0)\) in \(S_A\)? Yes, because \(0+0=0\). 2. Closure under Addition: Let \(\mathbf{u}=(x_1, y_1, z_1)\) and \(\mathbf{v}=(x_2, y_2, z_2)\) be in \(S_A\). Then \(x_1+y_1=0\) and \(x_2+y_2=0\). Their sum is \(\mathbf{u}+\mathbf{v} = (x_1+x_2, y_1+y_2, z_1+z_2)\). Check the condition: \((x_1+x_2) + (y_1+y_2) = (x_1+y_1) + (x_2+y_2) = 0+0=0\). So, it's closed under addition. 3. Closure under Scalar Multiplication: Let \(\mathbf{u}=(x,y,z) \in S_A\) and \(c\) be a scalar. \(c\mathbf{u} = (cx, cy, cz)\). Check the condition: \(cx+cy = c(x+y) = c(0) = 0\). So, it's closed under scalar multiplication. Thus, A is a subspace.
B. \( S_B = \{(x,y,z) : x-y=0\ \):} 1. Zero Vector: Is \((0,0,0)\) in \(S_B\)? Yes, because \(0-0=0\). 2. Closure under Addition: Let \(\mathbf{u}, \mathbf{v} \in S_B\). Then \(x_1-y_1=0\) and \(x_2-y_2=0\). For the sum, \((x_1+x_2) - (y_1+y_2) = (x_1-y_1) + (x_2-y_2) = 0+0=0\). Closed. 3. Closure under Scalar Multiplication: Let \(\mathbf{u} \in S_B\). For \(c\mathbf{u}\), \(cx-cy = c(x-y) = c(0) = 0\). Closed. Thus, B is a subspace.
C. \( S_C = \{(x,y,z) : x+y=1\ \):} 1. Zero Vector: Is \((0,0,0)\) in \(S_C\)? No, because \(0+0=0 \neq 1\). Since it doesn't contain the zero vector, C is not a subspace.
D. \( S_D = \{(x,y,z) : x-y=1\ \):} 1. Zero Vector: Is \((0,0,0)\) in \(S_D\)? No, because \(0-0=0 \neq 1\). Since it doesn't contain the zero vector, D is not a subspace.
Step 3: Final Answer:
Only the sets A and B satisfy the conditions for a subspace.