The set S = {(x, y) ∈ ℝ² : x + y = 1} is proposed as a subspace of ℝ². What is the correct verdict?
AIt is a subspace because it is a line and lines are subspaces of ℝ²
BIt is not a subspace because it fails closure under scalar multiplication: 0·(1, 0) = (0, 0) ∉ S
CIt is not a subspace because it contains uncountably many vectors
DIt is a subspace because any two vectors in S can be added together
The failure is immediately detected by the zero vector test: every subspace must contain the zero vector (since c·w = 0 when c = 0), but (0,0) does not satisfy x + y = 1. Also, addition fails: (1,0) and (0,1) are both in S but their sum (1,1) has 1+1=2 ≠ 1. Option A contains the key misconception — lines are subspaces only when they pass through the origin.
Question 2 Multiple Choice
To verify that a nonempty subset W of vector space V is a subspace, the minimum you need to check is:
AW contains the zero vector and all additive inverses of its elements
BW is closed under addition and closed under scalar multiplication
CW satisfies all eight vector space axioms applied to its elements
DW is closed under addition and contains the zero vector
Closure under addition and scalar multiplication is both necessary and sufficient. The zero vector follows automatically: if w ∈ W, then 0·w = 0 must be in W by scalar closure. Additive inverses follow: −w = (−1)·w must be in W. All other axioms are inherited from the ambient space V. Option D is the common shortcut error — you do not need to verify the zero vector separately; it is a consequence of scalar closure.
Question 3 True / False
Every subspace of ℝⁿ must contain the zero vector.
TTrue
FFalse
Answer: True
True. If W is a nonempty subspace and w ∈ W, then 0·w = 0 must be in W by closure under scalar multiplication. The zero vector is guaranteed — it is a theorem, not an extra axiom. A subset that does not contain 0 cannot be a subspace, which is why the fastest way to disprove a subspace claim is to check whether 0 is in the set.
Question 4 True / False
A nonempty subset of ℝ² that is closed under vector addition is expected to be a subspace.
TTrue
FFalse
Answer: False
False. Closure under addition alone is insufficient — you also need closure under scalar multiplication. Consider the set of all vectors with integer coordinates: {(m, n) : m, n ∈ ℤ}. This is closed under addition, but (0.5)·(1, 0) = (0.5, 0) is not in the set, so it fails scalar closure. Both conditions are required.
Question 5 Short Answer
Explain why a plane in ℝ³ that does not pass through the origin cannot be a subspace, using the closure conditions.
Think about your answer, then reveal below.
Model answer: A plane not through the origin fails closure under scalar multiplication: if p is any vector on the plane, then 0·p = 0 must belong to the plane by closure, but the zero vector is not on a plane that misses the origin. Equivalently, adding two vectors on the plane generally leaves the plane (their sum shifts to double the displacement from the origin). Subspaces must pass through the origin because scaling any vector by 0 must yield the zero vector, which must stay in the set.
The origin requirement is a theorem, not an arbitrary restriction. It distinguishes subspaces (vector-space-compatible subsets) from affine subspaces (translates of subspaces). A plane at height z = 1 is an affine subspace — it has the right shape but is displaced from the origin — and fails both closure conditions. Geometrically, any subset closed under scaling must include 0 (the limit as the scalar approaches 0), so it must pass through the origin.