A student proposes f(v) = v₁ + v₂ as a 'norm' on ℝ² (where v = (v₁, v₂)). Why does this fail?
AIt violates homogeneity: f(2v) ≠ 2f(v) in general
BIt violates positive definiteness: f(1, −1) = 0 but (1, −1) is not the zero vector
CIt violates the triangle inequality for most pairs of vectors
DIt is actually a valid norm — the ℓ¹ norm on ℝ²
f(1, −1) = 1 + (−1) = 0, yet (1, −1) is not the zero vector. A norm requires ‖v‖ = 0 only when v = 0 (positive definiteness). The fix is simple: use |v₁| + |v₂|, which IS the ℓ¹ norm. The key lesson is that signs must be removed — norms measure magnitude, not signed displacement.
Question 2 Multiple Choice
A norm ‖·‖ on a vector space V automatically induces a metric. What is that metric?
Ad(u, v) = ‖u‖ + ‖v‖
Bd(u, v) = ‖u − v‖
Cd(u, v) = ‖u‖ · ‖v‖
Dd(u, v) = |‖u‖ − ‖v‖|
The natural metric is d(u, v) = ‖u − v‖ — the norm of the difference, which measures how 'far apart' u and v are. The triangle inequality for the norm directly gives the triangle inequality for this metric: d(u, w) = ‖u − w‖ = ‖(u − v) + (v − w)‖ ≤ ‖u − v‖ + ‖v − w‖ = d(u, v) + d(v, w). This is the ladder: norm → metric → topology.
Question 3 True / False
In finite-dimensional vector spaces, the ℓ¹ and ℓ² norms produce different unit balls (diamond vs. circle), but they induce the same topology.
TTrue
FFalse
Answer: True
In finite dimensions, all norms are equivalent: they induce the same open sets, convergent sequences, and continuous functions — only the shape of the unit ball differs. This equivalence breaks down in infinite-dimensional spaces, where different norms can yield genuinely incompatible topologies. This is why the choice of norm becomes a central concern in functional analysis.
Question 4 True / False
Any metric on a vector space arises from a norm via d(u, v) = ‖u − v‖.
TTrue
FFalse
Answer: False
Not every metric on a vector space comes from a norm. The discrete metric (d(u,v) = 1 if u ≠ v, 0 if u = v) is a valid metric but cannot be expressed as ‖u − v‖ for any norm — a norm-induced metric must scale with scalar multiplication, e.g. d(2u, 0) = 2d(u, 0), which the discrete metric violates. Norms induce a special subclass of metrics compatible with the vector space structure.
Question 5 Short Answer
Explain in your own words why each of the three norm axioms — positive definiteness, homogeneity, and the triangle inequality — corresponds to a property we should demand of any reasonable notion of 'length.'
Think about your answer, then reveal below.
Model answer: Positive definiteness ensures only the zero vector has zero length (nothing else is 'nowhere'). Homogeneity ensures scaling a vector scales its length proportionally — stretching an arrow by 3 triples its length. The triangle inequality ensures that a direct path is never longer than a detour: going from u to w directly can't be longer than going via v.
The power of these axioms is their minimality — they capture exactly the intuitive content of 'length' with nothing extra. Any function satisfying all three can be used for analysis: you can define open balls, speak of convergence, and prove theorems. Functions that fail even one axiom lead to contradictions or unintuitive behavior (e.g., a 'length' that assigns zero to non-zero vectors makes it impossible to distinguish nearby points).