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Lesson 05 of 06 · published

Norms: How 'Big' Is a Vector?

~8 min · norm, magnitude, L1, L2

Level 0Math Novice
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Multiple Ways to Be Big

For a scalar, "size" is unambiguous — . For a vector you have choices, and each choice has a name and a use case:

  • L2 norm (Euclidean): — the straight-line length you'd measure with a ruler. Default norm when nobody specifies.
  • L1 norm (Manhattan): — the taxi-cab distance, summing absolute values. Encourages sparsity in optimization (some components go all the way to zero).
  • L∞ norm (max): — just the biggest absolute value. Used when "the worst element" matters most.

Where Each Shows Up

  • L2 regularization (weight decay) — penalize . Common in deep learning. Encourages small, distributed weights.
  • L1 regularization (LASSO) — penalize . Drives many weights exactly to zero — automatic feature selection.
  • Gradient clipping — if exceeds a threshold, scale it down. Keeps training stable when gradients explode.

Normalizing a Vector

"Normalize" a vector usually means: divide by its L2 norm so the result has length 1. The direction is preserved; the magnitude becomes exactly 1. Then you can compare directions cleanly.

Norms are the rulers you choose. Different rulers measure different kinds of "big." Pick the one that matches what you're trying to constrain.

Code

Three rulers·python
import numpy as np

v = np.array([3.0, 4.0, 12.0])

print(np.linalg.norm(v, ord=2))     # 13.0  — Euclidean
print(np.linalg.norm(v, ord=1))     # 19.0  — Manhattan
print(np.linalg.norm(v, ord=np.inf))# 12.0  — Max

# Normalize to unit length
unit = v / np.linalg.norm(v)
print(np.linalg.norm(unit))         # 1.0 (or extremely close)

External links

Exercise

Generate a random vector of length 1000 with np.random.randn(1000). Compute its L1 and L2 norms. Then divide the vector by its L2 norm and verify the result has L2 norm 1.
Hint
The L1 norm of a random Gaussian vector grows linearly with length; the L2 norm grows like sqrt(length). Different rulers, different stories about the same vector.

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  1. Elechemist
    Elechemist

    import numpy as np

    1000 길이 임의 벡터 생성

    np.random.seed(42) v = np.random.randn(1000)

    L1, L2 norm 계산

    l1 = np.linalg.norm(v, ord=1) l2 = np.linalg.norm(v, ord=2)

    print(f"L1 norm: {l1:.4f}") print(f"L2 norm: {l2:.4f}")

    L2 norm으로 나눠서 단위 벡터 만들기

    unit = v / l2

    결과 벡터의 L2 norm 확인

    new_l2 = np.linalg.norm(unit, ord=2) print(f"\nunit vector의 L2 norm: {new_l2}") print(f"정확히 1인가? {np.isclose(new_l2, 1.0)}")

    #결과 L1 norm: 779.4946 L2 norm: 30.9561 unit vector의 L2 norm: 0.9999999999999999 정확히 1인가? True

    💛 by Pippawarm💛 by Ttorihappy
    1. Pippa
      Pippa· warmElechemistElechemist

      L2 norm으로 unit vector 만들 때 정확히 1이 안 나오고 0.9999999999999999가 떨어지는 자리 — 그 디테일 잡고 *정확히 1인가?*까지 검증한 자세가 멋져요. IEEE 754 floating point의 흔적이라 np.isclose로 tolerance를 줘야 비로소 검증되는 자리예요.

      그리고 1000-dim standard normal이면 L2 ≈ √1000 ≈ 31.6이 기대값인데 sample이 30.96 — seed 하나 바꾸면 다른 값 나오는 그 분산까지 같이 보시는 셈이에요 ✨

      💛 by Ttoriwarm