Vector and Matrix Norms

Why It Matters: Norms are the foundation of measuring "size" and "distance" in vector spaces. They determine how we quantify error, define convergence, and control optimization. In machine learning, the choice of norm directly influences model behavior — L1 produces sparse solutions, L2 promotes smoothness, and the L∞ norm captures worst-case scenarios. Understanding norms is essential for regularization, numerical stability analysis, and distance-based algorithms.

What is a Norm

DfVector Norm

A norm is a function $\|\cdot\|: V \to \mathbb{R}_{\geq 0}$ from a vector space $V$ to the non-negative real numbers that satisfies four fundamental axioms:

Axiom	Property	Description
1	Non-negativity	$\\|\vec{x}\\| \geq 0$ for all $\vec{x} \in V$
2	Definiteness	$\\|\vec{x}\\| = 0 \iff \vec{x} = \vec{0}$
3	Homogeneity	$\\|\alpha \vec{x}\\| = \\|\alpha\\| \cdot \\|\vec{x}\\|$ for all scalars $\alpha$
4	Triangle Inequality	$\\|\vec{x} + \vec{y}\\| \leq \\|\vec{x}\\| + \\|\vec{y}\\|$ for all $\vec{x}, \vec{y} \in V$

A vector space equipped with a norm is called a normed vector space. The norm induces a natural distance function $d(\vec{x}, \vec{y}) = \|\vec{x} - \vec{y}\|$ , making it a metric space.

Vector Norms

Lp Norm Family

\\|\\vec{x}\\|_p = \\left(\\sum_{i=1}^{n} |x_i|^p\\right)^{1/p}

Here,

$\vec{x}$ =Vector in \mathbb{R}^n
$p$ =Parameter satisfying p \geq 1
$|x_i|$ =Absolute value of the i-th component

Norm	Formula	When to Use
L1 (Manhattan)	$\\|\vec{x}\\|_1 = \sum_{i=1}^{n} \|x_i\|$	Sparse solutions, feature selection (Lasso)
L2 (Euclidean)	$\\|\vec{x}\\|_2 = \sqrt{\sum_{i=1}^{n} x_i^2}$	Smooth solutions, general-purpose (Ridge)
L∞ (Max Norm)	$\\|\vec{x}\\|_\infty = \max_i \|x_i\|$	Worst-case analysis, adversarial robustness
Lp (General)	$\\|\vec{x}\\|_p = \left(\sum \|x_i\|^p\right)^{1/p}$	Interpolation between L1 and L∞
L0 (Pseudo-norm)	$\\|\vec{x}\\|_0 = \#\{i : x_i \neq 0\}$	Cardinality (non-convex, NP-hard to optimize)

Step-by-Step Example: Computing Vector Norms

📝Computing Vector Norms for x = [1, -2, 3, -4]

Given $\vec{x} = \begin{bmatrix} 1 \\ -2 \\ 3 \\ -4 \end{bmatrix}$ , compute all common norms.

Step 1: L1 Norm

\\|\\vec{x}\\|_1 = |1| + |-2| + |3| + |-4| = 1 + 2 + 3 + 4 = 10

Step 2: L2 Norm

\\|\\vec{x}\\|_2 = \\sqrt{1^2 + (-2)^2 + 3^2 + (-4)^2} = \\sqrt{1 + 4 + 9 + 16} = \\sqrt{30} \\approx 5.477

Step 3: L∞ Norm

\\|\\vec{x}\\|_\\infty = \\max(|1|, |-2|, |3|, |-4|) = 4

Step 4: L4 Norm (example of Lp)

\\|\\vec{x}\\|_4 = (1^4 + 2^4 + 3^4 + 4^4)^{1/4} = (1 + 16 + 81 + 256)^{1/4} = 354^{1/4} \\approx 4.34

💡Solution

Key Insight: For any vector, $\|\vec{x}\|_\infty \leq \|\vec{x}\|_2 \leq \|\vec{x}\|_1$ . The L∞ norm captures only the largest component, L2 averages all components, and L1 sums all magnitudes. As $p$ increases, the Lp norm converges to the L∞ norm.

Matrix Norms

Frobenius Norm

\\|A\\|_F = \\sqrt{\\sum_{i=1}^{m}\\sum_{j=1}^{n} a_{ij}^2} = \\sqrt{\\text{tr}(A^TA)} = \\sqrt{\\sum_{i=1}^{\\min(m,n)} \\sigma_i^2}

Here,

$A$ =Matrix of size m × n
$a_{ij}$ =Element in row i, column j
$\text{tr}$ =Trace (sum of diagonal elements)
$\sigma_i$ =Singular values of A

The Frobenius norm treats a matrix as a vector in $\mathbb{R}^{m \times n}$ and computes its Euclidean norm. It equals the square root of the sum of squared singular values.

Spectral Norm (Operator 2-Norm)

\\|A\\|_2 = \\sigma_{\\max}(A) = \\sqrt{\\lambda_{\\max}(A^TA)}

Here,

$\sigma_{\max}(A)$ =Largest singular value of A
$\lambda_{\max}(A^TA)$ =Largest eigenvalue of A^T A

The spectral norm measures the maximum "stretch" factor of a linear transformation. It equals the largest singular value.

Nuclear Norm

\\|A\\|_* = \\sum_{i=1}^{r} \\sigma_i = \\text{tr}(\\sqrt{A^TA})

Here,

$\sigma_i$ =Singular values of A
$r$ =Rank of A

The nuclear norm (also called the trace norm) is the convex envelope of the rank function over the unit spectral norm ball. It is used in matrix completion and low-rank approximation.

Comparison of Matrix Norms

Norm	Formula	Use Case
Frobenius	$\\|A\\|_F = \sqrt{\sum a_{ij}^2}$	General matrix similarity, reconstruction error
Spectral	$\\|A\\|_2 = \sigma_{\max}$	Stability analysis, condition number, Lipschitz constants
Nuclear	$\\|A\\|_* = \sum \sigma_i$	Matrix completion, low-rank recovery
L1 (entry-wise)	$\\|A\\|_{1,1} = \sum \|a_{ij}\|$	Sparse matrix recovery
L∞ (entry-wise)	$\\|A\\|_{\infty,\infty} = \max \|a_{ij}\|$	Bounded perturbations

Induced (Operator) Norms

DfInduced Matrix Norm

An induced norm (also called an operator norm) measures the maximum output norm given an input constrained to unit norm. The most common induced norms are:

Induced Norm	Definition	Formula
Induced 2-norm	$\\|A\\|_2 = \max_{\\|\vec{x}\\|=1} \\|A\vec{x}\\|_2$	$\sigma_{\max}(A)$
Induced 1-norm	$\\|A\\|_1 = \max_{\\|\vec{x}\\|=1} \\|A\vec{x}\\|_1$	$\max_j \sum_i \|a_{ij}\|$
Induced ∞-norm	$\\|A\\|_\infty = \max_{\\|\vec{x}\\|=1} \\|A\vec{x}\\|_\infty$	$\max_i \sum_j \|a_{ij}\|$

Example: Computing Matrix Norms

📝Matrix Norms for A = [[1, 2], [3, 4]]

Given $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ .

Frobenius Norm:

\\|A\\|_F = \\sqrt{1^2 + 2^2 + 3^2 + 4^2} = \\sqrt{30} \\approx 5.477

Spectral Norm: Compute $A^TA$ :

A^TA = \\begin{bmatrix} 1 & 3 \\\\ 2 & 4 \\end{bmatrix} \\begin{bmatrix} 1 & 2 \\\\ 3 & 4 \\end{bmatrix} = \\begin{bmatrix} 10 & 14 \\\\ 14 & 20 \\end{bmatrix}

\\lambda_{\\max} = \\frac{30 + \\sqrt{30^2 - 4(200-196)}}{2} = \\frac{30 + \\sqrt{880}}{2} \\approx 29.37

\\|A\\|_2 = \\sqrt{29.37} \\approx 5.42

Induced 1-norm:

\\|A\\|_1 = \\max(1+3, 2+4) = \\max(4, 6) = 6

Induced ∞-norm:

\\|A\\|_\\infty = \\max(1+2, 3+4) = \\max(3, 7) = 7

Norm Equivalence

ThNorm Equivalence in Finite Dimensions

For any two norms $\|\cdot\|_a$ and $\|\cdot\|_b$ on a finite-dimensional vector space $V$ (with $\dim(V) = n$ ), there exist constants $c_1, c_2 > 0$ such that for all $\vec{x} \in V$ :

c_1 \\|\\vec{x}\\|_a \\leq \\|\\vec{x}\\|_b \\leq c_2 \\|\\vec{x}\\|_a

Specific bounds for $\mathbb{R}^n$ :

Relationship	Bound
$\\|\vec{x}\\|_\infty \leq \\|\vec{x}\\|_2$	$\\|\vec{x}\\|_2 \leq \sqrt{n} \\|\vec{x}\\|_\infty$
$\\|\vec{x}\\|_2 \leq \\|\vec{x}\\|_1$	$\\|\vec{x}\\|_1 \leq \sqrt{n} \\|\vec{x}\\|_2$
$\\|\vec{x}\\|_\infty \leq \\|\vec{x}\\|_1$	$\\|\vec{x}\\|_1 \leq n \\|\vec{x}\\|_\infty$

Implication: In finite dimensions, all norms define the same topology — convergence in one norm implies convergence in all others. However, the constants matter: the L1 norm can be up to $n$ times larger than the L∞ norm. In infinite dimensions (function spaces), norms need NOT be equivalent.

Unit Ball: Geometric Interpretation

The unit ball of a norm is the set $B = \{\vec{x} : \|\vec{x}\| \leq 1\}$ . Its shape reveals the geometric character of the norm.

Norm	Unit Ball Shape	Geometry
L1	Diamond (rotated square in 2D)	Vertices at $(\pm 1, 0)$ and $(0, \pm 1)$
L2	Circle / Sphere	Smooth, rotationally symmetric
L∞	Square / Hypercube	Vertices at $(\pm 1, \pm 1)$
Lp (1<p<∞)	Rounded polygon	Interpolates between diamond and circle

The L1 unit ball's "pointy" vertices at the axes explain why L1 optimization produces sparse solutions — the optimal point is more likely to land on a vertex where some coordinates are exactly zero.

Norms in Optimization

Regularized Loss Function

\\min_{\\vec{w}} \\mathcal{L}(\\vec{w}) + \\lambda \\|\\vec{w}\\|_p^p

Here,

$\mathcal{L}(\vec{w})$ =Loss function (e.g., squared error)
$\lambda$ =Regularization strength
$\|\vec{w}\|_p^p$ =p-norm penalty (p=1 or 2 common)

Penalty	Name	Effect
$\lambda \\|\vec{w}\\|_1$	Lasso	Sparse solutions, automatic feature selection
$\lambda \\|\vec{w}\\|_2^2$	Ridge	Small weights, no feature selection
$\lambda \\|\vec{w}\\|_1 + \lambda_2 \\|\vec{w}\\|_2^2$	Elastic Net	Combines sparsity and smoothness
$\lambda \\|\vec{w}\\|_\infty$	Minimax	Bounded maximum coefficient

Distance Metrics

A norm $\|\cdot\|$ induces a distance metric $d(\vec{x}, \vec{y}) = \|\vec{x} - \vec{y}\|$ that satisfies:

Non-negativity: $d(\vec{x}, \vec{y}) \geq 0$
Identity: $d(\vec{x}, \vec{y}) = 0 \iff \vec{x} = \vec{y}$
Symmetry: $d(\vec{x}, \vec{y}) = d(\vec{y}, \vec{x})$
Triangle inequality: $d(\vec{x}, \vec{z}) \leq d(\vec{x}, \vec{y}) + d(\vec{y}, \vec{z})$

Distance	Formula	Use Case
Manhattan ( $L_1$ )	$d_1 = \sum \|x_i - y_i\|$	Grid-based movement, high-dimensional data
Euclidean ( $L_2$ )	$d_2 = \sqrt{\sum (x_i - y_i)^2}$	Geometric distance, clustering
Chebyshev ( $L_\infty$ )	$d_\infty = \max \|x_i - y_i\|$	Warehouse logistics, robotics

Python Implementation

import numpy as np

# --- Vector Norms ---
x = np.array([1, -2, 3, -4])

l1 = np.linalg.norm(x, ord=1)           # L1: 10.0
l2 = np.linalg.norm(x, ord=2)           # L2: sqrt(30) ≈ 5.477
linf = np.linalg.norm(x, ord=np.inf)    # L∞: 4.0
l4 = np.linalg.norm(x, ord=4)           # L4: 354^(1/4) ≈ 4.34

print(f"L1: {l1}, L2: {l2:.4f}, L∞: {linf}, L4: {l4:.4f}")

# --- Matrix Norms ---
A = np.array([[1, 2], [3, 4]])

frob = np.linalg.norm(A, ord='fro')     # Frobenius: sqrt(30) ≈ 5.477
spectral = np.linalg.norm(A, ord=2)     # Spectral: largest singular value
nuclear = np.linalg.norm(A, ord='nuc')  # Nuclear: sum of singular values

print(f"Frobenius: {frob:.4f}")
print(f"Spectral: {spectral:.4f}")
print(f"Nuclear: {nuclear:.4f}")

# --- Induced Norms ---
induced1 = np.linalg.norm(A, ord=1)     # Induced 1-norm: max column sum
induced_inf = np.linalg.norm(A, ord=np.inf)  # Induced ∞-norm: max row sum
print(f"Induced 1-norm: {induced1}")
print(f"Induced ∞-norm: {induced_inf}")

# --- Distance Computation ---
from scipy.spatial.distance import cdist, pdist

points = np.array([[0, 0], [1, 0], [0, 1], [1, 1]])
dist_l1 = cdist(points, points, metric='cityblock')   # Manhattan
dist_l2 = cdist(points, points, metric='euclidean')   # Euclidean
dist_linf = cdist(points, points, metric='chebyshev') # Chebyshev

# --- Regularization Comparison ---
from sklearn.linear_model import Lasso, Ridge

np.random.seed(42)
X = np.random.randn(100, 10)
true_coef = np.zeros(10)
true_coef[:3] = [3, -2, 1]  # Only 3 non-zero features
y = X @ true_coef + np.random.randn(100) * 0.1

lasso = Lasso(alpha=0.1).fit(X, y)
ridge = Ridge(alpha=0.1).fit(X, y)

print(f"Lasso coefficients: {np.round(lasso.coef_, 3)}")  # Sparse
print(f"Ridge coefficients: {np.round(ridge.coef_, 3)}")  # Small but non-zero

Applications in AI/ML

L1 Regularization (Lasso): The L1 norm penalty $\lambda \|\vec{w}\|_1$ drives some weights to exactly zero, performing automatic feature selection. This is critical in high-dimensional settings where only a subset of features matter (genomics, NLP feature selection).

L2 Regularization (Ridge): The L2 norm penalty $\lambda \|\vec{w}\|_2^2$ shrinks all weights toward zero but never sets them exactly to zero. It prevents overfitting and improves generalization. It is the default regularization in most linear models.

Adversarial Robustness: The L∞ norm measures the maximum perturbation allowed in adversarial examples. Models trained with the PGD adversarial method optimize $\max_{\|\delta\|_\infty \leq \epsilon} \mathcal{L}(x + \delta, y)$ .

Gradient Clipping: In deep learning, gradients are clipped by norm: if $\|\vec{g}\| > \tau$ , then $\vec{g} \leftarrow \tau \cdot \vec{g} / \|\vec{g}\|$ . This prevents exploding gradients and stabilizes training.

Matrix Completion (Netflix Prize): The nuclear norm $\|A\|_*$ is minimized to recover low-rank matrices from partial observations: $\min_{X} \|X\|_*$ subject to observed entries matching.

Common Mistakes

Mistake	Why It's Wrong	Correct Approach
Using L0 norm for optimization	L0 is non-convex, NP-hard	Use L1 as convex relaxation
Confusing $\\|A\\|_F$ with $\\|A\\|_2$	Frobenius sums all singular values, spectral takes the max	$\\|A\\|_2 \leq \\|A\\|_F \leq \sqrt{r} \\|A\\|_2$
Forgetting $\\|c\vec{x}\\| = \|c\| \\|\vec{x}\\|$	Homogeneity requires absolute value on scalar	$\\|-3\vec{x}\\| = 3\\|\vec{x}\\|$ , not $-3\\|\vec{x}\\|$
Assuming all norms are equal in infinite dimensions	Norm equivalence requires finite dimensions	In function spaces, different norms define different topologies
Using L2 norm for sparse feature selection	L2 shrinks but doesn't zero out features	Use L1 (Lasso) or Elastic Net
Ignoring norm when computing condition number	$\kappa(A) = \\|A\\| \cdot \\|A^{-1}\\|$ depends on the norm	Choose the norm appropriate for your error metric

Interview Questions

Q1: Why does L1 regularization produce sparse solutions while L2 does not?

💡Solution

Geometrically, the L1 unit ball is a diamond with vertices on the axes. The level curves of the loss function are more likely to intersect the L1 ball at a vertex, where some coordinates are exactly zero. The L2 ball is a circle — level curves typically intersect it at points where all coordinates are non-zero.

Q2: What is the relationship between the spectral norm and the Frobenius norm?

💡Solution

For any matrix $A$ : $\|A\|_2 \leq \|A\|_F \leq \sqrt{r} \cdot \|A\|_2$ , where $r = \text{rank}(A)$ . The spectral norm equals the largest singular value, while the Frobenius norm equals the root-sum-of-squares of all singular values.

Q3: When would you use the nuclear norm instead of the Frobenius norm?

💡Solution

Use the nuclear norm when you want to encourage low-rank structure in a matrix. The nuclear norm is the tightest convex relaxation of the rank function. Applications include matrix completion (e.g., recommender systems), denoising, and dimensionality reduction.

Q4: Prove that the L∞ norm is indeed a norm on $\mathbb{R}^n$ .

💡Solution

Non-negativity: $\max |x_i| \geq 0$ since each $|x_i| \geq 0$ .
Definiteness: $\max |x_i| = 0 \implies x_i = 0$ for all $i \implies \vec{x} = \vec{0}$ .
Homogeneity: $\|\alpha \vec{x}\|_\infty = \max |\alpha x_i| = |\alpha| \max |x_i| = |\alpha| \|\vec{x}\|_\infty$ .
Triangle inequality: $\|\vec{x} + \vec{y}\|_\infty = \max |x_i + y_i| \leq \max (|x_i| + |y_i|) \leq \max |x_i| + \max |y_i| = \|\vec{x}\|_\infty + \|\vec{y}\|_\infty$ .

Q5: What is the condition number of a matrix, and why does the norm matter?

💡Solution

The condition number is $\kappa(A) = \|A\| \cdot \|A^{-1}\|$ . It measures how sensitive the solution of $A\vec{x} = \vec{b}$ is to perturbations in $\vec{b}$ . A large condition number indicates an ill-conditioned problem. The value depends on the norm chosen — typically the spectral norm or the L∞ norm is used.

Q6: How do norms relate to convergence in optimization algorithms?

💡Solution

Convergence of an iterative algorithm $\vec{x}^{(k)} \to \vec{x}^*$ is defined with respect to a norm: $\|\vec{x}^{(k)} - \vec{x}^*\| \to 0$ . In finite dimensions, convergence in one norm implies convergence in all norms. However, the rate of convergence (and practical numerical behavior) can differ significantly between norms.

Practice Problems

Problem 1: Compute the L1, L2, L∞, and L4 norms of $\vec{x} = \begin{bmatrix} 3 \\ -4 \\ 0 \\ 5 \end{bmatrix}$ .

💡Solution

\\|\\vec{x}\\|_1 = 3 + 4 + 0 + 5 = 12

\\|\\vec{x}\\|_2 = \\sqrt{9 + 16 + 0 + 25} = \\sqrt{50} = 5\\sqrt{2} \\approx 7.07

\\|\\vec{x}\\|_\\infty = \\max(3, 4, 0, 5) = 5

\\|\\vec{x}\\|_4 = (81 + 256 + 0 + 625)^{1/4} = 962^{1/4} \\approx 5.57

Verify: $\|\vec{x}\|_\infty \leq \|\vec{x}\|_4 \leq \|\vec{x}\|_2 \leq \|\vec{x}\|_1$ : $5 \leq 5.57 \leq 7.07 \leq 12$ ✓

Problem 2: Verify the Cauchy-Schwarz inequality for $\vec{x} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$ and $\vec{y} = \begin{bmatrix} 4 \\ 5 \\ 6 \end{bmatrix}$ .

💡Solution

\\vec{x} \\cdot \\vec{y} = 1(4) + 2(5) + 3(6) = 32

\\|\\vec{x}\\|_2 = \\sqrt{1 + 4 + 9} = \\sqrt{14} \\approx 3.74

\\|\\vec{y}\\|_2 = \\sqrt{16 + 25 + 36} = \\sqrt{77} \\approx 8.77

|\\vec{x} \\cdot \\vec{y}| = 32 \\leq \\sqrt{14} \\cdot \\sqrt{77} = \\sqrt{1078} \\approx 32.83

Cauchy-Schwarz holds: $32 \leq 32.83$ ✓

Problem 3: Compute the Frobenius and spectral norms of $A = \begin{bmatrix} 2 & 0 \\\\ 0 & 3 \end{bmatrix}$ .

💡Solution

\\|A\\|_F = \\sqrt{4 + 0 + 0 + 9} = \\sqrt{13} \\approx 3.61

Since $A$ is diagonal, its singular values are $|2| = 2$ and $|3| = 3$ .

\\|A\\|_2 = \\sigma_{\\max} = 3

\\|A\\|_* = \\sigma_1 + \\sigma_2 = 2 + 3 = 5

Problem 4: Show that for any vector $\vec{x} \in \mathbb{R}^n$ : $\|\vec{x}\|_\infty \leq \|\vec{x}\|_2 \leq \sqrt{n} \|\vec{x}\|_\infty$ .

💡Solution

Lower bound: Let $j = \arg\max_i |x_i|$ . Then:

\\|\\vec{x}\\|_2 = \\sqrt{\\sum_i x_i^2} \\geq \\sqrt{x_j^2} = |x_j| = \\|\\vec{x}\\|_\\infty

Upper bound: Since $|x_i| \leq \|\vec{x}\|_\infty$ for all $i$ :

\\|\\vec{x}\\|_2 = \\sqrt{\\sum_i x_i^2} \\leq \\sqrt{\\sum_i \\|\\vec{x}\\|_\\infty^2} = \\sqrt{n \\|\\vec{x}\\|_\\infty^2} = \\sqrt{n} \\|\\vec{x}\\|_\\infty

Quick Reference

Concept	Formula	Key Property
L1 Norm	$\\|\vec{x}\\|_1 = \sum \\|x_i\\|$	Promotes sparsity
L2 Norm	$\\|\vec{x}\\|_2 = \sqrt{\sum x_i^2}$	Promotes smoothness
L∞ Norm	$\\|\vec{x}\\|_\infty = \max \\|x_i\\|$	Worst-case measure
Lp Norm	$\\|\vec{x}\\|_p = (\sum \\|x_i\\|^p)^{1/p}$	Interpolates L1–L∞
Frobenius	$\\|A\\|_F = \sqrt{\text{tr}(A^TA)}$	Matrix Euclidean norm
Spectral	$\\|A\\|_2 = \sigma_{\max}(A)$	Maximum stretch factor
Nuclear	$\\|A\\|_* = \sum \sigma_i$	Low-rank relaxation
Induced 1-norm	$\\|A\\|_1 = \max_j \sum_i \\|a_{ij}\\|$	Max column sum
Induced ∞-norm	$\\|A\\|_\infty = \max_i \sum_j \\|a_{ij}\\|$	Max row sum
Condition Number	$\kappa(A) = \\|A\\| \cdot \\|A^{-1}\\|$	Numerical stability

Cross-References

Vector Spaces: Foundation for defining norms
Eigenvalues and Singular Values: Used to compute spectral and nuclear norms
Inner Products: Cauchy-Schwarz inequality connects norms to inner products
Optimization: Regularization, gradient descent, constrained optimization
Machine Learning: Lasso, Ridge, Elastic Net, adversarial robustness
Numerical Linear Algebra: Condition numbers, stability analysis
Clustering: K-means uses Euclidean norm, K-medians uses L1
Dimensionality Reduction: PCA minimizes Frobenius norm reconstruction error

Relationship	Bound
$\\|\vec{x}\\|_\infty \leq \\|\vec{x}\\|_2$	$\\|\vec{x}\\|_2 \leq \sqrt{n} \\|\vec{x}\\|_\infty$
$\\|\vec{x}\\|_2 \leq \\|\vec{x}\\|_1$	$\\|\vec{x}\\|_1 \leq \sqrt{n} \\|\vec{x}\\|_2$
$\\|\vec{x}\\|_\infty \leq \\|\vec{x}\\|_1$	$\\|\vec{x}\\|_1 \leq n \\|\vec{x}\\|_\infty$