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L1 Norm — Definition, Formula & Examples

The L1 norm of a vector is the sum of the absolute values of all its components. It measures the total magnitude of a vector without allowing positive and negative entries to cancel each other out.

For a vector x=(x1,x2,,xn)Rn\mathbf{x} = (x_1, x_2, \ldots, x_n) \in \mathbb{R}^n, the L1 norm (also called the taxicab norm or Manhattan norm) is defined as x1=i=1nxi\|\mathbf{x}\|_1 = \sum_{i=1}^{n} |x_i|. It satisfies the three axioms of a norm: positive definiteness, absolute homogeneity, and the triangle inequality.

Key Formula

x1=i=1nxi=x1+x2++xn\|\mathbf{x}\|_1 = \sum_{i=1}^{n} |x_i| = |x_1| + |x_2| + \cdots + |x_n|
Where:
  • x\mathbf{x} = A vector in $\mathbb{R}^n$
  • xix_i = The $i$-th component of the vector
  • nn = The dimension (number of components) of the vector

How It Works

To compute the L1 norm, take each component of the vector, find its absolute value, then add all those absolute values together. The name "Manhattan norm" comes from a geometric interpretation: in a grid-based city like Manhattan, the L1 norm gives the distance you would walk along streets to get from the origin to a point, since you cannot cut diagonally. Unlike the more common L2 (Euclidean) norm, the L1 norm does not square the components, making it less sensitive to outliers.

Worked Example

Problem: Find the L1 norm of the vector x=(3,7,2,1)\mathbf{x} = (3, -7, 2, -1).
Step 1: Take the absolute value of each component.
3=3,7=7,2=2,1=1|3| = 3, \quad |-7| = 7, \quad |2| = 2, \quad |-1| = 1
Step 2: Sum all the absolute values.
x1=3+7+2+1=13\|\mathbf{x}\|_1 = 3 + 7 + 2 + 1 = 13
Answer: x1=13\|\mathbf{x}\|_1 = 13

Why It Matters

The L1 norm is central to sparse optimization problems in machine learning and compressed sensing, where minimizing x1\|\mathbf{x}\|_1 encourages solutions with many zero entries. In statistics, L1 regularization (Lasso regression) uses this norm to perform automatic feature selection in high-dimensional models.

Common Mistakes

Mistake: Summing the components without taking absolute values, getting a result that can be zero or negative.
Correction: Always apply absolute value to each component before summing. Without this step, positive and negative entries cancel, and the result is not a valid norm.