Skip to main content

Subsection1.2.4The vector $p$-norms

A vector norm is a measure of the magnitude of a vector. The Euclidean norm (length) is merely the best known such measure. There are others. A simple alternative is the 1-norm.

Definition1.2.4.1. Vector 1-norm.

The vector 1-norm, $\| \cdot \|_1 : \C^m \rightarrow \mathbb R \text{,}$ is defined for $x \in \C^m$ by

\begin{equation*} \| x \|_1 = \vert \chi_0 \vert + \vert \chi_1 \vert + \cdots + \vert \chi_{m-1} \vert = \sum_{i=0}^{m-1} \vert \chi_i \vert. \end{equation*}
Homework1.2.4.1.

Prove that the vector 1-norm is a norm.

Solution

We show that the three conditions are met:

Let $x, y \in \C^m$ and $\alpha \in \mathbb C$ be arbitrarily chosen. Then

• $x \neq 0 \Rightarrow \| x \|_1 > 0$ ($\| \cdot \|_1$ is positive definite):

Notice that $x \neq 0$ means that at least one of its components is nonzero. Let's assume that $\chi_j \neq 0 \text{.}$ Then

\begin{equation*} \| x \|_1 = \vert \chi_0 \vert + \cdots + \vert \chi_{m-1} \vert \geq \vert \chi_j \vert > 0 . \end{equation*}
• $\| \alpha x \|_1 = \vert \alpha \vert \| x \|_1$ ($\| \cdot \|_1$ is homogeneous):

\begin{equation*} \begin{array}{l} \| \alpha x \|_1 ~~~=~~~~ \lt \mbox{ scaling~a~vector-scales-its-components; definition} \gt \\ \vert \alpha \chi_0 \vert + \cdots + \vert \alpha \chi_{m-1} \vert \\ ~~~=~~~~ \lt \mbox{ algebra} \gt \\ \vert \alpha \vert \vert \chi_0 \vert + \cdots + \vert \alpha \vert \vert \chi_{m-1} \vert \\ ~~~=~~~~ \lt \mbox{ algebra} \gt \\ \vert \alpha \vert ( \vert \chi_0 \vert + \cdots + \vert \chi_{m-1} \vert ) \\ ~~~=~~~~ \lt \mbox{ definition} \gt \\ \vert \alpha \vert \| x \|_1. \end{array} \end{equation*}
• $\| x + y \|_1 \leq \| x \|_1 + \| y \|_1$ ($\| \cdot \|_1$ obeys the triangle inequality):

\begin{equation*} \begin{array}{l} \| x + y \|_1 \\ ~~~=~~~~ \lt \mbox{ vector~addition;~definition~of~1-norm} \gt \\ \vert \chi_0 + \psi_0 \vert + \vert \chi_1 + \psi_1 \vert + \cdots + \vert \chi_{m-1} + \psi_{m-1} \vert \\ ~~~\leq~~~~ \lt \mbox{ algebra} \gt \\ \vert \chi_0 \vert + \vert \psi_0 \vert + \vert \chi_1 \vert + \vert \psi_1 \vert + \cdots + \vert \chi_{m-1} \vert + \vert \psi_{m-1} \vert \\ ~~~=~~~~ \lt \mbox{ commutivity} \gt \\ \vert \chi_0 \vert + \vert \chi_1 \vert + \cdots + \vert \chi_{m-1} \vert + \vert \psi_0 \vert + \vert \psi_1 \vert + \cdots + \vert \psi_{m-1} \vert \\ ~~~= ~~~~ \lt \mbox{ associativity;~definition} \gt \\ \| x \|_1 + \| y \|_1. \end{array} \end{equation*}

The vector 1-norm is sometimes referred to as the "taxi-cab norm". It is the distance that a taxi travels, from one point on a street to another such point, along the streets of a city that has square city blocks.

Another alternative is the infinity norm.

Definition1.2.4.2. Vector $\infty$-norm.

The vector $\infty$-norm, $\| \cdot \|_\infty : \C^m \rightarrow \mathbb R \text{,}$ is defined for $x \in \C^m$ by

\begin{equation*} \| x \|_\infty = \max( \vert \chi_0 \vert, \ldots , \vert \chi_{m-1} ) = \max_{i=0}^{m-1} \vert \chi_i \vert. \end{equation*}

The infinity norm simply measures how large the vector is by the magnitude of its largest entry.

Homework1.2.4.2.

Prove that the vector $\infty$-norm is a norm.

Solution

We show that the three conditions are met:

Let $x, y \in \C^m$ and $\alpha \in \mathbb C$ be arbitrarily chosen. Then

• $x \neq 0 \Rightarrow \| x \|_\infty > 0$ ($\| \cdot \|_\infty$ is positive definite):

Notice that $x \neq 0$ means that at least one of its components is nonzero. Let's assume that $\chi_j \neq 0 \text{.}$ Then

\begin{equation*} \| x \|_\infty = \max_{i=0}^{m-1} \vert \chi_i \vert \ge \vert \chi_j \vert > 0. \end{equation*}
• $\| \alpha x \|_\infty = \vert \alpha \vert \| x \|_\infty$ ($\| \cdot \|_\infty$ is homogeneous):

\begin{equation*} \begin{array}{lcl} \| \alpha x \|_\infty \amp =\amp \max_{i=0}^{m-1} \vert \alpha \chi_i \vert \\ \amp =\amp \max_{i=0}^{m-1} \vert \alpha \vert \vert \chi_i \vert \\ \amp =\amp \vert \alpha \vert \max_{i=0}^{m-1} \vert \chi_i \vert \\ \amp =\amp \vert \alpha \vert \| x \|_\infty. \end{array} \end{equation*}
• $\| x + y \|_\infty \leq \| x \|_\infty + \| y \|_\infty$ ($\| \cdot \|_\infty$ obeys the triangle inequality):

\begin{equation*} \begin{array}{lcl} \| x + y \|_\infty \amp =\amp \max_{i=0}^{m-1} \vert \chi_i + \psi_i \vert \\ \amp \leq\amp \max_{i=0}^{m-1} ( \vert \chi_i \vert + \vert \psi_i \vert ) \\ \amp \leq\amp \max_{i=0}^{m-1} \vert \chi_i \vert + \max_{i=0}^{m-1} \vert \psi_i \vert \\ \amp = \amp \| x \|_\infty + \| y \|_\infty. \end{array} \end{equation*}

In this course, we will primarily use the vector 1-norm, 2-norm, and $\infty$-norms. For completeness, we briefly discuss their generalization: the vector $p$-norm.

Definition1.2.4.3. Vector $p$-norm.

Given $p \geq 1 \text{,}$ the vector $p$-norm $\| \cdot \|_p : \C^m \rightarrow \mathbb R$ is defined for $x \in \C^m$ by

\begin{equation*} \| x \|_p = \sqrt[p]{\vert \chi_0 \vert^p + \cdots + \vert \chi_{m-1} \vert^p} = \left( \sum_{i=0}^{m-1} \vert \chi_i \vert^p \right)^{1/p}. \end{equation*}

The proof of this result is very similar to the proof of the fact that the 2-norm is a norm. It depends on Hölder's inequality, which is a generalization of the Cauchy-Schwartz inequality:

We skip the proof of Hölder's inequality and Theorem 1.2.4.4. You can easily find proofs for these results, should you be interested.

Remark1.2.4.6.

The vector 1-norm and 2-norm are obviously special cases of the vector $p$-norm. It can be easily shown that the vector $\infty$-norm is also related:

\begin{equation*} \lim_{p \rightarrow \infty} \| x \|_p = \| x \|_{\infty}. \end{equation*}
Ponder This1.2.4.3.
Consider Homework 1.2.3.3. Try to elegantly formulate this question in the most general way you can think of. How do you prove the result?
Ponder This1.2.4.4.

Consider the vector norm $\| \cdot \|: \Cm \rightarrow \mathbb R \text{,}$ the matrix $A \in \mathbb C^{m \times n}$ and the function $f: \Cn \rightarrow \mathbb R$ defined by $f( x ) = \| A x \| \text{.}$ For what matrices $A$ is the function $f$ a norm?