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Subsection 1.2.2 What is a vector norm?

A vector norm extends the notion of an absolute value to vectors. It allows us to measure the magnitude (or length) of a vector. In different situations, a different measure may be more appropriate.

Definition 1.2.2.1. Vector norm.

Let \(\nu: \Cm \rightarrow \mathbb R \text{.}\) Then \(\nu \) is a (vector) norm if for all \(x, y \in \Cm \) and all \(\alpha \in \mathbb C \)

  • \(x \neq 0 \Rightarrow \nu( x ) > 0 \) (\(\nu \) is positive definite),
  • \(\nu( \alpha x ) = \vert \alpha \vert \nu( x )\) (\(\nu \) is homogeneous), and
  • \(\nu(x + y ) \leq \nu( x ) + \nu( y ) \) (\(\nu \) obeys the triangle inequality).

Homework 1.2.2.1.

TRUE/FALSE: If \(\nu: \Cm \rightarrow \mathbb R \) is a norm, then \(\nu( 0 ) = 0 \text{.}\)

Hint

From context, you should be able to tell which of these \(0\)'s denotes the zero vector of a given size and which is the scalar \(0\text{.}\)

\(0 x = 0 \) (multiplying any vector \(x \) by the scalar \(0 \) results in a vector of zeroes).

Answer

TRUE.

Now prove it.

Solution

Let \(x \in \Cm \) and, just for clarity this first time, \(\vec{0} \) be the zero vector of size \(m \) so that \(0 \) is the scalar zero. Then

\begin{equation*} \begin{array}{l} \nu( \vec 0 ) \\ ~~~= ~~~~\lt 0 \cdot x = \vec 0 \gt \\ \nu( 0 \cdot x ) \\ ~~~=~~~~ \lt \nu(\cdots) \mbox{ is homogeneous } \gt \\ 0 \nu( x ) \\ ~~~ = ~~~~\lt \mbox{ algebra } \gt \\ 0 \end{array} \end{equation*}
Remark 1.2.2.2.

We typically use \(\| \cdot \| \) instead of \(\nu( \cdot ) \) for a function that is a norm.

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