The norm is the “length” of a vector, defined generally using the inner product as:
\begin{equation} \|v\| = \sqrt{\langle v,v \rangle} \end{equation}
additional information
properties of the norm
- \(\|v\| = 0\) IFF \(v=0\)
- \(\|\lambda v\| = |\lambda|\|v\|\)
Proof:
- By definition of an inner product, \(\langle v,v \rangle = 0\) only when \(v=0\)
- See algebra:
\begin{align} \|\lambda v\|^{2} &= \langle \lambda v, \lambda v \rangle \\ &= \lambda \langle v, \lambda v \rangle \\ &= \lambda \bar{\lambda} \langle v,v \rangle \\ &= |\lambda |^{2} \|v\|^{2} \end{align}
motivating the norm using actual numbers
In linear algebra, the norm of a vector in a real vector space is defined as follows:
\begin{equation} \| x\| = \sqrt{{{x_1}^{2} + \dots + {x_n}^{2}}} \end{equation}
Note that, given the definition of dot product, \(\| x \|^{2} = x \cdot x\).
The norm in complex vector space requires taking the absolute value (for \(a+bi\), \(|a+bi| = \sqrt{{a^{2}+b^{2}}}\)) of each slot. That is, for Euclidean Inner Product spaces:
\begin{equation} \|z\| = \sqrt{|z_1|^{2} + \dots |z_{n}|^{2}} \end{equation}
otherwise, simply squaring the complex number (giving us \(a^{2}-b^{2}\)) may very well yield negative numbers, which means we’d have an imaginary norm!