Section 2.6 Scalar Products
Some norms come from a richer structure called scalar product, defined as follows.Definition 2.6.1. Scalar Product.
Let \(V\) be a vector space over \(\bR\text{.}\) A scalar product on \(V\) is a bilinear map \(\langle\cdot,\cdot\rangle:V\times V\to\bR\) such that:- \(\langle v,v \rangle\geq0\) for all \(v\in V\text{;}\)
- \(\langle v,v \rangle=0\) if and only if \(v=0\text{;}\)
- \(\langle u,v \rangle=\langle v,u \rangle\) for all \(u,v\in V\text{.}\)
\begin{equation*}
\langle u,v \rangle=\overline{\langle v,u \rangle}.
\end{equation*}
Every space with a scalar product is a normed vector space (see Exercise 2.7.5) with respect to the norm
\begin{equation*}
\|v\| = \langle v,v \rangle^{1/2}.
\end{equation*}
For instance, the Euclidean norm \(\|\cdot\|_2\) in dimension \(n\) comes from the Euclidean scalar product
\begin{equation*}
\langle u,v \rangle = u^1v^1+\dots+u^nv^n.
\end{equation*}
Does each norm come from a scalar product? The answer is negative because each norm coming from a scalar product satisfies the parallelogram equality
\begin{equation*}
2\|u\|^2+2\|v\|^2 = \|u+v\|^2 + \|u-v\|^2\text{ for all }u,v\in V.
\end{equation*}
A direct check shows that this equality does not hold for the Taxicab and Chebychev norms. The inverse also holds: