Let \(V\) be an inner product space. The distance between two vectors \(\alpha\) and \(\beta\) in \(V\) is defined by \[ d(\alpha, \beta) = ||\alpha - \beta||. \] Show that
\(d(\alpha, \beta) \geq 0\);
\(d(\alpha, \beta) = 0\) if and only if \(\alpha = \beta\);
\(d(\alpha, \beta) = d(\beta, \alpha)\);
\(d(\alpha, \beta) \leq d(\alpha, \gamma) + d(\gamma, \beta)\).
Solution
By definition, \[ d(\alpha,\beta)=\|\alpha-\beta\|. \]
1. The distance is nonnegative and vanishes exactly on the diagonal.
Every norm is nonnegative, so \[ d(\alpha,\beta)=\|\alpha-\beta\|\ge0. \] Moreover, \[ d(\alpha,\beta)=0 \iff \|\alpha-\beta\|=0 \iff \alpha-\beta=0 \iff \alpha=\beta. \] This proves (a) and (b).
2. The distance is symmetric.
Since norms are absolutely homogeneous, \[ d(\beta,\alpha) =\|\beta-\alpha\| =\|-(\alpha-\beta)\| =|-1|\,\|\alpha-\beta\| =d(\alpha,\beta). \]
3. The distance satisfies the triangle inequality.
Because \[ \alpha-\beta=(\alpha-\gamma)+(\gamma-\beta), \] the norm triangle inequality gives \[ \begin{aligned} d(\alpha,\beta) &=\|\alpha-\beta\|\\ &\le\|\alpha-\gamma\|+\|\gamma-\beta\|\\ &=d(\alpha,\gamma)+d(\gamma,\beta). \end{aligned} \] Thus \(d\) is a metric [@HK71].