1. Databases
2. Proof database
3. For a positive definite metric if the triangle inequality is an equality the two vectors are colinear of same direction

For a positive definite metric if the triangle inequality is an equality the two vectors are colinear of same direction

Theorem : Iff two vectors $x, y \in V$ are colinear, such that there exists a $\lambda > 0$ such that $x = \lambda y$, then, for our norm $\| x \| = \sqrt{x^\top M x}$,

$$\begin{equation} \| x + y \| = \| x \| + \| y \| \end{equation}$$

Proof : If $x$ and $y$ are colinear, then

$$\begin{eqnarray} \| x + y \| &=& \| \lambda y + y \|\\ &=& \| (1 + \lambda) y \|\\ &=& |1 + \lambda| \| y \| \end{eqnarray}$$

and

$$\begin{eqnarray} \| x \| + \| y \| &=& \| \lambda y \| + \| y \|\\ &=& |\lambda| \| y \| + \| y \|\\ &=& (1 + |\lambda|) \| y \|\\ &=& |1 + \lambda| \| y \| \end{eqnarray}$$

Conversely,

$$\begin{eqnarray} \| x + y \| - \| x \| - \| y \| &=& \sqrt{(x + y)^\top M (x + y)} - \sqrt{x^\top M x} - \sqrt{y^\top M y}\\ &=& \sqrt{x^\top M x + x^\top M y + y^\top M x + y^\top M y} - \sqrt{x^\top M x} - \sqrt{y^\top M y}\\ &=& 0 \end{eqnarray}$$