For two functions $f(x): \mathbb R^n \mapsto \mathbb R^m$, $g(x): \mathbb R^n \mapsto \mathbb R^m$. An usual way to define their distance may be $$ \|f-g\| = \sqrt{\int_x \|f(x)-g(x)\|^2 dx}. $$ I think we can prove this distance is a metric. Now suppose I allow linear invariance, i.e.:
If there exists a matrix $W \in \mathbb R^{m \times m}$ such that for any $x$, $f(x) = W g(x)$, then the distance between $f$ and $g$ is zero.
How could I modify the definition of distance above to take care of this invariance? Can we still prove that the modified distance is also a metric?
Following on from the intuition in projective spaces, the following would make a good metric for your new space
$$d(f,g) = \cos^{-1}\left[\int_{\Bbb{R}^n}\hat{f}\cdot \hat{g}\:dx\right] = \cos^{-1}\left[\frac{\int_{\Bbb{R}^n}f\cdot g\:dx}{\sqrt{\int_{\Bbb{R}^n}||f||^2\:dx}\sqrt{\int_{\Bbb{R}^n}||g||^2dx}}\right]$$
where $\hat{f}$ and $\hat{g}$ are the chosen representatives of their equivalence classes with unit norm in the original space.