Rate of vectors becoming collinear with respect to different inner products I am convinced that the rate with which two vectors become collinear is asymptotically independent of the inner product that we use to define the angle between the two vectors. However, I don't have an elegant proof for this, but believe that it should be a standard result on the relation between different inner products on finite dimensional vector spaces.
Given a finite dimensional real vector space $V$ with its standard topology inherited form $\mathbb{R}^n$ and two continuous functions $v: [0,\infty)\to V$ and $w: [0,\infty)\to V$ that become asymptotically collinear in the sense that the angle
\begin{align}
\theta(t)=\cos^{-1}\left(\frac{\langle v(t),w(t)\rangle_g}{\lVert v(t)\rVert_g\,\lVert w(t)\rVert_g}\right)
\end{align}
with respect to the inner product $g$ on $V$ approaches zero as $t\to\infty$, THEN IT IS TRUE that the angle $\tilde{\theta}(t)$ with respect to another inner product $\tilde{g}$ has the same asymptotic behavior, that is
\begin{align}
 \lim_{t\to\infty}\frac{\tilde\theta(t)}{\theta(t)}=1\,.
\end{align}
Essentially, I believe that vectors that become continuously collinear do it asymptotically in the same way with respect to all possible inner products. Obviously, that should only apply to finite dimensional vector spaces where one can bound the relation between different inner products.
Do you agree? Is there a standard proof for this or theorem that I can quote? Do you have an elegant way of showing it? I studied it explicitly for 2d and I might have some ideas towards a proof, but it's not very elegant. Moreover, I assume that somebody will have shown this before me, unless I'm wrong and it doesn't hold.
 A: Seems to me that this is not true. And if it ain't true, don't try to prove it.
Simple example, in $\Bbb R^2$, consider a scalar product $x^\top(\begin{smallmatrix}1&0\\0&\lambda\end{smallmatrix})y$ with $\lambda>0$, which of course for $\lambda=1$ gives the standard inner product. I'll consider the angle between $v=(1,0)$ and $w=(1,y)$, for small$~y$. Small angles are well approximated by their sine, so $\arccos(t)\approx\sqrt{1-t^2}$ for small$~t$. Now we have $(v,w)=1=\|v\|=1$ independently of$~\lambda$, and $\|w\|=\sqrt{1+\lambda y^2}$. Then we are computing
$$
  \arccos\left(\frac1{\sqrt{1+\lambda y^2}}\right)
  \approx\sqrt{1-\frac1{1+\lambda y^2}}
  \approx\sqrt{\lambda y^2} = \sqrt\lambda\, y,
$$
so the angle is asymptotically proportional to$~y$ by the square root of the coefficient $\lambda$ used in the inner product. Comparing this for different values of $\lambda$, the ratio of the angles does not tend to$~1$ as $y\to0$.
A: Alright, it is enough to say that the first inner product is the ordinary dot product, while the second amounts to a symmetric positive definite matrix $A.$ All eigenvalues are real, the iner product is just $u^T A v,$ where both $u,v$ are written as column vectors. 
Let $u \cdot u = 1,$ $w \cdot w = 1,$ with $u \cdot w = 0.$ Then define
$$ v = u \cos t + w \sin t. $$ We se that $u \cdot v = \cos t.$
Now we want the ratio 
$$ \frac{u^T A v}{\sqrt {u^T A u} \; \; \; \sqrt {v^T A v}}. $$
For convenience, name
$$ \alpha = u^TAu, \; \; \beta = u^T A v, \; \; \gamma = v^T A v $$
Note that, if $m$ is the smallest eigenvalue of $A$ and $M$ its largest, that $\alpha \geq m$ and $|\beta|, \gamma \leq M.$ I get that your ratio above is
$$ \frac{\cos t + \left( \frac{\beta}{\alpha} \right) \sin t}{ \sqrt{\cos^2 t + 2\left( \frac{\beta}{\alpha} \right) \cos t \sin t + \left( \frac{\gamma}{\alpha} \right)  \sin^2 t \; \;}}$$  
This approaches $1,$ and the difference from $1$ can be bounded in terms of $M$ and $m,$ independent of the actual direction of $w.$  
