Does there exist an angle reducing linear transformation in $\mathbb{R}^2$

Question

I have this ill posed question which has been bothering me for a while now.

Let $x,y$ be arbitrary vectors in $\mathbb{R}^2$.

Does there exists a non-trivial linear transform $T$ where $w = T(x)$ always lies in the lesser of the two regions* in between $x$ and $y$?

(Two regions: one is the one where $w$ lies as shown in the picture, the other is where its reflection lies)

Is there a generalization to $\mathbb{R}^n$?

Answer 1

Your question is completely unclear to me; two vectors do not partition the space into two regions, let alone into one that is less that the other (whatever that could mean).

But if you are looking for $f$ such that the angle between $f(v)$ and $f(w)$ is always (strictly) less than that between $v$ and $w$ then look no further; it does not exist. (If you allow equality of angles, then you can take the identity of a scalar multiplication; not really thrilling.)

This is simply because flat angles (as between $v$ and $-v$) are always preserved by a linear transformation. An even if you would exclude flat angles from your condition of decreasing angles, just divide a flat angle into a finite number of smaller angles; if they are all becoming smaller, then so is their sum, but this sum has to remain a flat angle.

Answer 2

This is impossible, for the following reason. If you consider unit vectors $x$ and $y,$ as in your picture, we can reflect $y$ across the line determined by $x$ to arrive at $y'.$ Now $T(x)$ needs to lie in the "lesser" region determined by $x$ and $y,$ so it has to lie above the line determined by $x,$ at a point like $w.$ But $T(x)$ also needs to lie below the line determined by $x,$ at a point like $w',$ since it needs to lie in the "lesser" region determined by $x$ and $y'.$ Since $w$ and $w'$ have a line between them, they certainly cannot be equal, so $T(x)$ cannot be defined properly.

To make this more rigorous, we can consider the effect of such a transformation on the inner product between vectors. Your condition amounts to saying that $\langle T(x),T(y)\rangle\geq\langle x,y\rangle$ for all unit vectors $x$ and $y.$ Then it must be true for $x$ and $-y,$ and using properties of the inner product and linearity of $T$: $$-\langle T(x),T(y)\rangle=\langle T(x),T(-y)\rangle\geq\langle x,-y\rangle=-\langle x,y\rangle,$$ so we also have for all $x$ and $y$ that $\langle T(x),T(y)\rangle\leq\langle x,y\rangle.$ Then these inner products are equal for all unit vectors $x$ and $y,$ which ensures that $T$ is an orthogonal transformation, i.e., a rotation or reflection, and by the same argument, it is clearly impossible for this inequality to be strict for all unit vectors $x$ and $y.$

These arguments can be easily generalized to $\mathbb{R}^{n},$ and in particular the second argument makes no reference to the fact that this is happening in the plane (besides when it says that $T$ must be a rotation or a reflection).