Linear transformation which preserves angle

Question

Given a linear transformation $L:V->W$. $L$ follows a property that for every perpendicular vectors $(u , v)$ in $V$ , $(T(u),T(v))$ is perpendicular in W. Formally, $<u,v>=0$ implies $<L(u),L(v)>=0$. (here we consider V and W to be real vector space of n dimensions)

We need to prove that transformation L preserves angle. In other words, for every pair of vectors $(u,v)$ making angle $x$ in V space, pair $(T(u),T(v))$ also makes angle $x$ in W.

My approach- I have proved that orthogonal matrix transformations preserves angle. Following on the same line, I realised the fact that - If $L$ changes the length of all vectors with the same multiple, then the proof is fairly straightforward. Which i believe is the case here. Visually, I proved the fact for two and three dimensions but am unable to proceed.

Answer 1

Let $(e_i)_{i\in I}$ be an orthonormal basis of $V$ and let $i,j\in I$. Define $u=e_i-e_j$ and $v=e_i+e_j$. Then,

$$ 0=\langle u,v\rangle=\langle Lu,Lv\rangle= \langle Le_i,Le_i\rangle-\langle Le_j, Le_j\rangle+2 \Im\langle Le_i,Le_j\rangle=\|Le_i\|^2-\|Le_j\|^2, $$ so $i\mapsto \|Le_i\|$ is a constant map.

Hence, for general $u=\sum_{i\in I} \langle u,e_i\rangle e_i$ and $v=\sum_{i\in I} \langle v,e_i\rangle e_i$, we have

$$ \langle Lu,Lv\rangle= \sum_{i\in I} \langle u,e_i\rangle \overline{\langle v,e_i\rangle} \|Le_i\|^2=\|Le_1\|^2\langle u,v\rangle, $$ Applying this observation to the case $u=v$ yields that $\|Lu\|=\|Le_1\| \|u\|$.

All in all, we get that $L$preserves angles unless it is the $0$-map.

Answer 2

To see that a scalar multiple of two orthogonal vectors is still orthogonal we can use the bilinearity of the inner product as $\langle ku,kv \rangle=k^2\langle u,v \rangle = 0$ for any scalar $k$ with $u$ and $v$ orthogonal.

Answer 3

You know that $L(v) \cdot L(w) = \lambda^2 (v \cdot w)$. You also know that $||L(v)|| = \lambda ||v||$. You need to proceed to show that these statements are equivalent (L changes all vectors with the same multiple).