Let $A\colon\,\mathbb R^2 →\mathbb R^2$ be a linear map, and suppose there exists $c > 0$ such that $\langle Av, Aw\rangle = c \langle v,w\rangle$ for all $v, w ∈ \mathbb R^2$
Show that $A$ preserves the angles between vectors.
My attempt:
I have to show that $A$ is conformal.
Since $A$ is a linear map, we can write $Av = A(v.1) = v.A(1) = v$ and $Aw = A(w.1) = w.A(1) = w$.
And so, I would have $\langle Av,Aw\rangle = \langle v,w\rangle = c\langle v,w\rangle$.
Hence $\langle v,w\rangle = c\langle v,w\rangle$, so the angle between $v$ and $w$ hasn't changed. The inner product has just been multiplied by a constant $c$ which has dilated and rotated $\langle v,w\rangle$.
Thus $A$ is conformal.
Is my attempt correct? Any other solutions to this problem? Thank you.
No – a linear map $A$ takes vectors into vectors – expression $A(1)$ is not defined, because 1 is not a vector.
The angle between two vectors $v, w$ can be calculated using its cosine
$$\cos\angle(v,w) = \frac{\langle v, w\rangle}{\sqrt{\langle v, v \rangle \langle w, w\rangle }}$$
Just express $\cos\angle(Av,Aw)$ using the above definition and the property of map $A$.