Show a transformation is Conformal

Question

The question I've been given is as follows

A linear map $T:X→Y$ is said to be conformal when it preserves orthogonality;

$$∀x, ̃x∈X,〈x, ̃x〉= 0⇐⇒ 〈Tx,T ̃x〉= 0.$$

Show that this is the case if and only if $T^∗T=λI$ for some $λ >0$ . Moreover, if the underlying field is real,show that angles between vectors $cos(θ)=〈x,y〉/‖x‖‖y‖$ are preserved

Intuitively this makes sense to me, but I'm not sure how to prove. I think I need to use the paralleogram law to relate the inner product and the norm, something like below.

$$\langle Tx, T\tilde x\rangle = 0 = \langle T^*Tx, \tilde x \rangle$$ $$4\langle T^*Tx, \tilde x \rangle = ||T^*Tx + \tilde x||^2 - ||T^*Tx - \tilde x||^2 +i(||iT^*Tx + \tilde x||^2 - ||iT^*Tx - \tilde x||^2)$$ $$4\langle T^*Tx, \tilde x \rangle = ||T^*T||^2(||x + \tilde x||^2 - ||x - \tilde x||^2)+i||T^*T||^2(|ix + \tilde x||^2 - ||ix - \tilde x||^2)$$ But I don't know where to go from here. We were also given the following hints, but I don't think these prove useful until later on.

[Hint: Consider $〈x−y,x+y〉$ for orthogonal unit vectors x and y and use that $H= span(x)⊕span(x)^⊥$. Note that $〈w,x〉=〈z,x〉$ for all $x∈H$ implies that $w=z$. Why?]

Answer 1

since \begin{align*} \forall x,\tilde{x} & \in X,\left\langle x,\tilde{x}\right\rangle =0\Leftrightarrow\left\langle Tx,T\tilde{x}\right\rangle =0 \end{align*} then \begin{align*} \left\langle Tx,T\tilde{x}\right\rangle & =\left\langle T^{*}Tx,\tilde{x}\right\rangle \\ \left\langle T^{*}Tx,\tilde{x}\right\rangle & =\lambda\left\langle x,\tilde{x}\right\rangle \\ \left\langle T^{*}Tx-\lambda x,\tilde{x}\right\rangle & =0\\ \left\langle \left(T^{*}T-\lambda I\right)x,\tilde{x}\right\rangle & =0 \end{align*} and therefore, since this is true for all $\tilde{x}$ \begin{align*} \left(T^{*}T-\lambda I\right)x & =0 \end{align*} and because it must be true for all $x$: \begin{align*} T^{*}T & =\lambda I \end{align*}

Answer 2

Here is an extended hint. You still have to fill in the details. Using the original hint we have for orthogonal vectors $$ \|x\|=\|y\| \ \Leftrightarrow \|Tx\|=\|Ty\|. $$ Fix a unit vector $\tilde x$, then $T\tilde x\ne0$. Define $Sx:= \frac1{\|T\tilde x\|}Tx$. Then $S$ is norm preserving, i.e., $\|Sx\|=\|x\|$ for all $x$, where one uses $H = span(x) \oplus span(x)^\perp$. By these polarization identities, $S$ preserves the inner product, i.e., $\langle Sx,Sy\rangle = \langle x,y \rangle $ for all $x,y$. This is equivalent to $T^*T = \|T\tilde x\|^2 Id$.