In $\mathbb{C}^n$, with inner product $\langle z,w \rangle = \sum_{j} z_j\overline{w}_j$. Let $T$ a linear operator such that $\langle T(z), T(w) \rangle = 0$, if $\langle z,w \rangle = 0$. Prove that $T = cU$, where c is a scalar and $U$ is unitary. I do not know where to start. I guess I need the matrix representation of $T$ but I do not know. In some exercisa like this I prove that the matrix representation was a diagonal matrix with different $c$ and then I proved that every $c$ was the same... But in this exercise I do not see how to start.
Any hint? Thanks in advance.
Let $\ z,w\in\mathbb{C}^n\ $ with $\ \langle z,w\rangle\ne0\ $, and let $$ v=z-\|w\|^{-2}\langle z,w\rangle w\ . $$ Then $\ \langle v,w\rangle=0\ $, and therefore \begin{align} 0&=\langle Tv,Tw\rangle\\ &=\langle Tz-\|w\|^{-2}\langle w,z\rangle Tw, Tw\rangle\\ &=\langle Tz,Tw\rangle-\|w\|^{-2}\langle w,z\rangle\langle Tw, Tw\rangle\ . \end{align} Since $\ \langle z,w\rangle\ne0\ $, this gives $$ \frac{\|Tw\|^2}{\|w\|^2}=\frac{\langle Tz,Tw\rangle}{\langle z,w\rangle}\ , $$ and since the expression left side of this equation is real, so is the one on the right side, and so we have \begin{align} \frac{\|Tw\|^2}{\|w\|^2}&=\frac{\langle Tz,Tw\rangle}{\langle z,w\rangle}\\ &=\frac{\langle Tw,Tz\rangle}{\langle w,z\rangle}\\ &=\frac{\|Tz\|^2}{\|z\|^2}\ , \end{align} where last identity follows by reversing the roles of $\ z\ $ and $\ w\ $ in the preceding argument, since $\ \langle w,z\rangle=\overline{\langle z,w\rangle}\ne0\ $.
Now suppose that $\ z,w\in\mathbb{C}^n\ $, with $\ z\ne0\ne w\ $, but $\ \langle z,w\rangle=0\ $. Let $\ u=z+w\ $. Then $\ \langle z,u\rangle=\|z\|^2\ne0\ $ and $\ \langle w,u\rangle=\|w\|^2\ne0\ $. The above argument now gives us \begin{align} \frac{\|Tu\|^2}{\|u\|^2}&=\frac{\|Tz\|^2}{\|z\|^2}\\ &=\frac{\|Tw\|^2}{\|w\|^2}\ . \end{align} It follows that $\ \frac{\|Tz\|^2}{\|z\|^2}=\frac{\|Tw\|^2}{\|w\|^2}\ $ for all non-zero $\ z,w\in\mathbb{C}^n\ $. That is, there's a scalar constant $\ c\ $ such that $\ \frac{\|Tz\|^2}{\|z\|^2}=c^2\ $ for all non-zero $\ z\in\mathbb{C}^n\ $. If $\ c=0\ $ then so is the operator $\ T\ $, and we have $\ T=cU\ $ for any unitary operator. Otherwise, we have $$ \|c^{-1}Tz\|= \|z\| $$ for all $\ z\in\mathbb{C}^n\ $, which is a sufficient condition for the operator $\ c^{-1}T\ $ to be unitary.