Proving that operator is linear if it is unitary and surjective 
Let $X$ be unitary space. Show that surjective operator $T: X\to X$ such that $\langle Tx, Ty\rangle = \langle x, y \rangle \forall x, y \in X$ is linear.


The only thing I was able to conclude (trivial stuff) is that
$$
\langle Tx, Ty\rangle = \langle x, y \rangle \implies \langle Tx, Tx\rangle = \langle x, x \rangle \implies \sqrt{\langle Tx, Tx\rangle} = \sqrt{\langle x, x \rangle} \\
\implies \| Tx\| = \|x\|
$$
so $T$ is isometry.
To show that it is linear I want to have:
$$
T(ax) = aT(x) \\
T(x+y) = T(x)+T(y)
$$
 A: Here is an interpretation of the answers so far given, breaking down the argument, and highlighting an
important intermediate result.
First fact. Let $T$ be a surjective isometry.  Then $T$ admits an adjoint operator.  In other words,
for every $y$ in $X$,  there exists a (necessarily unique) $z$ such that
$$
  \langle T(x), y\rangle  =   \langle x, z\rangle , \quad\forall x\in  X.
  $$
Proof (@Udalricus.S)  Given $y$,  find $z$ such that $T(z)=y$.  Then
$$
  \langle T(x), y\rangle  =   \langle T(x), T(z)\rangle  = \langle x, z\rangle .
  \tag*{$\square$}
  $$
Incidentally,  the vector $z$ referred to above is often denoted by $T^*(y)$,  so the highlited expression
in the statement  becomes the familiar one:
$$
  \langle T(x), y\rangle  =   \langle x, T^*(y)\rangle .
  $$
Second  fact. Let $T:X\to X$ be a function (not necessarily linear,  unitary or surjective) and suppose that there exists another
function $T^*:X\to X$,  such that
$$
  \langle T(x), y\rangle  =   \langle x, T^*(y)\rangle \quad\forall x, y\in  X.
  $$
Then $T$ is linear.
Proof.  (@Chrystomath) Given $x_1$ and $x_2$ in $X$,  we have for all $y$ that
$$
  \langle T(x_1+x_2), y\rangle  =
  \langle x_1+x_2, T^*(y)\rangle  = $$$$ =
  \langle x_1, T^*(y)\rangle  +  \langle x_2, T^*(y)\rangle  =
  \langle T(x_1), y\rangle  +   \langle T(x_2), y\rangle  = $$$$ =
  \langle T(x_1) +   T(x_2), y\rangle .
  $$
Since $y$ is arbitrary, we deduce that $T(x_1+x_2)=T(x_1) +   T(x_2)$.  Similarly one proves that
$T(\lambda x)=\lambda T(x)$.  QED
As an added bonus, should $X$ be a Hilbert space, one can show by means of the closed graph Theorem that
the map $T$ of the second fact above is moreover bounded!
A: For example, use the following: Let $x,y\in X$. By surjectivity, there is a $z\in X$ such that $y=Tz$. Then
$$\langle T(\alpha x),y\rangle = \langle T(\alpha x), Tz\rangle =\langle \alpha x, z\rangle=\alpha\langle x, z\rangle = \alpha\langle Tx, y\rangle$$
Then we have for all $y$
$$\langle (T(\alpha x)-\alpha Tx),y\rangle=0$$
By non-degeneracy, it follows that $T(\alpha x)-\alpha Tx=0$. I think the second property can be shown in a similar way.
A: \begin{align}
\langle T(x+y),T(z)\rangle&=\langle x+y,z\rangle\\
&=\langle x,z\rangle +\langle y,z\rangle\\
&=\langle T(x),T(z)\rangle+\langle T(y),T(z)\rangle\\
&=\langle T(x)+T(y),T(z)\rangle
\end{align}
Since $T$ is onto, $T(z)$ can be any vector, hence $T(x+y)=T(x)+T(y)$.
Same proof can be modified for $T(\alpha x)=\alpha T(x)$.
