Proving that a linear operator is unitary Let $V$ be a finite dimensional complex vector space with a scalar product and let $u$ be an unitaty vector in $V$. Let $H_u: V \to V$ be the linear operator defined by $H_u(v)=v-2\langle v, u \rangle u$ for all $v \in V$. I want to prove that $H_u$ is unitary, i.e., $\langle H_u(v), H_u(w)\rangle = \langle v, w \rangle$ for all $v, w \in V$.
My attempt:
Since, $\langle u, u \rangle=1$,
\begin{eqnarray}
\langle H_u(v), H_u(w)\rangle & = & \langle v-2\langle v, u \rangle u, w-2\langle w, u \rangle u \rangle \\
& = & \langle v, w \rangle-2\langle w, u \rangle\langle v, u \rangle -2 \langle v, u \rangle\langle u, w \rangle +4\langle v, u \rangle\langle w, u \rangle \\
& = & \langle v, w \rangle-2\langle w, u \rangle\langle v, u \rangle -2 \langle v, u \rangle\langle u, w \rangle +2\langle v, u \rangle\langle w, u \rangle +2\langle v, u \rangle\langle w, u \rangle \\
& = & \langle v, w \rangle-2\langle w, u \rangle\langle v, u \rangle -2 \langle v, u \rangle\langle u, w \rangle +2\langle v, u \rangle\langle w, u \rangle +2\langle v, u \rangle\langle w, u \rangle \\
& = & \langle v, w \rangle -2 \langle v, u \rangle\langle u, w \rangle +2\langle v, u \rangle\langle w, u \rangle \\
& = & \langle v, w \rangle -2 \langle v, u \overline{\rangle\langle w, u \rangle} +2\langle v, u \rangle\langle w, u \rangle,
\end{eqnarray}
where the bar means complex conjugate. I don't know what to do because $-2 \langle v, u \overline{\rangle\langle w, u \rangle} +2\langle v, u \rangle\langle w, u \rangle$ should be zero. What can I do? Thanks in advance.
 A: I think your first step is wrong. It should be
\begin{align*}
\langle H_u(v), H_u(w)\rangle &= \langle v-2\langle v, u \rangle u, w-2\langle w, u \rangle u \rangle \\
&= \langle v, w \rangle - 2\langle v, \langle w, u \rangle u \rangle - 2 \langle \langle v,u \rangle u, w \rangle +4 \langle\langle v, u \rangle u, \langle w, u \rangle u \rangle \\
&= \langle v, w \rangle - 2 \overline{\langle w, u \rangle} \langle v, u \rangle - 2 \langle v,u \rangle \langle u, w \rangle + 4 \langle v, u \rangle \overline{\langle w, u \rangle} \langle u, u \rangle \\
&= \langle v, w \rangle - 2 \overline{\langle w, u \rangle} \langle v, u \rangle - 2 \langle v,u \rangle \langle u, w \rangle + 4 \langle v, u \rangle \overline{\langle w, u \rangle} \\
&= \langle v, w \rangle - 2 \langle v,u \rangle \langle u, w \rangle + 2 \langle v, u \rangle \overline{\langle w, u \rangle} \\
&= \langle v, w \rangle - 2 \langle v,u \rangle \langle u, w \rangle + 2 \langle v, u \rangle \langle u, w \rangle \\
&= \langle v, w \rangle.
\end{align*}
A: Another option would be to represent the linear map in an orthonormal basis, and prove that its matrix is unitary. For instance, take any orthonormal basis that whose first vector is $u$, call it $\mathcal{B}=(u,e_2,\dots,e_n)$. Then
$$\mathrm{Mat}(u;\mathcal B)=\begin{pmatrix}-1\\&1\\&&\ddots\\&&&1\\&&&&1\end{pmatrix}$$
is unitary, and you are done.
