Prove that $U$ is a self adjoint unitary operator Let $W$ be the finite dimensional subspace of an inner product space $V$ and $V=W\oplus W^\perp $.
Define $U:V \rightarrow V$ by $U(v_1+v_2)=v_1-v_2$ where $v_1\in W$ and $v_2 \in W^\perp$.
Prove that $U$ is a self adjoint unitary operator.  
I know I have to show that $\parallel U(x) \parallel=\parallel x \parallel $ but can't proceed from this stage.

We can verify easily that 
$$U^2=\mathrm{Id}_V$$
so $U $ is the orthogonal symmetry relative to subspace $W $.
By the Pythagorean theorem we have
$$||U(x)||^2=||v_1-v_2||^2=||v_1||^2+||v_2||^2=||x||^2$$
