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.

• Hint: notice your operator is really of the form $\mbox{diag}\{1,-1\}$ in that orthogonal decomposition, where "$1$" means identity operator on the appropriate subspace. Hint #2: a self-adjoint unitary must be a $2 \times 2$ operator matrix of that form by the spectral theorem. – Michael Jun 1 '13 at 18:50
• @Michael: I don't think $W$ is necessarily one-dimensional. – Najib Idrissi Jun 1 '13 at 18:52
• For the unitary/isometry part: Pythagoras. What are $\|v_1+v_2\|^2$ and $\|v_1-v_2\|^2$ when $v_1$ and $v_2$ are orthogonal? – Julien Jun 1 '13 at 18:57
• Possible duplicate of Self - adjoint and Unitary operator – Maurice P May 7 at 18:39

$\langle U(x),U(x)\rangle = \langle U(v_1+v_2) , U(v_1+v_2)\rangle = \langle v_1 - v_2, v_1 - v_2\rangle = \langle v_1,v_1\rangle + \langle v_2,v_2\rangle = \langle x,x\rangle$ where last two equalities comes frome the fact that $\langle v_1,v_2\rangle = 0$
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 $$||U(x)||^2=||v_1-v_2||^2=||v_1||^2+||v_2||^2=||x||^2$$