Let $V$ be a finite dimensional real inner product space and $L: V \rightarrow V$ be a self-adjoint isometry. Show that there exists a subspace $U$ of $V$ such that $L(x +y) = x -y$ whenever $x \in U$ and $y \in U^{\perp}$.
Here is my solution (it is a problem from a qual I took an hour ago since the posting time and I will appreciate any critique or an alternatively better argument):
The primary decomposition theorem says that if $p(t) = p_{1}(t) ... p_{k}(t)$ is a polynomial so that $\text{gcd}(p_{1}, ..., p_{k}) = 1$ and $p(L) = 0$, where $L$ is a linear map on a finite dimensional vector space $V$, then $V = \ker p_{1}(L) \oplus \ker p_{2}(L) \oplus ... \oplus \ker p_{k}(L)$.
Note that $L: V \rightarrow V$ is self-adjoint $\implies L^{*} = L$ and $L$ being isometry $\implies L$ is unitary so that $ LL^{*} = I$. It follows that $I = LL^{*} = L^{2}$ so that $L$ satisfies the polynomial $p(t) = (t-1)(t+1)$.
By the primary decomposition theorem, $V = \ker (L -I) \oplus \ker (L +I)$. Let $U = \ker (L -I)$. I did assume without showing that $U^{\perp} = \ker (L + I)$. Let $v = x +y$ be written uniquely, then $L(v) = L(x) + L(y) = x - y$.