Given $W$ is a real inner product space of dimension $2k$. Assume $T$ is an orthogonal transformation on $W$ such that $T^2=-I$ (i.e.negative identity matrix). Show there exists a subspace $P\subset W$ of dimension $k$.
My idea: By the definition of orthogonal projection, any vector in an inner product space admits a unique representation $v=v_1 +v_2$ where $v_1 \in E\subset V$ and $v_2 \in E^{\perp}$. If it is a good starting point, could you give me a hint to proceed? In particular, I am wondering how to build connections between $T$ and $M$. Also, I know since $\dim W=2k$, the corresponding complex inner product space would have dimension $k$. But how does this help? (I literally do not understand the hint.)
Hint: Consider $P=\{x+T(x), x\in W\}$ and $Q=\{x-T(x),x\in W\}$,
$T(x+T(x))=T(x)+T^2(x)=T(x)-x=-(x-T(x))$ implies that $T(P)\subset Q$ and $dim P\leq dim Q$ since $T$ is injective, $T(x-T(x))=T(x)-T^2(x))=x+T(x)$ implies that $T(Q)\subset P$ and $dimQ\leq dim P$ since $T$ is injective, $x={{x+T(x)}\over 2}+{{x-T(x)}\over 2}$ implies that $W=P+Q$
$\langle T(x)+x,T(x)-x\rangle=\langle T(x),T(x)\rangle-\langle T(x),x\rangle+\langle x,T(x)\rangle-\langle x,x\rangle=0$ implies that $P$ and $Q$ are orthogonal. We deduce that $P\cap Q=\{0\}$ and since $W=P\oplus Q$ since $dimP=dimQ$, we deduce that $dimP=k$.