Why do I have to show this subspace is an invariant subspace? Consider a vector space $V \cong \mathbb{R}^n$ with an operator $I \in O(n)$ satisfying the property $I^2 = -Id_{V}$. See Linear Complex Structure for context. I want to show that $V$ has real dimension $2n$ (even dimension).
To do this, I was given the hint of working with $W = span\{x,Ix\}$ for a vector $x \in V$. It can be shown easily that $W$ is an invariant subspace of $I$ and $x \; \bot \; Ix$. Next I was told to note that $V = W \oplus W^{\perp}$ and to show that $W^{\perp}$ is also an invariant subspace of $I$.
This is where I am confused. I do not understand why we should show that $W^{\perp}$ is also an invariant subspace. I see that $W$ has degree $2$ and $W^{\perp}$has degree $n-2$, but i'm not really sure what I should be examining with the directsum $W \oplus W^{\perp}$.
 A: Let $\vec{0}\neq x\in V$ be arbitrary and 
$$W = \text{span}(\{x,Ix\}).$$ 
W is invariant
Since $I$ is orthogonal, we get $x\perp Ix$. Since $I^2x=-Ix\neq \vec{0}$, we have
$$\text{dim}(W)=2. $$ 
The set $W$ is invariant with respect to $I$ since 
$$IW=I\left(\text{span}(\{x,Ix\})\right) = \text{span}(\{Ix,I^2x\}) = W.$$ 
$W^\perp$ is invariant
Consider $y\in W^\perp$ i.e. $y\perp x,\ y\perp Ix$. We get
$$\langle Iy,x\rangle=\langle I^2y,Ix\rangle=-\langle y,Ix\rangle=0$$
and
$$\langle Iy,Ix\rangle=\langle I^2y,I^2x\rangle=\langle y,x\rangle=0$$
Therefore $Iy\in W^\perp$. 
Dimension argument
If $W^\perp=\{0\}$, the dimension of $V$ is $2$ and therefore even. If not, choose any $\vec{0}\neq x_2\in W^\perp$ define 
$$V_2=W^\perp,\ I_2=I|_{V_2},\ W_2=\text{span}(\{x_2,Ix_2\})$$
We can apply the same algebra on $(V_2, I_2, W_2, x_2)$ as we did previously on $(V,I,W,x)$. Note that the invariance of $W^\perp=V_2$ lets us view $I_2$ as an operator
$$I_2: V_2\rightarrow V_2$$
We can therefore repeat the prevoius step to get
$$V = W \oplus W_2 \oplus W_3 \oplus ... \oplus W_k $$
$$\text{dim}(W_i) = 2,\ i=1,...,k$$
for some $k\in \mathbb{N}$. Now we can conclude
$$\text{dim}(V) = 2k$$
which is even.
A: Note that if $I$ satisfies $I^2 = -\mathrm{Id}_V$ then
$$ \det(I^2) = \det(I)^2 = \det(-\mathrm{Id}_V) = (-1)^{\dim V} $$
which immediately implies that $\dim V$ must be even. You don't even need to assume that $I$ is orthogonal.
A: The reason is matters that $W^{\perp}$ is invariant is that you can treat it as a vector space in its own right that has a linear transformation $I:W^{\perp}\to W^{\perp}$ with $I^2=-Id_{W^{\perp}}$. Since $W^{\perp}$ has strictly smaller dimension than $V$, we can use this fact in an induction proof to conclude that $\dim W^{\perp}$ is even. Now $\dim V=\dim W^{\perp}+\dim W$. Since $\dim W=2$, $\dim V$ is even.
