# 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 $W \oplus W^{\perp}$.

• Induction, induction, induction! – Matt Samuel Feb 7 '16 at 21:42
• @MattSamuel More precisely $\underbrace{\text{Induction}, \text{Induction},\ldots, \text{Induction}}_n$. – Hagen von Eitzen Feb 7 '16 at 21:44
• When you know, that $W^\perp$ is invariant, you can choose $y\in W^\perp$ and show that $W_2=$span$\{y,Iy\}$ is invariant. Then $W_2^\perp$ is invariant... – Gregor de Cillia Feb 7 '16 at 21:50
• Note that the restriction of $I$ to $W^\perp$ satisfies $I^2=-Id$. – Omnomnomnom Feb 7 '16 at 21:54
• @GregordeCillia Maybe I am not fully understanding what Invariant means. To be an invariant subspace of $I$, this means that $IW^{\perp} \subseteq W^{\perp}$ correct? I am trying to figure out the hint of why induction answers my question of why we care about invariance. – LinearGone Feb 7 '16 at 22:04

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.

• Thanks for asking Gregor. Please tell me why we care about $W^{\perp}$ being invariant. What does being invariant tell us? I feel that if I understood this, all will be clear. – LinearGone Feb 7 '16 at 22:33
• Use indirect logic: If the dimension of $V$ would be 3, we would get $\text{dim}(W_2)=1$ which is impossible since $I^2 \neq 0$. If it would be 5, we would get $\text{dim}(W_3)=1$ and basically the same contradiction – Gregor de Cillia Feb 7 '16 at 22:40
• So we have $V = W_{1} \oplus W_{1}^{\perp}$. If $W_{1}^{\perp}$ is non-empty, choose some $x_{2} \in W_{1}^{\oplus}$. Then $W_{1}^{\perp} = W_{2} \oplus W_{2}^{\perp}$. Continuing this process, we eventually get $$V = W_{1} \oplus W_{2} \oplus \cdots W_{k} \oplus W_{k}^{\perp}$$ Each $W_{j}$, $j \leq k$, has dimension $2$. How do we know what $dim(W_{k}^{\perp})$ is? – LinearGone Feb 8 '16 at 1:20

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.

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.