Show that $V=\text{Fix}(\phi )\oplus \text{Eig}(-1, \phi)$

Question

Let $\mathbb{K}$ be a field with $1_{\mathbb{K}}+1_{\mathbb{K}}\neq 0_{\mathbb{K}}$.

Let $V$ be a $\mathbb{K}$-vector space and let $\phi:V\rightarrow V$ be a inear map with $\phi^2=\text{id}_V$.

I want to show that $$V=\text{Fix}(\phi )\oplus \text{Eig}(-1, \phi)$$

It holds that $\text{Fix}(\phi )=\left \{v\in V\mid \phi (v)=v\right \}$ and $\text{Eig}(-1, \phi )=\left \{v\in V\mid \phi (v)=(-1)v\right \}=\left \{v\in V\mid \phi (v)=-v\right \}$.

Let $v\in V$.

To get the desiredresult do we have to start with $v=v+(v-v)$ ? Then applying $\phi$ we get $\phi (v)=\phi (v)+\phi (v-v)$. Does this help us? Or do we show that in a completely other way?

Answer 1

Well, we want to write each $v \in V$ as $u+w$, where $u \in \operatorname{Fix}(\phi)$ and $w \in \operatorname{Eig}(-1,\phi)$, so, suppose that $v = u+w$ and we try to say who is $u$ and $v$ with this information. Note that $$\phi(v) = \phi(u) + \phi(w) = u-w$$ and then $v + \phi(v) = (u+w)+(u-w) = 2u$ and $v - \phi(v) = (u+w)-(u-w) = 2w$, so, put $$u = \frac{v+\phi(v)}{2} \quad \textrm{and} \quad w = \frac{v-\phi(v)}{2}.$$ Verify that this works, i.e., that $u \in \operatorname{Fix}(\phi)$ and $w \in \operatorname{Eig}(-1,\phi)$. This shows $V = \operatorname{Fix}(\phi) + \operatorname{Eig}(-1,\phi)$. Now I leave to you that $\operatorname{Fix}(\phi) \cap \operatorname{Eig}(-1,\phi) = \{0\}$.