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 $phi:Vrightarrow V$ 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 $text{Fix}(phi )=left {vin Vmid phi (v)=vright }$ and $text{Eig}(-1, phi )=left {vin Vmid phi (v)=(-1)vright }=left {vin Vmid phi (v)=-vright }$.
Let $vin V$.
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?

azif00 · Answer

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 $operatorname{Fix}(phi) cap operatorname{Eig}(-1,phi) = {0}$.

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

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