Why is $\phi$ diagonalizable if $\phi \circ \phi =id_V$? V is a finite-dimensional $\mathbb{Q}$- vector space with $\phi: V \rightarrow V$
Why does it follow that $\phi$ is diagonalizable if $\phi \circ \phi = id_{V}$?
My ideas so far:
I do know that if i can show that the sum of the eigenvalues is equal to the dimensions of the matrix that it therefore has to be diagonalizable, as well as that if i can show that there exists an invertable matrix M that results in this $M^{-1}*\phi*M$ being diagonal then $\phi$ has to be diagonalizable.
I also think there was a way to go about this by showing that there is a Basis of V taht consists of eigenvectors but im not sure about how that works.
 A: Assume that the characteristic of the field (such as $\mathbb Q$) is not $2$. Here is a simple way to see it. As has been noted, all the eigenvalues are either $1$ or $-1$. Let $V_+$ be the subspace of $V$ spanned by eigenvectors of eigenvalue $1$, and $V_-$ be similarly defined (a priori, $V_{\pm}$ could be $\{0\}$). One now proceeds to show that $$V=V_+\oplus V_-.$$ If $v\in V_+\cap V_-$, then $\phi(v)=v=-v,$ hence $v= 0$. This shows that $$V_+\cap V_-=\{0\}.\qquad (1)$$ For any vector $v\in V$, consider $$v_+:=v+\phi(v)~{\rm and ~}v_-:=v-\phi(v).\qquad (2)$$ One has $$\phi(v_+)=\phi(v+\phi(v))=\phi(v)+\phi^2(v)=v+\phi(v)=v_+\in V_+$$and $$\phi(v_-)=\phi(v-\phi(v))=\phi(v)-\phi^2(v)=-(v-\phi(v))=-v_-\in V_-.$$ It then follows from (2) that $v=\frac 1 2(v_++v_-)\in V_++V_-$, hence from (1), one gets a direct sum $$V=V_+\oplus V_-.$$ Since there is a basis consisting of eigenvectors, $\phi$ can be diagonaized.
A: We have\begin{align}
\phi\circ\phi & = id_V\\
\phi\circ\phi - id_V & = 0\\
(\phi + id_V)(\phi - id_V) & = 0,\\
\end{align}
so the minimal polynomial for $\phi$ divides $(x + 1)(x - 1)$. Have you learned yet how to characterize a diagonalizable linear operator by its minimal polynomial? Look for such a theorem in your textbook.
