TITLE: proof that $\phi\circ\phi=id$ implies the existence of a diagonal matrix
QUESTION [5 upvotes]: As exam preparation we were trying to proof the following task:
Let $V=\mathbb{R}^2$ and let $\phi$ be an endomorphism of $V$ with $\phi \circ \phi = id$ and $\phi \neq id$ and $\phi \neq -id$. Proof that this implies the existence of a basis $B=(b_1,b_2)$ of $V$ with $\phi(b_1) = b_1$ and $\phi(b_2)=-b_2$ 
Unfortunately we aren't able to solve that task and would very much appreciate some proofs and and a "how to" of how to approach such problems.
Thanks for your help.

REPLY [4 votes]: An (elementary) algebraic approach that leads to this conclusion might be as follows. Since $\phi \neq id$ there is some $x$ for which $\phi(x) - x \neq 0$. Applying $\phi$ to this vector gives
$$\phi(\phi(x) - x) = \phi \circ \phi(x) - \phi(x) = x - \phi(x)$$
So $x - \phi(x)$ is a nonzero vector $\phi$ takes to negative itself. Similarly, since $\phi$ is not $-id$, there is some $y$ for which $\phi(y) + y$ is nonzero, and applying $\phi$ analogously to this vector one gets back itself. These two vectors are not multiples of each other since $\phi$ has opposite behaviors on the two vectors. Hence they are a basis.
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