proof that $\phi\circ\phi=id$ implies the existence of a diagonal matrix 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.
 A: 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.
A: My approach would be to write the eigenvalues-eigenvectors equatio:n it's inmediate that the eigenvalues must be +1 or -1. What remains it to see that the eigenvectors are orthogonal and that from the alternatives (1,1) (1,-1) (-1,-1) only the second is possible.
A: The polynomial $X^2-1=(X-1)(X+1)$ annihilates $\phi$, and splits over $\mathbb{R}$, and with simple roots. Thus $\phi$ is diagonalizable. However, $\phi$ cannot be identity or minus identity, then there exists a basis $B=(b_1,b_2)$ of $V=\mathbb{R}^2$ with $\phi(b_1)=b_1$ and $\phi(b_2)=−b_2$.
