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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.

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    @ftiaronsem: This may not be worth a full answer (because to some extent it only reformulates other answers), but another possible approach is to consider $P=\frac{1}{2}(\phi+I)$. From the properties of $\phi$ it follows that $P^2=P$, $P\neq0$, and $P\neq I$. Take $x$ such that $Px\neq0$ and let $b_1=Px$. Since $Pb_1=b_1$, $\phi b_1=b_1$. Since $P\neq I$, there is a $y$ such that $Py\neq y$. Let $b_2=Py-y\neq 0$. Then $Pb_2=0$, which implies $\phi b_2=-b_2$. Both $x$ and $y$ are very easy to find in a concrete example.2011-01-23

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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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    Thank you very very much. Now I understand of how to approach and how to solve this problem. This is really nice.2011-01-22
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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.

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    The eigenvectors need not be orthogonal (with respect to the standard inner product). E.g., consider $\phi(x,y)=(x+y,-y)$.2011-01-23
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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 $\phi(b_2)=−b_2$.

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    @ftiaronsem: There is something called the "minimal polynomial" which is the monic polynomial of least degree that annihilates the matrix; it is easy to check that it will divide *any* polynomial that annihilates the matrix. In this case, you know that the minimal polynomial divides $X^2-1$; the minimal pol. cannot be $X-1$ or $X+1$, because then the matrix would be id or -id. So the minimal polynomial is $X^2-1$. And by the Cayley-Hamilton Theorem, the minimal polynomial divides the characteristic polynomial, so $X^2-1$ divides the char pol; since the char pol is degree 2, you are done.2011-01-23