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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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Your intuition should be that phi is reflection across an axis, and you should be thinking of ways to verify this. Pick a nonzero vector v such that phi v \neq \pm v. Then phi^2 v = v. So if we write w = phi v (the reflection of v across this hypothetical axis), then phi v = w and phi w = v. That is totally consistent with the reflection hypothesis; actually, it implies it. Can you see, geometrically if not algebraically, what to do from here? – Qiaochu Yuan Jan 22 '11 at 13:52
Thanks for your comment. Now I understand that $\phi$ is a reflection. So the basis which would have the required properties, would consist of the vector $(v -\phi(v))$ and a vector being orthogonal to this one. $v-\phi(v)$ would correspond to $b_2$ and the orthogonal vector would correspond to $b_1$. Is this correct so far? However I can't figure an algebraic proof for this... – ftiaronsem Jan 22 '11 at 14:47
Try drawing out what v and w would be if phi were a reflection and see if that gives you any ideas. (You've already figured out that v - w is important. What's another important vector here?) You shouldn't rely too much on your geometric intuition, since you don't have an inner product to work with, but you should at least be able to figure out what the second element of the basis is. – Qiaochu Yuan Jan 22 '11 at 14:53
Right. Simple proof with exhibition of $b_1=v+\phi(v)$ and $b_2=v-\phi(v)$, and justifying it is a basis of the space. – Wok Jan 22 '11 at 15:02
Thanks Qiaochu Yuan. Woks answer showed me how to proof the problem and your comments showed me the path to this solution. I really appreciate that. – ftiaronsem Jan 22 '11 at 22:11
up vote 4 down vote accepted

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. – ftiaronsem Jan 22 '11 at 22:09

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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While I agree that it's a good idea to be aware of how the machinery applies to this problem, I think for this particular problem it's enlightening to find the basis more or less directly. – Qiaochu Yuan Jan 22 '11 at 14:54
thanks for this answer. Unfortunatelly I can't follow you saying it is immediate that the eigenvalues must be +1 or -1. According to some considerations of mine, this has to be true. However I am unable to see an algebraic proof for that. – ftiaronsem Jan 22 '11 at 21:57
@ftiaronsem: For p eigenvector and a eigenvalue we have, A(p) = a p Applying the transformaton to both sides: we get 1 = a^2 – leonbloy Jan 22 '11 at 22:12
ohh, ok. I see that now, thank you very much. – ftiaronsem Jan 22 '11 at 22:33
The eigenvectors need not be orthogonal (with respect to the standard inner product). E.g., consider $\phi(x,y)=(x+y,-y)$. – Jonas Meyer Jan 23 '11 at 6:11

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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Thanky you very much for this answer. I asume that $X^2-1$ is the charecteristic polynom of the Matrix describing $\phi$. How did you get figure that out. I remember my prof saying that the charactersitic polynomial is: $(-1)^n\lambda^n+(-1)^{n-1}trace(A)\lambda^{n-1}+ ... + det(A)$, where A is the matrix describing $\phi$. Why is $trace(A)=0$ in this case? – ftiaronsem Jan 22 '11 at 22:02
@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. – Arturo Magidin Jan 23 '11 at 5:57

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