# Find all endmorphisms $f$ of a real vector $V$ space such that $f\circ f=\operatorname{id}_V$

Find all endmorphisms $f$ of a real vector $V$ space such that $f\circ f=\operatorname{id}_V$

The problem is trivial if $V$ is a $\mathbb{C}$-vector space. Yet here $V$ is a vector space over $\mathbb{R}$. And we do not have Jordan blocks here. Suppose $\dim(V)=n$.

My idea is to complexify $V$. i.e. let $V'=V\bigotimes \mathbb{C}$, $f'=f\bigotimes \operatorname{id}_{\mathbb{C}},$the tensor over $\mathbb{R}$. Since $f'\circ f'=\operatorname{id}_{V'}$, we can find a basis $\{w_i\bigotimes c_i|i=1,2,\ldots,n\}$ for $V'$ $(c_i\neq 0)$such that $f'(w_i\bigotimes c_i)=-w_i\bigotimes c_i(i=1,2,\ldots,k)$ and $f'(w_i\bigotimes c_i)=w_i\bigotimes c_i(i=k+1,\dots,n)$. On the other hand, $f'(w_i\bigotimes c_i)=f(w_i)\bigotimes c_i$. So $f(w_i)=-w_i(i=1,2,\ldots,k)$ and $f(w_i)=w_i(i=k+1,\ldots,n)$.

If $\{w_i\}$ are $\mathbb{R}$-linear dependent, i.e. $\sum a_iw_i=0(a_i\in \mathbb{R})$, then $\sum \frac{a_i}{c_i}w_i\bigotimes c_i=0$, contradicting the fact that $\{w_i\bigotimes c_i\}$ is a basis for $V'$.

So in this way we find a basis $\{w_i\}$ for $V$ such that $f$ has the very ideal form. And thus all such endmorphisms $f$ such that $f\circ f=\operatorname{id}_V$ has this form in some basis.

Is the above proof right? Are there any flaws? Or do you have some better ideas? Can you give me some hints for it?

Thank you very much!

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## 1 Answer

Well, given any $v \in V$, $f$ takes $f(v$) to $v$. So you can take any basis $v_1,...,v_n$, select an even number of vectors $v_1,...,v_{2m}$ and just stipulate $f(v_{2i -1}) = v_{2i}$ and $f(v_{2i}) = v_{2i-1}$ for $i \leq m$, and then stipulate that $f(v_j) = v_j$ for $j > 2m$. This extends to an endomorphism. And every endomorphism with $f^2 = I$ is of this form, just start out with any $v_1$. See if $f(v_1) = v_1$; if not let $v_2 = f(v_1)$ and then pick any $v_3$ not in the span of $v_1$ and $v_2$, and then check if $f(v_3) = v_3$, etc etc.

In matrix form this means $f = V^{-1}MV$, where $M$ is the direct sum of 2-dimensional $[0,1;1,0]$ blocks and one dimensional $[1]$ blocks.

EDIT: Something occurred to me.. in the above $f(v_{2i-1} + v_{2i}) = v_{2i-1} + v_{2i}$, while $f(v_{2i-1} - v_{2i}) = v_{2i} - v_{2i-1} = -(v_{2i-1} - v_{2i})$. So replacing $v_{2i-1}$ and $v_{2i}$ by $w_{2i-1} = v_{2i} - v_{2i-1}$ and $w_{2i} = v_{2i-1} + v_{2i}$ there is actually always a basis of vectors $w_1,...,w_n$ such that $f(w_j) = \pm w_j$ for all $j$. Hence in matrix form, $f = V^{-1}MV$, where now $W$ is a diagonal matrix with $1$s and $-1$s on the diagonal.

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:thank you very much!A smart answer! – user14242 Nov 27 '11 at 1:50