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I'm attempting to prove that the characteristic polynomial of an involution in $\mathbb{R}^n$, i.e. a linear transformation that satisfies $f^2=I$ will always factor into linear factors.

We know that $p_{f^2}(\lambda)=(1-\lambda)^n$ where $p_f$ is the characteristic polynomial of the linear operator $f$. That's as far as I've gotten with useful thoughts (except a dozen of dead ends).

After some contemplation, I've tried googling the problem and found out all of the proofs are using the minimal polynomial, which hasn't been introduced in my course, therefore I'm guessing there is some more elementary way to prove this.

Thanks for any help!

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4 Answers

up vote 1 down vote accepted

Suppose not. Then the characteristic polynomial $p_f(x)$ of $f$ will contain an irreducible factor of degree at least $2$, and hence $p_f(x)$ will have a non-real complex root $\lambda$. So, viewing $f$ now as an element of $M_n(\mathbb{C})$, let $v \in \mathbb{C}^n$ be a non-zero eigenvector for the eigenvalue $\lambda \in \mathbb{C} \setminus \mathbb{R}$. What is $f^2 v$?

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Thanks! So ${\lambda}^2=1$ thus $\lambda$ cannot have an imaginary part, is that correct? –  Dahn Jahn Mar 27 '13 at 20:33
    
The only square roots of unity are $1$ and $-1$, so... –  Branimir Ćaćić Mar 27 '13 at 20:34
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Another solution:

$(A-\lambda I)(A+\lambda I)=(1-\lambda^2) I$. Hence $\det(A-\lambda I)$ divides $\det((1-\lambda^2) I) = (1-\lambda^2)^n$.

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This is very nice. –  Branimir Ćaćić Mar 27 '13 at 20:34
    
Indeed, this is beautiful! Hope one day I'll be coming up with answers like this myself. –  Dahn Jahn Mar 27 '13 at 20:42
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@Branimir Ćaćić: Thank you/Hvala –  Boris Novikov Mar 27 '13 at 20:45
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Since $f^2=I$ it follows that the minimal polynomial of $f$ is $x-1$ or $x+1$ or $(x-1)(x+1)$.
The characteristic polynomial of $f$ has the same roots as the minimal polynomial of $f$. Therefore it factors into linear terms.

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You can do this without mentioning the minimal polynomial, although morally the argument below is the same as the one (given in another answer) that does.

If $v$ is an eigenvector of (the complexified) $f$ with (complex) eigenvalue$~\lambda$, then $f^k(v)=\lambda^kv$, and so by linearity $P(f)(v)=P(\lambda)v$ for any polynomial $P$. This implies that any polynomial equation satisfied by $f$ must also be satisfied by$~\lambda$; in the case $f^2=I$ implies that $\lambda^2=1$ for all eigenvalues$~\lambda$. This means that all (complex) roots of the characteristic polynomial are roots of $X^2-1$, and these roots are $+1,-1$, both real. Since a characteristic polynomial always splits over the complex numbers, this $p_f$ splits over the reals too.

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