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let $A\in M{_n}\ (\mathbb R)$ s.t $A^2=I$ such that $A\neq I$, $A\neq -I$ how prove $-(n-1)\le tr A\le n-1.$ Thanks in advance

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Note that $(A-I)(A+I)=0$. – Sigur Feb 14 '13 at 20:11
See here: – Sigur Feb 14 '13 at 20:21
up vote 4 down vote accepted

This question is easy to solve, but you can give better bounds. As $A^{2}=I$ so $A$ satisfies the equation $x^{2}-1=0$. And since its minimal polynomial $p_{A}(x)$ should divide $x^{2}-1$, so $p_A(x)$ can be one of $x^{2}-1,\,x-1,\,x+1$. If $p_{A}(x)=x-1$ then $A-I=0$ and $A=I$, which is a contradiction. And with same reason $p_{A}(x)\neq x+1$. So $p_{A}(x)=x^{2}-1$. By Cayley-Hamilton Theorem, the characteristic polynomial $f_{A}(x)$ of $A$ should take the form of $(x-1)^{a_{1}}(x+1)^{a_{2}}$ where $a_{1},a_{2}\geq1$ and $a_{1}+a_{2}=n$. So $1$ and $-1$ are $A$'s eigenvalues with multiplicities of $a_{1}$ and $a_{2}$ accordingly. Now pay attention: as the minimal polynomial is decomposable into distinct linear factors over $\mathbb{R}$, $A$ is diagonalizable. From this and the above discussion about eigenvalues, $A$ is similar to a diagonal matrix with $a_{1}$ elements "$1$" and with $a_{2}$ elements "$-1$" on the diagonal. We know that similar matrices have the same trace, so the trace of $A$ is equal to $a_{1}(1)+a_{2}(-1)=a_{1}-a_{2}$ but we know that \begin{align*} &1\leq a_{1}\leq n-1,\\ &1\leq a_{2}\leq n-1. \end{align*} The second inequality yields $1-n\leq -a_{2}\leq -1$, so, $$1+(1-n)\leq a_{1}-a_{2}\leq (n-1)-1.$$ And the tighter bounds for trace of $A$ are: $$-(n-2)\leq \operatorname{tr}(A)\leq n-2.$$

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The eigenvalues $\lambda_1$, $\dots$, $\lambda_n$ of $A$ are $+1$ or $-1$, and they are not all equal.

It follows that in $\operatorname{tr} A=\lambda_1+\dots+\lambda_n$ there are at least one $+1$ and at most $n-1$, and at least one $-1$ and at most $n-1$. Your result follows.

My first sentence above depends on knowing that $A$ is diagonalizable. This is immediate given a little theory: we know that the minimal polynomial of $A$ divides $t^2-1$, so it has simple roots, and this tells us that $A$ is diagonalizable.

A silly, hands-on, alternative way to see the matrix is diagonalizable is the following. Let $x$ be a non-zero vector. Then either $x$ is an eigenvector of $A$ or $x=\tfrac12(x+Ax)+\tfrac12(x-Ax)$ is a sum of two eigenvectors. It follows that the set of eigenvectors of $A$ spans $\mathbb R^n$, and that set must then contain a basis of $\mathbb R^n$: it follows that $A$ is diagonalizable.

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This is probably trivial, but how do you prove that if all the eigenvalues of $A$ are 1 and $A^2=I$, then $A=I$? – user60725 Feb 14 '13 at 20:41
@skrivbord I don't think it is that obvious. I don't see a way to prove it in less than one line. Anyway, the way I see it you need to first prove that $A$ is diagonalizable and then you can conclude. – Git Gud Feb 14 '13 at 20:42
@Mariano Suárez-Alvar: can you explain more the end of second line please(there are at least one +1 and at most n−1, and at least one −1 and at most n−1) – Maisam Hedyelloo Feb 14 '13 at 20:49
@MaisamHedyelloo, I suggest you consider all possible sums of $6$ ones and minus ones, not all of which are equal, to see what happens. – Mariano Suárez-Alvarez Feb 14 '13 at 20:51
Your result follows... Well, in fact what follows is the stronger result with the $\pm(n-2)$ bounds. – Did Feb 15 '13 at 8:06

The minimal polynomial of $A$ is $(x-1)(x+1)\Rightarrow $ the characteristic polynomial of $A$ is $$f(x)=(x-1)^k(x+1)^{n-k}=x^n+((n-k)x^kx^{n-k-1}-kx^{k-1}x^{n-k})+\ldots \\=x^n+a_{n-1}x^{n-1}+\ldots, \ \text{where} \ 1\leq k\leq n-1.$$ Now $|\text{tr}(A)|=|a_{n-1}|=|(n-k)-k|=|n-2k|$.

So we have the stronger result that $|\text{tr}(A)|\leq |n-2|$
(and $\text{tr}(A)\in\{2-n,4-n,\ldots,n-4,n-2\}$).

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