Need help with this: Let $A^2 = I$, and all eigenvalues of $A$ are $1$. Prove that $A = I$. ($A$ is over the complexes)

I thought that because $A^2=I$, then $A$ is reversible and $A^{-1} = A$, and there are only two matrices that do this: the identity matrix and the zero matrix.
But it's only intuition and I couldn't prove that.

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    $\begingroup$ If a matrix (over the reals or complexes) satisfies $A^n=I$ then it is diagonalisable. $\endgroup$ – Lord Shark the Unknown May 7 '17 at 7:48
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    $\begingroup$ @LordSharktheUnknown: diagonalisable over $\mathbb{C}$, you mean? How about a 1/4 rotation? $\endgroup$ – Joppy May 7 '17 at 7:58
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    $\begingroup$ "there are only two matrices that do $A^{-1}=A$: the identity and the zero matrix". Definitely not the zero matrix. To find other matrices, take the identity and negate whichever elements you wish. This gives you at least $2^n$ such $n\times n$ matrices. $\endgroup$ – Teepeemm May 8 '17 at 2:09

We have $(A+I)(A-I)=0$. Since $-1$ is not an eigenvalue, $A+I$ is invertible and so $A-I=0$.

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    $\begingroup$ I don't see exactly why $-1$ not being an eigenvalue would make $A + I$ invertible. Could you explain? $\endgroup$ – AJFarmar May 7 '17 at 14:33
  • $\begingroup$ @AJFarmar One of equivalent definitions of $c$ being an eigenvalue is that $A-cI$ is nonsingular. $\endgroup$ – Wojowu May 7 '17 at 14:53
  • $\begingroup$ @Wojowu, you mean "is singular". $\endgroup$ – lhf May 7 '17 at 15:41
  • $\begingroup$ @lhf Of course :) $\endgroup$ – Wojowu May 7 '17 at 15:46

If all the eigenvalues of $A$ are equal to 1, then its characteristic polynomial is $(-1)^n(x-1)^n$ and its minimal polynomial is of the form $(x-1)^m$, for some $m\le n$. Since the polynomial $p(x)=x^2-1$ annihilates $A,$ then then the minimal polynomial DIVIDES the polynomial $p$, and consequently the minimal polynomial of $A$ is $q(x)=x-1$.

Therefore, $A-I=0$.


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