TITLE: Let $A^2 = I$. Prove that $A = I$ if all eigenvalues of $A$ are $1$
QUESTION [12 upvotes]: 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.

REPLY [31 votes]: We have $(A+I)(A-I)=0$. Since $-1$ is not an eigenvalue, $A+I$ is invertible and so $A-I=0$.