TITLE: Can a matrix A with the property $A=A^{-1}$ only have the eigenvalues -1 and 1?
QUESTION [6 upvotes]: If a matrix A has the property $A=A^{-1}$, are the only possible eigenvalues 
 1 and -1 ?
How can the matrices with integer values and the property $A=A^{-1}$ be
 characterized ?
I found out that if A has the property $A=A^{-1}$, then $-A$, $A^T$ and
 $B^{-1}AB$ for any invertible B also have the property. 
I also think that the theorem of caley-hamilton is useful for my
 problem.

REPLY [6 votes]: Without making explicit use of characteristic polynomials or the relationship between the eigenvalues of $A$ and $A^{-1}$ we can obtain the result just by using the definition of eigenvalue.
Suppose $v$ is an eigenvector for $A$ with corresponding eigenvalue $\lambda$. Then 
$$v = Iv = A^{-1}Av = A^2v = A(\lambda v) = \lambda(Av) = \lambda^2v.$$
It follows that $\lambda^2 = 1$ so $\lambda = \pm 1$.