Can a matrix A with the property $A=A^{-1}$ only have the eigenvalues -1 and 1?

Question

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.

Answer 1

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 $\lambda = \pm 1$.

Answer 2

Set $p(t) = t^2 -1$. Since $A=A^{-1}$ it holds $A^2=I$, which implies $p(A)=0$. The eigenvalues of $A$ are roots of $p$, so your conclusion is valid.

Answer 3

The eigenvalues of $A$ and $A^{-1}$ are reciprocals. Since $A=A^{-1}$ we have that for any eigenvalue $\lambda = 1/\lambda$ which only works for $1$ and $-1$.