Relationship between trace and determinant of a non-diagonal $3 \times 3$ involutory matrix

Question

Let $A$ be a $3 \times 3$ real, non-diagonal matrix with $A^{-1}=A$. Show that $\mbox{tr}(A) = -\det(A) = \pm 1$.

Answer 1

Hints: $\det(A)^{2}=\det(A^{2})=1$ so $\det (A) =\pm 1$. The only eigen values are $\pm 1$ because $A^{2}=I$. Trace is the sum of the eigen values. You only have to rule out the possibility of all eigen values being $1$ or all of them $-1$. But in the first case Cayley Hamilton Theorem gives $(A-I)^{3}=0$ which simplifies to $A=I$ (since $A^{2}=I$)! The second case is similar.