Trace of $A$ if $A =A^{-1}$ Suppose $I\neq A\neq -I$, where $I$ is the identity matrix and $A$ is a real $2\times2$ matrix. If $A=A^{-1}$ then what is the trace of $A$? I was thinking of writing $A$ as $\{a,b;c; d\}$ then using $A=A^{-1}$ to equate the positions but the equations I get suggest there is an easier way.

Let $J = V^{-1}A V$ be a Jordan normal form of $A$. Then $J = \mathbb{diag}(J_{\lambda_1,m_1},...,J_{\lambda_k,m_k})  $, where $J_{\lambda,i}$ is a Jordan block with eigenvalue $\lambda$ and size $i \times i$.
Then from $A^2=I$ we have $J^2 = I$, from which we get $J_{\lambda_l,m_l}^2 = I$ for each block. It follows from this that $\lambda_l^2 = 1$, or equivalently, $\lambda_l = \pm 1$.
A brief computation shows that if $m_l>1$, then $[J_{\lambda_l,m_l}^2]_{12} = 2 \lambda_l \neq 0$, from which it follows that $m_l = 1$ for all $l$, and consequently, $J$ is diagonal with entries $\pm 1$.
Consequently, $A^2 = I$ iff $A$ has the form $V \mathbb{diag}(\pm 1,..., \pm 1)V^{-1}$.
Since the trace is invariant to similarity transformations, and the dimension is $2$, there are three possible values corresponding to the values $\{(\pm  1) + (\pm 1) \}$. However the values $\{-2, 2\}$ correspond to the matrices $-I$ and $I$ respectively, eliminating them leaves $0$ as the only possible value.
