Trace of a matrix $A$ with $A^2=I$ Let $A$ be a complex-value square matrix with $A^2=I$ identity.
Then is the trace of $A$ a real value?
 A: Another way.
$$A^2-I=0,$$
Since minimal polynomial divide this polynomial (why) then it must be either
$(x-1)(x+1)$ or $(x-1)$ or $(x+1)$.
hence all the eigenvalues are reals (even in set $\{1,-1\}$) (why) and hence the result follows.
A: Yes. Moreover, it must be an integer. 
Let $v$ be an eigenvector of A, then:
$Av=\lambda v$
for some lambda. Multiplying by A we get:
$A^2v=Iv=v=\lambda^2v$
Which means that any eigenvalues $\lambda_n$ satisfy $\lambda_n^2=1$ and thus $\lambda_n=\pm1$. Recall that the trace of a matrix is the sum of the eigenvalues of the matrix, which directly implies the result. 
A: $$A=\begin{pmatrix}a&b\\c&d\end{pmatrix}\implies A^2=\begin{pmatrix}a^2+bc&b(a+d)\\c(a+d)&d^2+bc\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}\implies$$
$$\begin{align}a^2+bc=1=d^2+bc\implies a=\pm d\end{align}$$
But we also have that
$$b(a+d)=0=c(a+d)\implies \begin{cases}a=-d\\or\\b=c=0\end{cases}$$
In the first case
$$a=-d\implies \text{Tr.}\,A=0$$
In the second case:
$$b=c=0\implies a=\pm d=\pm 1\implies \;\text{Tr.}\,A=0\;\;or\;\;\text{Tr.}\,A=\pm2$$
