Eigenvalues of a 2x2 matrix A such that $A^2$=I

If $v$ is an eigenvector of $A$ with eigenvalue $\lambda$, then $$v=Iv=A^2v=A(Av)=A(\lambda v)=\lambda(Av)=\lambda^2 v.$$ Thus, if $\lambda$ is an eigenvalue of $A$ and $A^2=I$ then $\lambda^2=1$. This gives only two possibilities for $\lambda$, $\pm 1$.

Notice, that we never assumed that $A$ is $2\times 2$. Indeed, if $A$ is any square matrix and $A^2=I$ then the only possible eigenvalues of $A$ are $\pm 1$.

Let $A$ be an arbitrary $2\times 2$ matrix, and $a,b,c,d\in\mathbb{R}$ such that: $$A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$$ Then we have: $$A^2=I\Longrightarrow\begin{pmatrix}a^2+bc&ab+bd\\ac+cd&bc+d^2\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}$$ Therefore, any matrix of the form: $$A=\begin{pmatrix}a&b\\c&-a\end{pmatrix}\text{ or }A=\begin{pmatrix}\pm1&0\\0&\pm1\end{pmatrix}$$ will satisfy the condition $A^2=I$. In the first case its eigenvalues are then given by: $$\begin{vmatrix}a-\lambda&b\\c&-a-\lambda\end{vmatrix}=0\Longrightarrow \lambda^2=a^2+bc=1$$ Therefore $\lambda=\pm 1$. The second case also gives us $\lambda=\pm1$.