Matrix with $A^2=I$ If we have a field $K$ with $char(K)\ne2$ and a matrix $A \in K^{2 \times 2}$ with $A^2=I$. Is it sufficient enough to conclude that because of $$\begin{align} A^2=I &\iff A^2-I=0 \\ &\iff A^2-I^2=0 \\ &\iff(A-I)(A+I)=0 \end{align}$$ The eigenvalues of A must be either only $-1$ or $1$ or $-1$ and $1$ together. So the only possible matrix would be I,-I and some $A$ with $A=SDS^{-1}$ with $D=\begin{bmatrix}
1 & 0 \\
0 & -1  \\
\end{bmatrix}$ and $S\in Gl_n(K)$ ?
I've seen a solution to this problem where they basically use that $A^2=\begin{bmatrix}
a & b \\
c & d  \\
\end{bmatrix}^2=I$ and therefore$a^2+bc=1=cb+d^2, b(a+d)=0=c(a+d)$  and differentiate between the two cases where $(a+d)\ne 0$ and $(a+d)= 0$ and find the same matrices.
Is my solution valid or can I not simply "read of" the eigenvalues ?
 A: If you want to be sure that you covered all cases and if you also want to make the line of argument clearer, use the minimal polynomial.
The minimal polynomial $\mu_A$ of a matrix $A$ divides each polynomial $p$ that satisfies $p(A)=0.$ In our case, we can set $p(t)= t^2-1,$ because we know that $A^2-I=0.$
Therefore, there are only three cases: $\mu_A(t)=t^2-1$ or $\mu_A(t)=t-1$ or $\mu_A(t)=t+1.$ This is because $t^2-1,$ $t-1$ and $t+1$ are the only monic dividers of $t^2-1$ with a degree of at least $1.$
In the first case, $\mu_A(t)=t^2-1,$ the degree of the minimal polynomial matches the size of the matrix, the minimal polynomial must be identical with the characteristic polynomial, and we have two distinct eigenvalues $1$ and $-1,$ each of which has an algebraic multiplicity of $1.$ The matrix $A$ is diagonalizable in this case. There is some $S\in\mathrm{GL}_2(K)$ such that $A=S\begin{bmatrix} 1 & 0 \\ 0 & -1\end{bmatrix}S^{-1}.$
In the second case, we have $\mu_A(t)=t-1,$ which means $A-I=0$ or $A=I.$
In the third case, we have $\mu_A(t)=t+1,$ which means $A+I=0$ or $A=-I.$
A: Infact, all the three possibilities for the minimal polynomial viz. $t-1$, $t+1$ or $(t-1)(t+1)$ implies that matrix $A$ is always diagonalizable due to the fact that the minimal polynomial contains linear factors only. Hence the existence of some invertible $S$ s.t. $A=SDS^{-1}$, where $D$ is $diag(1,1)$ or $diag(-1,-1)$ or $diag(1,-1)$ is always guaranteed.
