Number of matrices whose square is identity How many matrices are such that $A^2 =I$, where $A$ is a $2\times2$ matrix and $I$ is a $2\times2$ identity matrix? 
I can only think of the identity and it negative are they more? Is it an application of Cayley-Hamilton theorem. I have seen a similarly post by I cannot follow it. Could someone answer in simple and understandable terms.
 A: You can compute this manually if you want:
$$\left[\begin{array}{cc} a & b \\c & d \end{array}\right]^2=\left[\begin{array}{cc} a^2+bc & b(a+d) \\c(a+d) & bc+d^2 \end{array}\right]$$
To get the identity matrix, either $a=-d$, so $a^2+bc=1$ (and these can be picked freely, leaving plenty of options) or $b=0$ and $c=0$, so $a=\pm1, d=\pm 1$.
More conceptually, you're asking this: "What linear transformation, applied twice, brings you back to where you started?" You could swap the $x$ and $y$ axes:
$$\left[\begin{array}{cc} 0 & 1 \\1 & 0 \end{array}\right]$$
flip the space around the $x$ axis:
$$\left[\begin{array}{cc} -1 & 0 \\0 & 1 \end{array}\right]$$
Or a number of other things! Just think of a transformation that is undone by applying it again, and find the matrix that corresponds to it.
A: Hint:$A=\left(\begin{array}{cc}1&a\\ 0&-1 \end{array}\right)$ check that $A^2=I$
A: Let's call $q$ the minimum polynomial such that $q(A)=0$. It's a well known fact that $q$ must divide $t^2-1$. This means that it could be:


*

*$t-1$

*$t+1$

*$(t-1)(t+1)$


In each one of this cases A is diagonalizable. That's because $q$ is the minimun polynomial which nullify $A$ (So, as example, $A-I$ is enough to nullify the generalized eigenspace for $1$).
So we have just 3 Jordan forms.
$\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}$
$\begin{pmatrix}-1 & 0 \\ 0 & -1\end{pmatrix}$
$\begin{pmatrix}-1 & 0 \\ 0 & 1\end{pmatrix}$
To answer your question you have those 3 Conjugacy classes.
Notes: 
This argument can easily be generalized.
In $\mathbb{F_2}$ there is just identity.
A: The possible eigenvalues are $\pm 1$.  With the exception of $I$ and $-I$ the matrix will be a reflection.  Therefore in a suitable basis it is given by $$\begin{pmatrix} 1 &0 \cr 0  & -1 \end{pmatrix}\  . $$  All other solutions are conjugate to this matrix. e.g.  $$\begin{pmatrix}11 & -20 \cr 6 & -11 \end{pmatrix}$$
