Proof verification for Identity matrices So I have the following question:
Analyze the following 'Claim' (which may or may not be true) and the corresponding 'Proof', by writing 'TRUE' or 'FALSE' (together with the reason) for each step. [Note: $I_n$ is the $n \times n$ identity matrix.]
Claim: Let $A$ be any $n \times n$ matrix satisfying $A^2=I_n$. Then either $A=I_n$ or $A=-I_n$.
'Proof'.
Step 1: $A$ satisfies $A^2-I_n = 0$ (True or False)
True.
My reasoning: Clearly, this is true. $A^2=I_n$ is not always true, but because it is true, I should have no problem moving the Identity matrix the the LHS.
Step 2: So $(A+I_n)(A-I_n)=0$ (True or false)
True.
My reasoning: Because $I_n$ is the identity matrix, there should be no issues with factoring just like normal algebra.
Step 3: $A+I_n=0$ or $A-I_n=0$
I'm not sure about this part. I'm very tempted to say this is fine but I am not sure how I can justify this, if I even can.
Therefore $A=-I_n$ or $A=I_n$. (End of 'Proof'.)
Is what I am doing right so far or am I messing up somewhere?
 A: Consider any diagonal matrix with diagonal elements $\pm 1$. Show that$A^{2}=I_n$. you get $2^{n}$ matrices whose square is $I_n$.
A: Furthermore if you want a concrete example of a matrix whose square is the identity but not itself a simple matrix consider for example this one: 
$$\begin{bmatrix}\frac{1}{2} & \frac{3}{4} \\ 1 & -\frac{1}{2}\end{bmatrix}$$
These matrices are called involuntory
A: *

*Rather than saying that moving the identiy to the LHS, it is due to we add $-I$ to both sides.

*We have $A^2-I=(A-I)(A+I)$, we just have to expand the right hand side to verify that.

*In matrices, $AB=0$ doesn't imply that $A=0$ or $B=0$. For example $$\begin{bmatrix} 2 & 0 \\ 0 & 0\end{bmatrix}\begin{bmatrix} 0 & 0 \\ 0 & -2\end{bmatrix}= \begin{bmatrix} 0 & 0 \\ 0 & 0\end{bmatrix}$$

*In particular, 
$$\left(\begin{bmatrix} 1 & 0 \\ 0 & -1\end{bmatrix}+\begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}\right)\left(\begin{bmatrix} 1 & 0 \\ 0 & -1\end{bmatrix}-\begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}\right)= \begin{bmatrix} 0 & 0 \\ 0 & 0\end{bmatrix}$$
that is we cant' conclude that $(A+I)(A-I)=0$ implies $A+I=0$ or $A-I=0$ as well.
