Is the Matrix Diagonalizable if $A^2=4I$ I have two question:

Let $A$ be a non-scalar matrix, $A_{k \times k} \in \Bbb R$, and $A^2=4I$. Is the matrix A, always diagonalizable in $\Bbb R$?
Answer

I know that the answer is yes:
$$A^2=4I \rightarrow A^2-4I=0$$
Then by the Cayley Hamilton theorem I know that the matrix satisfies the equation above.
Now I don't know how to explain the fact that the characteristic polynomial is $P_A=(\lambda-2)(\lambda+2)$, then the characteristic polynomial has two different roots, and no more. what's the reason for it?
Second Question - Irrelevant to the first question
When we say that a matrix is diagonalizable if it has different linear roots, it means that for if I have the following characteristic polynomial (for example) $(t-1)^2(t-2)$ then the matrix is not diagonalizable since the root 1 appears twice in the characteristic polynomial?

I'll try to answer without using the minimal and characteristic polynomials. 
Consider the Jordan normal form of $A$,
$$
A = Q J Q^t,
$$
with $Q$ orthogonal, and split $J$ into its diagonal part $D$ plus its strictly upper triangular part $U$, namely
$$
J = D + U.
$$
Then,
$$
4I = A^2 = Q J Q^t QJ Q^t = Q J^2 Q^t \Longrightarrow J^2 = 4 I.
$$
Exploiting the splitting, we obtain
$$
4I = (D + U)(D+U) = D^2 + DU + UD + U^2.
$$
Notice that $D^2$ is diagonal, whereas $DU + UD + U^2$ is upper triangular, with zeros on the main diagonal. Hence, 
$$
D^2 = 4I \Longrightarrow D = \pm2I.
$$
Moreover, the upper triangular part of $4I$ is zero, hence
$$
0 = DU + UD + U^2 = \pm 4 U + U^2 = U ( U \pm 4 I).
$$
This is possible only for $U = 0$.
