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?

The matrix $A$ is a root of the polynomial $t^2-4$, hence the minimal polynomial of $A$, a divisor of $t^2-4$, has only simple roots. This is is equivalent to $A$ being diagonalisable.
Answer to the second question: a matrix over a field $K$ is diagonalisable over $K$ if and only if its minimal polynomial splits  over $K$ into a product of distinct linear factors.
