Consider the matrix $$A = \frac{1}{4} \begin{bmatrix} 17 & -6 & -9 & -12 \\ 15 & 26 & 69 & 36 \\ 3 & 6 & 29 & 12 \\ -12 & -12 & -48 & -16 \end{bmatrix}. $$ I am given that $(A-I)(A-2I)(A-5I)=0$. I have to explain why this matrix is diagonalizable and then find the minimal polynomial of this matrix.
For the diagonalizability, I was thinking that since the matrix has 3 distinct eigenvalues, then the minimal polynomial would have 3 distinct linear factors, making this matrix diagonalizable. However I'm not sure how to show this and I'm also not sure if this is necessarily correct.
As for finding the minimal polynomial, I was given a hint that says to use the trace but I don't see the relationship between the two.
Since $(A-I)(A-2I)(A-5I)=0$ the minimal polynomial $m(A)$ is a divisor of $(x-1)(x-2)(x-5)$, hence it has only simple roots, which is a necessary and sufficient condition of diagonalisability.
Edit: concerning the minimal polynomial, we know the eigenvalues are among $\{1,2,5\}$. To determine it, you only have to determine which, among $A-I, A-2I, A-5I$ has a rank less than $4$.