A is diagonalizable if and only if its minimal polynomial is a product of distinct monic linear factors I have to prove that a matrix A is diagonalizable if and only if its minimal polynomial is a product of distinct monic linear factors.
I have already proved it in one direction, meaning if f it's minimal polynomial is a product of distinct monic linear factors it's diagonalizable.
I can't find a way to prove the other direction, meaning if A is diagonalizable so its minimal polynomial is a product of distinct monic linear factors.
Thank you so much for the help!

Suppose A is diagonalizable with basis $\{ v_k \}$ and corresponding eigenvalues $\{ \lambda_k \}$. If $\{ \mu_k \}$ are the distinct eigenvalues, then you can check that $(A-\mu_1 I)(A-\mu_2 I)\cdots(A-\mu_n I)$ annihilates every eigenvector and, hence, must be the $0$ matrix. So the minimal polynomial $q$ divides $p(\lambda)=(\lambda-\mu_1)(\lambda-\mu_2)\cdots(\lambda-\mu_k)$. The minimal polynomial divides this product, and, if there were any factor omitted from the minimal polynomial, you could easily argue that $q(A)$ would not annihilate the eigenvectors corresponding to the eigenvalue $\mu_j$ of the missing factor. So $q=p$.
