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!

Let $\{v_1,\cdots,v_n\}$ be an eigenvector basis. If the minimal polinomial $m_A$ had a factor $(x-\lambda_i)^k$ with $k\geq 2$ it means that the polinomial $p(x)=\frac{m_A}{(x-\lambda_i)}$ does not become null in $A$, thus, there exists $v=\sum a_jv_j$ such that $p(A) \cdot v \neq 0$.
But $p(A) \cdot v=\sum a_j p(A) \cdot v_j=0$ because each $v_j$ is an eigenvector for which there is a factor $(x-\lambda_j)$ in $p$ that makes $p(A)v_j =0$. 
This is a contradiction.
