A is diagonalizable if and only if its minimal polynomial is a product of distinct monic linear factors

Question

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 its 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!

Answer 1

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 $q=p$.

Answer 2

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.