Show that Matrix $M$ is diagonizable if the characteristic polynomial is irreducible

Question

Let $K$ be a field with $K \subset \mathbb C$, $ M \in M_{n}(K)$ so that $\chi_{M} \in K[X]$ is irreducible. Prove that $M \in M_{n}(\mathbb C)$ is diagonalizable.

My path so far: Since $\chi_{M}$ irreducible that means $M$ has a maximum of one eigenvector with maximum multiplicity of $1$. Let $\lambda$ be this eigenvalue. The eigenspace $V_{\lambda}$ therefore can only be one-dimensional?! So how can $\dim_{K}(V_{\lambda})=n$?

Answer 1

When we're working over a subfield of $\mathbb C$, no irreducible polynomial has multiple roots in $\mathbb C$. And if all roots of the characteristic polynomial are simple, then the matrix is diagonalizable.