Let $M \in M_{3}(\mathbb R)$ such that $\chi_{M}$ only has one root in $\mathbb R$. Show that M is diagonalizable in $\mathbb C$.
My steps so far:
Let $\lambda \in \mathbb R$ be the root. That means $\chi_{M}=(X-\lambda)f$, whereby $f$ irreducible in $\mathbb R$ and $\deg f = 2 $. Since, however, $\mathbb C$ is algebraically closed, it means that every polynomial is reducible into degree $1$ factors. So that means $M$ has either 2 or 3 eigenvalues in $\mathbb C$. In either case,
Algebraic multiplicity = Geometric multiplicity.
Therefore diagonalizable in $\mathbb C$.
In your argument, if $M$ had only two eigenvalues then one of would have algebraic multiplicity 2, but there's no reason why the geometric multiplicity would be $2$ as well.
We need to assume that $\chi_M$ has only one real root of degree $1$, otherwise the result is not true.
Here's a hint: Let $\lambda$ be the real root of the characteristic polynomical $\chi_M$, so $\chi_M(x)=(x-\lambda)^{I}p(x)$, where $p(x)$ is real of degree $2$. The roots of $p(x)$ are not real, so prove that if $\alpha$ is a root then $\overline{\alpha}$ is another root. $3$ distinct roots $\iff$ $3$ distinct eigenvalues $\Rightarrow$ diagonalizable.
Example for degree of the real root $=3$: $$M=\begin{bmatrix}1&1&0\\0&1&0\\0&0&1\end{bmatrix}$$ is not diagonalizable.
Note that if $\lambda$ had degree $2$ in $\xi_M$, then the same argument as in the solution, writing $\xi_M(x)=(x-\lambda)^2p(x)$ would yield $p(x)$ of degree $1$, so it would have a real root, which would need to be equal to $\lambda$ from the assumption in the exercise, so $\lambda$ would actually have degree $3$, a contradiction.