Prove that characteristic polynomial = minimal polynomial Let $M \in M_{3}(K)$ whereby $K$ is a field, such that $\chi_{M}$ only has one root in $K$. Show: $\chi_{M}=\mu_{M}$ whereby $\mu_{M}$ is the minimal polynomial. Steps I thought about: Let $\lambda \in K$ be this root. This means $\chi_{M}=(X-\lambda)^{3}= (X^{3}-3\lambda X^{2}+3\lambda^{2}X+\lambda^{3}Id)$. By the Cayley-Hamilton we get: $0=\chi_{M}(M)=M^{3}-3\lambda M^{2}+3\lambda^{2}M+\lambda^{3}Id$ What now?

As Delta-u has noted, you are interpreting the statement wrongly. The problem states that if one of the roots of $\chi_{M}$ is in $K$, and the two other roots are not in $K$ (but of course in some extension field of $K$), then $\chi_{M} = \mu_{M}$.

The assumption implies that $$ \chi_{M} = (x - \lambda) f, $$ where $\lambda \in K$, and $f \in K[x]$ is a monic quadratic polynomial, which is irreducible over $K$.

Now you should know that $\mu_{M}$ divides $\chi_{M}$, and every root of $\chi_{M}$ is also a root of $\mu_{M}$.

Thus $\mu_{M}$ can be either $x - \lambda$ or $\chi_{M}$. But if $\mu_{M} = x - \lambda$, then $M$ is the scalar matrix $$ \begin{bmatrix} \lambda & 0 & 0\\\ 0 & \lambda & 0\\\ 0 & 0 & \lambda\\\\\end{bmatrix}, $$ so that $\chi_{M} = (x - \lambda)^{3}$, a contradiction.