TITLE: How to find a real $3\times3$ matrix that has no cubic root
QUESTION [5 upvotes]: How does one find a real $3\times3$ matrix that does not have a cubic root? If given a matrix without a cubic root, how can one prove that it does not have a cube root?

REPLY [4 votes]: Let $A$ be a nilpotent matrix of order $3$. For example,
$$
      A = \begin{pmatrix}0 & 1 & 0 \\ 0 & 0 &1 \\ 0 & 0 & 0\end{pmatrix}
$$
That is, $A \ne 0$, $A^{2} \ne 0$, $A^{3}=0$. Suppose $A=B^{3}$. Then $B^{9}=0$, which means that the minimal polynomial for $B$ is a power of $\lambda$. However, $B^{6} \ne 0$, which is impossible because of the Cayley Hamilton Theorem. $A$ also has no square root.