How to find a real $3\times3$ matrix that has no cubic root

Question

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?

Answer 1

Let $A=\begin{bmatrix}0&1&0\\0&0&0\\0&0&0\end{bmatrix},\;$ and let $B$ be a matrix with $B^3=A$.

If $\lambda$ is an eigenvalue of $B$, then $\lambda^3$ is an eigenvalue of $A$, so $\lambda^3=0$ and therefore $\lambda=0$.

Then $B=P^{-1}JP$ where $J=\begin{bmatrix}0&1&0\\0&0&0\\0&0&0\end{bmatrix}, \;J=\begin{bmatrix}0&0&0\\0&0&1\\0&0&0\end{bmatrix},\;J=\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}$, or $J=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}$.

Then $A=B^3=P^{-1}J^{3}P=P^{-1}0P=0$, which gives a contradiction;

so $A$ doesn't have a cube root.

Answer 2

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.