The problem is:
Is the statement true or false: Every 3x3 Matrices has the cube root.
After looking through all questions relevant to the problem in this forum, I realize that the statement is false (Or at least, people said that). However, all the proofs are unclear.
So I post the question here, again. Please help me.
Thanks in advance.
The matrix $$A:=\left[\matrix{0&1&0\cr 0&0&0\cr 0&0&0\cr}\right]$$ is triangular, hence its eigenvalues are the diagonal elements $0$, $0$, $0$. If $\lambda\in{\mathbb C}$ is an eigenvalue of a potential third root $B$ of $A$ then $\lambda^3=0$, hence $B$ has eigenvalues $0$, $0$, $0$ as well. But such a $B$ has Jordan form $${\rm jord}(B)=\left[\matrix{0&\epsilon_1&0\cr 0&0&\epsilon_2\cr 0&0&0\cr}\right]\ ,\qquad\epsilon_1,\epsilon_2\in\{0,1\} ,$$ which would imply $B^3=0$, hence $B^3\ne A$. It follows that $A$ has no third root.