Prove the following, for every $2 \times 2$ matrix $A$ with complex entries, there exist $2 \times 2$ matrix $B$ with complex entries, such that $$B^3=A^2$$ (working in the field of complex numbers)
My attempt so far, take the Jordan decomposition of $A$ to get $$A=UJU^{-1}$$ then $$A^2=UJ^2 U^{-1}$$
I have $J=D+S$ where $D$ is diagonal and $A$ is nilpotent, for $2 \times 2$ matrices the only nilpotent matrix is $$\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$$ Then $$J^2=(D+S)^2=D^2+DS+SD$$ Now chose $G$ such that $$G^3=J^2$$ (can I do that?) then finally I have $$B^3=UG^3U^{-1} =UJ^2U^{-1}=A$$.
I am stuck at this point.
Is this the correct approach? What should I do next?
Working with Jordan forms basically does it.
If $J = \operatorname{diag}(a,d)$ is diagonal, then just set $G = \operatorname{diag}((a^2)^{1/3}, (d^2)^{1/3}).$
If $J$ isn't diagonal, then $J = aI + S$ where $S=\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$, so $$J^2 = a^2I + 2aS.$$ Now, setting $G = (a^2)^{1/3}I + xS$ and setting equal the coefficients of $S$ in $G^3 = J^2$ gives $3((a^2)^{1/3})^2x = 2a$, which may be solved for $x$ if $a\neq 0$ and any $x$ works if $a=0$.