Prove the following equation is solvable in $M_m(\mathbb C)$. Prove the following equation is solvable in $M_m(\mathbb C)$ for any $n, l, m \in \mathbb N^*$: $${X^n}+{X^l}-{I_m}=\left( \begin{matrix}
   1 & 0 & 0 & \cdots  & 0  \\
   2 & 1 & 0 & \cdots  & 0  \\
   3 & 2 & 1 & \cdots  & 0  \\
   \vdots  & \vdots  & \vdots  & \ddots  & \vdots   \\
   m & m-1 & m-2 & \cdots  & 1  \\
\end{matrix} \right),$$where $I_m$ is the unit matrix of order $m$.
I'm not sure how to start, but I guess the problem may be quite hard. Can anyone give a hint or a detailed solution? Thanks.
In addition, I wonder whether this equation is solvable in $M_m(\mathbb R)$ or even $M_m(\mathbb Q)$.
 A: Let $N_{m\times m}=\left( \begin{matrix}
   0 & 1 & 0 & \cdots  & 0  \\
   0 & 0 & 1 & \cdots  & 0  \\
   0 & 0 & 0 & \cdots  & 0  \\
   \vdots  & \vdots  & \vdots  & \ddots  & \vdots   \\
   0 & 0 & 0 & \cdots  & 0  \\
\end{matrix} \right)$. Consider $A=Id+N$.
Notice that $A^n=Id+nN+B$ and $A^l=Id+lN+C$, where $B$ and $C$ are upper triangular matrices with coefficients in $\mathbb{Q}$, whose main diagonals and the second main diagonals (above the main diagonal) are zero, since they are linear combinations of $N^2,\ldots,N^n$ and $N^2,\ldots,N^l$, respectively. 
Therefore, $A^n+A^l-2Id=(n+l)N+B+C$. 
Notice that the main diagonal of $(n+l)N+B+C\in M_m(\mathbb{Q})$ is zero, so this matrix is nilpotent. Notice also that the $m-1$ cofficients of the second main diagonal of $(n+l)N+B+C$ are all equal to $n+l$. Thus, the rank of $(n+l)N+B+C$ is $m-1$ and since this matrix is nilpotent, the jordan form of $(n+l)N+B+C$ is $N$. 
Thus, exists $P\in M_m(\mathbb{Q})$ such that $P((n+l)N+B+C)P^{-1}=N$
So, $N=PA^nP^{-1}+PA^lP^{-1}-2Id=(PAP^{-1})^n+(PAP^{-1})^l-2Id$.
Notice that $(PAP^{-1})^n+(PAP^{-1})^l-Id=Id+N$.
Now, let us call the matrix on the right side of your equation $E$. The jordan form of $E$ is $Id+N$. Since $E\in M_m(\mathbb{Q})$ and the eigenvalues are rationals (all equal to 1) then exists $Q\in M_m(\mathbb{Q})$ such that $QEQ^{-1}=Id+N$.
Finally, $(PAP^{-1})^n+(PAP^{-1})^l-Id=QEQ^{-1}$ and  $(Q^{-1}PAP^{-1}Q)^n+(Q^{-1}PAP^{-1}Q)^l-Id=E$. Notice that $Q^{-1}PAP^{-1}Q\in M_m(\mathbb{Q})$.
