Given matrix $A$, decide the existence of a $k \times k$ matrix $X$ such that $X^n=A$

Question

Let $R$ be a commutative ring with an identity element $1$. Is there a $k\times k$ matrix $X$ over $R$ such that $$ X^n= \overbrace{ \begin{pmatrix} 1 & 1 & 0 & \cdots & 0\\ -1 & -1 & 0 &\cdots & 0 \\ 0 & 0 & 0 & \ldots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & 0 &\cdots & 0 \end{pmatrix} }^{\mathrm{A}}\ ? $$ If $k\leq n$, then the answer is no, since $A^2=O$, so $X^{2n}=O$. If the minimal polynomial of $X$ os $m(t)$, then $m(t)\mid t^{2n}$ and since $\deg m(t)\leq k$, it follows that $X^k=O$, so $X^n=O$, a contradiction. What can be said when $n<k$ ?

Answer 1

Let

• $J_m$ be the $m\times m$ (upper triangular) nilpotent Jordan block,
• $E_{ij}$ be the matrix with a $1$ at the $(i,j)$-th position and zeroes elsewhere,
• $P=P^{-1}$ be the permutation matrix for the transposition $\pmatrix{2&n+1}$, and
• $S=\pmatrix{-1&-1\\ 1&0}$, with $S^{-1}=\pmatrix{0&1\\ -1&-1}$ also lies inside $M_k(R)$.

Then \begin{aligned} A&=\pmatrix{1&1\\ -1&-1}\oplus0_{(k-2)\times(k-2)}\\ &=\left(S\oplus I_{k-2}\right)\left(J_2\oplus0_{(k-2)\times(k-2)}\right)\left(S^{-1}\oplus I_{k-2}\right)\\ &=\left(S\oplus I_{k-2}\right)PE_{1,n+1}P^{-1}\left(S^{-1}\oplus I_{k-2}\right)\\ &=\left(S\oplus I_{k-2}\right)P\left(J_{n+1}^n\oplus0_{(k-n-1)\times(k-n-1)}\right)P^{-1}\left(S^{-1}\oplus I_{k-2}\right)\\ &=\left[\left(S\oplus I_{k-2}\right)P\left(J_{n+1}\oplus0_{(k-n-1)\times(k-n-1)}\right)P^{-1}\left(S^{-1}\oplus I_{k-2}\right)\right]^{\,n}.\\ \end{aligned}