Inverse of a triangular matrix

Question

Let $A$ be an $n \times n$ upper triangular matrix with integer entries and all diagonal entries are $1$

Prove / Disprove: $A^{-1}$ is not a power of $A$ unless $A$ is the identity matrix

I only know $A^{-1}$ exists and of course, it has all integer entries.

Any help?

Answer 1

Denote $A=I+N$ where $N$ is some nilpotent non-zero upper triangular matrix and suppose $A^{-1}=(I+N)^k$.

Then it should be $(I+N)^{k+1}=I$ and then with the use of of binomial formula $a_1N+a_2N^2+\dots+ a_{k+1}N^{k+1} =0 $

$N(a_1I+a_2N +\dots+ a_{k+1}N^ k) =0$

but $B=a_1I+a_2N +\dots+ a_{k+1}N^ k$ is full rank matrix and $N$ in nonzero matrix so it is impossible to be $NB=0$.

So $N$ must be $0$.

Answer 2

Hint: work out the $2 \times 2$ and $3 \times 3$ cases by hand (i.e., assume your matrix is $$ A = \begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix} $$ and its inverse is $$ A^{-1} = \begin{bmatrix} 1 & c \\ 0 & 1 \end{bmatrix} $$ and see what you can conclude about $a$ and $c$.

Answer 3

Hint: It is enough to prove that $A^m=I$ implies $A=I$. Write $A=I+N$ with $N$ nilpotent. Then $A^m$ is easy to compute using the binomial theorem. Argue that $A^m=I$ implies $N=0$ by considering the diagonals of $N$. For instance, the secondary diagonal of $A^m$ is $m$ times the secondary diagonal of $N$.