TITLE: Which of the following properties must be true.
QUESTION [5 upvotes]: Consider a matrix $A = (a_{ij})_{n×n}$ with integer entries such that $a_{ij} = 0$ for $i>j$ and $a_{ii} = 1$ for $i = 1,2...,n$.
Which of the following properties must be true ?

$A^{-1}$ exists and it has integer entries.

$A^{-1}$ exists and it has some entries that are not integers.

$A^{-1}$ is a polynomial function of $A$ with integer coefficients.

$A^{-1}$ is not a power of $A$ unless $A$ is the identity matrix.


My Attempt:
Here we see that $|A|=1$ and $A^{-1} = \frac{adj(A)}{|A|} = adj(A)$. So option 1 is true and option 2 is false.
Also $ch_A(x) = (x-1)^n$ and we know that every square matrix satisfy its characteristic polynomial. So $(A-I)^n = O \implies n_{C_0}A^n(-I)^0+n_{C_1}A^{n-1}(-I)^1+n_{C_2}A^{n-2}(-I)^2+n_{C_3}A^{n-3}(-I)^3+...+n_{C_{n-1}}A(-I)^{n-1}+n_{C_n}A^0(-I)^n = 0$
There will be two cases.
Case 1: $n = 2m , m \in \Bbb N$. Therefore
$n_{C_0}A^n-n_{C_1}A^{n-1}+n_{C_2}A^{n-2}-n_{C_3}A^{n-3}+...-n_{C_{n-1}}A+n_{C_n}I = 0$ Now multiplying by $A^{-1}$ on both sides, we get $n_{C_0}A^{n-1}-n_{C_1}A^{n-2}+n_{C_2}A^{n-3}-n_{C_3}A^{n-4}+...-n_{C_{n-1}}I+n_{C_n}A^{-1} = 0 \implies n_{C_0}A^{n-1}+n_{C_1}A^{n-2}-n_{C_2}A^{n-3}+n_{C_3}A^{n-4}+...+n_{C_{n-1}}I = A^{-1}$
Case 2:
$n = 2m+1 , m \in \Bbb N$. Therefore $n_{C_0}A^{n-1}+n_{C_1}A^{n-2}-n_{C_2}A^{n-3}+n_{C_3}A^{n-4}+...-n_{C_{n-1}}I = A^{-1}$
We know that $n_{C_r} \in \Bbb Z$. Hence option 3 is correct.
If I take $A = I$ then above cases can be written as
Case 1: $n = 2m , m \in \Bbb N$.
$n_{C_0}I^{n-1}+n_{C_1}I^{n-2}-n_{C_2}I^{n-3}+n_{C_3}I^{n-4}+...+n_{C_{n-1}}I = A^{-1}$. So
$I(n_{C_0}+n_{C_1}-n_{C_2}+n_{C_3}+...+n_{C_{n-1}}) = A^{-1} \implies kI = A^{-1}$, where $k =(n_{C_0}+n_{C_1}-n_{C_2}+n_{C_3}+...+n_{C_{n-1}} \in \Bbb Z$
We see clearly that $A^{-1}$ is powers of $A$ and $A$ is not powers of $A$ when $A = I$. So option 4 is true. Please correct me if I'm wrong anywhere. Thanks.

REPLY [4 votes]: I am fine with your work for 1, 2, 3. If you are interested, one might argue a bit faster for 3 by staing in the realm of polynomials for longer: The polynomial $p(x):=1-(1-x)^n=1-(-1)^nch_A(x)\in\Bbb Z[x]$ has the property $p(0)=0$, hence can be written as $p(x)=xq(x)$ with $q\in\Bbb Z[x]$. As $ch_A(A)=0$, it follows that $p(A)=I$, so $Aq(A)=I$.
For 4, you are not done yet. Just because the characteristic polynomial, which is some polynomial that annihilates $A$, gives you a non-power expression, does not show the claim yet.

 The minimal polynomial $m_A(x)$  of $A$ must be a divisor of $ch_A(x)=(x-1)^n$. It follows that $m_A(x)=(x-1)^r$ for some $r\le n$.  As $A\ne0$, clearly, $r\ge1$. Assume $A^{-1}=A^m$ for some $m$. Then $A^{m+1}=I$ and $m_A(x)$ must also be a divisor of $x^{m+1}-1$. As $1$ is only a simple root of $x^{m+1}-1$, this implies $r\le 1$. So $r=1$ and $m_A(x)=x-1$, and finally $A=I$.