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Consider a matrix $A=(a_{ij})_{ n ×n }$ with integer entries such that $a_{ij}=0$ for $i>j$ and $a_{ii}=1$ for $i=1,…,n$. then which of the followings are true?

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

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

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

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

By the given conditions $A$ is the upper triangular matrix with diagonal elements $1$.so eigenvalues are $1$.so their product=determinant of $A =1.$ So 1 is true. Inverse of the identity matrix is itself with has all integer entries so 2 is false. But I have no idea about (3) and (4) can anyone help me please.

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very very sorry .there is a i correct it. – bibi Dec 13 '12 at 15:04

3) Edit: I misread the problem statement as "the entries of $A^{-1}$ are polynomials in the entries of $A$, and gave the hint as "note that $A^{-1} = \frac{1}{\det A}\mathrm{adj}A$." Obviously this hint does not work with the real problem statement. Here is a corrected hint:

Since $A$ is a triangular matrix with all diagonal entries equal to $1$, what is the characteristic polynomial of $A$? What do you get if you multiply the characteristic equation by $A^{-1}$ on both sides?

4) ($A^{-1}$ is definitely a power of $A$! The power is $-1$, isn't it?) If $A^{-1}=A^k$ for some nonnegative integer $k$, then $A^{k+1}=I$. Hence the polynomial $x^{k+1}-1$ annihilates $A$, i.e. the minimal polynomial of $A$ divides $x^{k+1}-1$. Meanwhile, the minimal polynomial of $A$ divides the characteristic polynomial of $A$ (Cayley-Hamilton theorem). So, what is this minimal polynomial?

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then both 3 and 4 are i correct? – bibi Dec 13 '12 at 15:44
Yes, they are both correct. – user1551 Dec 13 '12 at 16:09
@bibi I misread question 3 and gave a wrong hint! The statement is still true, but for a different reason. See my new edit. – user1551 Dec 13 '12 at 17:13

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