Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
very very sorry .there is a mistake.now i correct it. –  bibi Dec 13 '12 at 15:04
add comment

1 Answer 1

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?

share|improve this answer
    
then both 3 and 4 are right.am 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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.