Determinant proof from abstract algebra

[brydustin]

Homework Statement

Let A be a a square n*n matrix. Prove that A^-1 has only integer enteries if and only if the determinant of A is + or -1.

Homework Equations

general knowledge of determinants

The Attempt at a Solution

Proof:

=>
Suppose that det(A) = 1 (without losing generality, i.e. if det(A) = -1, then det(-A) = -det(A) = 1, easy enough).
So if det(A) = 1, then it follows that det(A^-1) = 1 because
1 = det( I ) = det(AIA^-1) = det(A) *det(I) * det(A^-1) = 1*1*det(A^-1) = 1, which implies det(A^-1) = 1 (or -1 if det(A) = -1).
Now what?

<=
Suppose that all the enteries of the matrix A^-1 are integer,...
I have no clue what to do now...

 

[lanedance]

what about
|0.5 0|
|0 2|

 

[brydustin]

Sorry I did a slight misquote of the website from where I pulled the problem:http://ocw.mit.edu/courses/mathematics/18-701-algebra-i-fall-2007/assignments/
first question on first assignment.

Basically, assuming that A is a square matrix with integer entries only; then prove that A^−1 has integer entries if and only if the determinant of A is ±1.

Sorry about that: so your matrix would not be a counter-example because neither it nor its inverse has only integer. Equivalently, if one matrix is all integers with a determinant of +-1 then the inverse of this matrix will also be an integer only matrix with the same determinant.

what about
|0.5 0|
|0 2|

 

[lanedance]

--> A^-1 has only integer entries
so as A^-1 exists, then the determinant of A is non-zero. Think of the effect of elementary row operations on the determinant to reduce the matrix to the identity...

 

[lanedance]

also det(-A) = -det(A) is not always true

 

[micromass]

Maybe you can use that $$det(A^{-1})=det(A)^{-1}$$...

 

[HallsofIvy]

Maybe you can use that $$det(A^{-1})=det(A)^{-1}$$...

Exactly. If both A and $A^{-1}$ have integer entries, then their determinants are integers. Find integers, m and n, such that m= 1/n!

Going the other way, if the determinant of A is -1 or 1, use the fact that the entries of $A^{-1}$ are the cofactors of A divided by det(A).

 

[brydustin]

Yeah... I see it now, almost... it seems obvious but I may be missing a piece...

Because the determinant operation is only multiplication and subtraction it preserves the integers (from A). On the other hand, we require that det( $A^{-1}$) =+-1 because 1 = det (I) = det(AIA^-1) = det( A)*det(I)*det( $A^{-1}$) = (+-1)(1)(det $A^{-1}$)
Now that the det of $A^{-1}$ is +-1 I still don't know why we require the entries to be integers. I see how having integer entries would result in an integer determinant, but I don't see the converse.

Going the other way around, if $A^{-1}$ has integer values then clearly its determinant will be an integer.
Now if we use the fact that det(A)^-1 = det(A^-1) then why should that make det(A) = +-1?
Let d = det(A^-1) where d is an integer. There is nothing that requires det(A)^-1 to be an integer. There are some holes in the logic we have... but we're nearly there...

Exactly. If both A and $A^{-1}$ have integer entries, then their determinants are integers. Find integers, m and n, such that m= 1/n!

Going the other way, if the determinant of A is -1 or 1, use the fact that the entries of $A^{-1}$ are the cofactors of A divided by det(A).