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Math Help - Matrices with integer entries

  1. #1
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    Matrices with integer entries

    If both (c_{ij}) and (c_{ij})^{-1} have integer entries, show that |c_{ij}| and |(c_{ij})^{-1}| are in \mathbb{Z}. I can't figure this out .

    ps this is pretty urgent if anyone can give me a pointer real quick

    Thanks.
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    You could easily do it by induction, by evaluating the determinant along any row or column.

    It should be pretty obvious, because the determinant only uses sums and products of the entries of the matrix, no divisons.
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  3. #3
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    Induction Proof.

    Base Case: Let A_2 =<br />
\left( \begin{array}{*{20}{cc}}<br />
a & { b} \\<br />
{ c} & d \\<br />
\end{array} \right)<br />
such that a,b,c,d are integers.
     Det(A_2) = ad-bc , and since the set of integers is closed under addition and multiplication, we know we will be getting an integer determinant.

    Now, let  A_3 = <br />
\left( \begin{array}{*{20}{ccc}}<br />
a & { b} & c\\<br />
{ d} & e & f \\<br />
g & h & i \\<br />
\end{array} \right)<br />
, such that a,b,c,d,e,f,g,h,i are integers.
    Using cofactor expansion, we can write  Det[A_3]
    as the sum of three 2x2 determinants multiplied by integer coefficients.
    Now, since we showed that a 2x2 integer matrix has an integer determinant, and again, the set of integers is closed under multiplication and addition,  Det(A_3) is too clearly an integer.

    Induction Step:

    Consider an NxN integer matrix. Assume that any square integer matrix smaller than NxN has an integer determinant. The determinant of the NxN integer matrix is the sum of N (N-1)x(N-1) integer determinants times their integer coefficients. Again, since the set of integers is closed under addition and multiplication, the NxN integer matrix has an integer determinant, for N>=2.
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  4. #4
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    Yeah it is obvious, since a determinant is just a polynomial of its terms so if all terms are integers, then obviously the determinant will be an integer. I wasn't thinking straight last night

    It becomes clear when seeing the Leibniz formula for the determinant. We have that for an n \times n-matrix A=(a_{ij}), then

    \det(A) = \sum_{\sigma \in S^n} \mathrm{sgn}(\sigma) \prod_{i=1}^n a_{i,\sigma(i)},

    where S^n is the symmetric group of n elements.
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