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Math Help - Algebra.....units

  1. #1
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    Exclamation Algebra.....units

    Find all the units in M2(Z2)

    Thanks guys!!!
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  2. #2
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    Quote Originally Posted by suedenation
    Find all the units in M2(Z2)

    Thanks guys!!!
    You need to find all the matrices, of form
    \left( \begin{array}{cc}a&b\\c&d\end{array} \right) where a,b,c,d\in\mathbb{Z}_2 such as it has an inverse.
    There are 16 possible such matrices, check each one if it is invertible.

    If I am not making a mistake, then I think that,
    \left( \begin{array}{cc}0&1\\1&0\end{array} \right)
    and,
    \left( \begin{array}{cc}1&0\\0&1\end{array} \right)
    and,
    \left( \begin{array}{cc}0&1\\1&1\end{array} \right)
    and,
    \left( \begin{array}{cc}1&1\\1&0\end{array} \right)
    and,
    \left( \begin{array}{cc}1&1\\0&1\end{array} \right)
    and,
    \left( \begin{array}{cc}1&0\\1&1\end{array} \right)
    Are the only units in this ring.
    Last edited by ThePerfectHacker; February 28th 2006 at 05:53 PM.
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  3. #3
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    I was thinking of a more elegant way of solving this. There is a theorem in Linear Algebra which states that a square matrix is invertible (a unit) if and only if its determinant is non-zero.
    The problem is I do not know if this is going to work here, possibly because \mathbb{Z}_2 is not a field.

    But let us assume, thus we need that,
    \left| \begin{array}{cc}a&b\\c&d\end{array} \right|\not =0
    Since there are only two elements in the set by elimination we have that,
    \left| \begin{array}{cc}a&b\\c&d\end{array} \right|=1
    Thus, by evaluating the determinant,
    ad-bc=1
    But remember in this ring since it has only two elements, that -x=x for all x.
    Thus,
    ad+bc=1
    Now, just observe the possibilities for a,b,c,d which make this equation true. Since there are only 16 of them it is easy.

    1)If a=0 then, b,c\not=0 thus, b,c=1 and d can be anything (either 1 or 0) this gives us two matrices,
    \left(\begin{array}{cc}0&1\\1&0\end{array}\right) and \left(\begin{array}{cc}0&1\\1&1\end{array}\right).

    2)If a=1 then , d=1 would mean that either b or c must be zero. Also, if d=0 would mean that both a and b cannot be zero (thus they are both one). This gives us four matrices,
    \left(\begin{array}{cc}1&0\\0&1\end{array}\right) and, \left(\begin{array}{cc}1&1\\0&1\end{array}\right) and,
    \left(\begin{array}{cc}1&0\\1&1\end{array}\right) and, \left(\begin{array}{cc}1&1\\1&0\end{array}\right)

    Now there is no need to observe what happens with b,c,d because we know what matrix would it be depending on the value of a.
    Last edited by ThePerfectHacker; March 1st 2006 at 06:55 AM.
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  4. #4
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    Quote Originally Posted by ThePerfectHacker
    There is a theorem in Linear Algebra which states that a square matrix is invertible (a unit) if and only if its determinant is non-zero.
    The problem is I do not know if this is going to work here, possibly because \mathbb{Z}_2 is not a field.
    It works fine, because \mathbb{Z}_2 actually is a field.

    Here's another way which generalises to any finite field GF(q) and any dimension n.

    A square matrix is invertible if and only if its rows are linearly independent. Take each row in turn. The first row must be non-zero so there are q^n-1 possibilities. The second row must not the in the subspace generated by the first row, which has dimension one, so there are q^n-q possiblities. The (r+1)-th row must not be in the subspace generated by the previous r rows, linearly independent by induction, so of dimension r giving q^n-q^r possibilities. The overal number of possiblities is thus (q^n-1)(q^n-q)\cdots(q^n-q^r)\cdots(q^n-q^{n-1})

    In the case q=2, n=2 this is (2^2-1)(2^2-2^1) = 6.
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  5. #5
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    Quote Originally Posted by rgep
    It works fine, because \mathbb{Z}_2 actually is a field.

    Here's another way which generalises to any finite field GF(q) and any dimension n.

    A square matrix is invertible if and only if its rows are linearly independent. Take each row in turn. The first row must be non-zero so there are q^n-1 possibilities. The second row must not the in the subspace generated by the first row, which has dimension one, so there are q^n-q possiblities. The (r+1)-th row must not be in the subspace generated by the previous r rows, linearly independent by induction, so of dimension r giving q^n-q^r possibilities. The overal number of possiblities is thus (q^n-1)(q^n-q)\cdots(q^n-q^r)\cdots(q^n-q^{n-1})

    In the case q=2, n=2 this is (2^2-1)(2^2-2^1) = 6.
    Thanks for the tip!
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