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Math Help - Showing a subset does not form a subgroup (explaination wanted)

  1. #1
    Member elninio's Avatar
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    Showing a subset does not form a subgroup (explaination wanted)

    If not sure how to show that a subset does not form a subgroup. Here is the example:

    Let G=GL_2(<b>R</b>). Show that the subset S of G defined by
     <br />
S={[<br />
a b<br />
c d<br />
]|b=c}<br />
    of symmetric 2x2 matrices deos not form a subgroup of G.

    I'm not sure where to start. I understand the three conditions:
    i)ab is an element of S
    ii) e is an element of S
    iii)a^-1 is an element of S
    But i'm not used to physically working with matrices to prove these. How do I show that at least one of these is not satisfied for S?
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  2. #2
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    Quote Originally Posted by elninio View Post
    If not sure how to show that a subset does not form a subgroup. Here is the example:

    Let G=GL_2(<b>R</b>).
    You can't use HTML tags inside LaTex. Use [ math ]G= GL_2(\bold{R})[ /math ] (without the spaces) to get G= GL_2(\bold{R}).

    Show that the subset S of G defined by
     <br />
S={[<br />
a b<br />
c d<br />
]|b=c}<br />
    use "[ math ]\begin{bmatrix}a & b \\ c & d\end{bmatrix}[ math ]" to get " \begin{bmatrix}a & b \\ c & d\end{bmatrix}"

    of symmetric 2x2 matrices deos not form a subgroup of G.

    I'm not sure where to start. I understand the three conditions:
    i)ab is an element of S
    ii) e is an element of S
    iii)a^-1 is an element of S
    But i'm not used to physically working with matrices to prove these. How do I show that at least one of these is not satisfied for S?
    As for (ii), the identity is \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix} wjhich is a symmetric matrix.

    As for (iii), the inverse of \begin{bmatrix} a & b \\ b & c\end{bmatrix}. a "general" symmetric matrix, can be shown to be \frac{1}{ac- b^2}\begin{bmatrix}c & -b \\ -b & a\end{bmatrix}, also a symmetric matrix.

    So I recommend you focus on (i). Is the product of two symmetric matrices symmetric? What is \begin{bmatrix} a & b \\ b & c\end{bmatrix}\begin{bmatrix}d & e \\ e & f\end{bmatrix}?
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  3. #3
    Member elninio's Avatar
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    Pardon me, I was away all day.

    To answer your i),
    \begin{bmatrix}da+eb & ea+fb \\ db+ec & eb+fc\end{bmatrix}

    Can I say that this does not satisfy i) since the product of two symmetric matrices is not a symmetric matrix?
    Does the question imply that my symmetric matrices should be abbd and accd? Since b=c, wouldnt this imply a symetric product matrix? Or am I completely on the wrong page on this one...
    Last edited by elninio; October 10th 2009 at 07:04 PM.
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    Quote Originally Posted by elninio View Post
    Pardon me, I was away all day.

    To answer your i),
    \begin{bmatrix}da+eb & ea+fb \\ db+ec & eb+fc\end{bmatrix}

    Can I say that this does not satisfy i) since the product of two symmetric matrices is not a symmetric matrix?
    Yes, that's the whole point. Since, in general, ea+fb\ne db+ec, the set of symmetric matrices is not closed under multiplication.

    Does the question imply that my symmetric matrices should be abbd and accd? Since b=c, wouldnt this imply a symetric product matrix? Or am I completely on the wrong page on this one...
    I'm not sure I understand your question. Being a symmetric matrix only means that the number in "first column, second row" is the same as the number in "second column, first row" for that particular matrix. It does NOT imply that all symmetric matrices have those same numbers.

    More specifically, both \begin{bmatrix}2 & 3 \\ 3 & 1\end{bmatrix} and \begin{bmatrix}3 & 2 \\ 2 & 1\end{bmatrix} are symmetric matrices and their product is \begin{bmatrix}2 & 3 \\ 3 & 1\end{bmatrix}\begin{bmatrix}3 & 2 \\ 2 & 1\end{bmatrix}= \begin{bmatrix}12 & 7 \\ 11 & 7\end{bmatrix} which is NOT a symmetric matrix.
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