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Math Help - Isomorphic Binary Structure

  1. #1
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    Isomorphic Binary Structure

    Problem: Determine whether the given map \phi is an isomorphism of the first binary structure with the second.

    <\mathbb{R}^{2x2},\cdot> with <\mathbb{R},\cdot> where \phi(A) is the determinant of matrix A.

    My attempt:

    I know that I need to complete four steps to prove isomorphism.

    1) Define the function that gives isomorphism S to S'.
    2) Show that \phi is one-to-one.
    3) " " is onto S'.
    4) Show that \phi(x*y) = \phi(x) * \phi(y) for all x,y \in S.

    For step one,
    \phi : \mathbb{R}^{2x2} \rightarrow \mathbb{R}
    det(x) for all x \in \mathbb{R}^{2x2} is in \mathbb{R}, so \phi(x) \in \mathbb{R}.

    For step two,
    If \phi(x) = \phi(y), det(x) = det(y). This is false, since determinants are not unique. det(x) can equal det(y) while x is not equal to y. Therefore, this map is not isometric since it is not one-to-one.

    Then I stop there since it's not one-to-one, there's no need to prove the rest. But am I right? And if so, how would I prove that it is or isn't onto?
    Last edited by tangibleLime; September 20th 2011 at 06:56 PM.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Re: Isomorphic Binary Structure

    Quote Originally Posted by tangibleLime View Post
    Problem: Determine whether the given map \phi is an isomorphism of the first binary structure with the second.

    <\mathbb{R}^{2x2},\cdot> with <\mathbb{R},\cdot> where \phi(A) is the determinant of matrix A.

    My attempt:

    I know that I need to complete four steps to prove isomorphism.

    1) Define the function that gives isomorphism S to S'.
    2) Show that \phi is one-to-one.
    3) " " is onto S'.
    4) Show that \phi(x*y) = \phi(x) * \phi(y) for all x,y \in S.

    For step one,
    \phi : \mathbb{R}^{2x2} \rightarrow \mathbb{R}
    det(x) for all x \in \mathbb{R}^{2x2} is in \mathbb{R}, so \phi(x) \in \mathbb{R}.

    For step two,
    If \phi(x) = \phi(y), det(x) = det(y). This is false, since determinants are not unique. det(x) can equal det(y) while x is not equal to y. Therefore, this map is not isometric since it is not one-to-one.

    Then I stop there since it's not one-to-one, there's no need to prove the rest. But am I right? And if so, how would I prove that it is or isn't onto?
    Yes, you can just stop there. It's onto, if you're curious since \det\left(\begin{smallmatrix}-1 & 0\\ 0 & 1\end{smallmatrix}\right)=-1 and \det(\lambda I)=\lambda^2 for any positive number \lambda. So, either x\in\mathbb{R} is positive in which case \det(\sqrt{x}I)=x otherwise it's negative and so \det(\sqrt{|x|}I\left(\begin{smallmatrix}-1 & 0\\0 & 1\end{smallmatrix}\right)=x.
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  3. #3
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    Re: Isomorphic Binary Structure

    Thanks! It's good to know I was on track. And thanks for the clarification of the onto portion.
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