# Isomorphic Binary Structure

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• Sep 20th 2011, 05:42 PM
tangibleLime
Isomorphic Binary Structure
Problem: Determine whether the given map $\displaystyle \phi$ is an isomorphism of the first binary structure with the second.

$\displaystyle <\mathbb{R}^{2x2},\cdot>$ with $\displaystyle <\mathbb{R},\cdot>$ where $\displaystyle \phi(A)$ 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 $\displaystyle \phi$ is one-to-one.
3) " " is onto S'.
4) Show that $\displaystyle \phi(x*y) = \phi(x) * \phi(y)$ for all $\displaystyle x,y \in S$.

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

For step two,
If $\displaystyle \phi(x) = \phi(y)$, 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?
• Sep 20th 2011, 07:25 PM
Drexel28
Re: Isomorphic Binary Structure
Quote:

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

$\displaystyle <\mathbb{R}^{2x2},\cdot>$ with $\displaystyle <\mathbb{R},\cdot>$ where $\displaystyle \phi(A)$ 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 $\displaystyle \phi$ is one-to-one.
3) " " is onto S'.
4) Show that $\displaystyle \phi(x*y) = \phi(x) * \phi(y)$ for all $\displaystyle x,y \in S$.

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

For step two,
If $\displaystyle \phi(x) = \phi(y)$, 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 $\displaystyle \det\left(\begin{smallmatrix}-1 & 0\\ 0 & 1\end{smallmatrix}\right)=-1$ and $\displaystyle \det(\lambda I)=\lambda^2$ for any positive number $\displaystyle \lambda$. So, either $\displaystyle x\in\mathbb{R}$ is positive in which case $\displaystyle \det(\sqrt{x}I)=x$ otherwise it's negative and so $\displaystyle \det(\sqrt{|x|}I\left(\begin{smallmatrix}-1 & 0\\0 & 1\end{smallmatrix}\right)=x$.
• Sep 21st 2011, 08:50 AM
tangibleLime
Re: Isomorphic Binary Structure
Thanks! It's good to know I was on track. And thanks for the clarification of the onto portion.