1. ## 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?

2. ## Re: Isomorphic Binary Structure

Originally Posted by tangibleLime
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 $\det(\sqrt{|x|}I\left(\begin{smallmatrix}-1 & 0\\0 & 1\end{smallmatrix}\right)=x$.

3. ## Re: Isomorphic Binary Structure

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