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?

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

Re: Isomorphic Binary Structure

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