Isomorphism: matrix determinant

kala
Messages
21
Reaction score
0
Determine whether the given map \varphi is an isomorphism of the first binary structure with the second.
< M2(R ), usual multiplication > with <R, usual multiplication> where \varphi(A) is the determinant of matrix A.

The determinant of the matrix is ad-bc, so \varphi(A)=ad-bc.
For this to be an isomorphism, I have to show that the function is one to one, onto and preserves the operations.
I'm having trouble getting this to work. Any suggestions?
 
Physics news on Phys.org
kala said:
I'm having trouble getting this to work.
Maybe it can't...
 
According to the book it is suppose to be an isomorphism, the question says it is. I can get it to be one to one and onto, but i am having trouble with it preserving the operations.
 
You just said that you must prove the function is "one to one". That is that two different matrices, such as
\begin{bmatrix}2 &amp; 1 \\ 1 &amp; 1\end{bmatrix}
and
\begin{bmatrix}3 &amp; 2\\ 1 &amp; 1\end{bmatrix}
must not have the same determinant. Is that true?

I think you should to reread that problem.
 
kala said:
the question says it is.
No it doesn't. The question is asking you whether or not it is.
 
Oh duh... That was stupid... It doesn't have to be. Thanks
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

Similar threads

Back
Top