[B]hi, im new to the forum

how do you show that GL(2,R) is a group under matrix mulitplication?

and then show that F(matrix)= [ {a b} {-b a}

, a,b are elements of R, and a,b dont equal 0]

thanks

Printable View

- Dec 6th 2009, 04:02 PMsaambregroup matrix multiplication
[B]hi, im new to the forum

how do you show that GL(2,R) is a group under matrix mulitplication?

and then show that F(matrix)= [ {a b} {-b a}

, a,b are elements of R, and a,b dont equal 0]

thanks

- Dec 6th 2009, 05:35 PMlvleph
Can you use $\displaystyle \det(AB) = \det(A)\det(B) = 0 \Leftrightarrow \det(A) = 0 \text{ or } \det(B) = 0 $?

- Dec 7th 2009, 12:59 AMsaambrena
na we cant. We havent been taught that

- Dec 7th 2009, 01:07 AMSwlabr
You have four things to show:

Closure: $\displaystyle det(A) \neq 0$, $\displaystyle det(B) \neq 0$. We shall do this last of all...

Associativity: You know this holds from your linear algebra course (there is nothing special about multiplication in GL, it is just the same as normal matrix multiplication which is associative), but there is nothing wrong with going through it one more time. $\displaystyle (AB)C = \ldots = A(BC)$.

Identity: You know what this is...

Inverses: Again, you know what the inverse of a matrix with non-zero determinant is. Just look up your linear algebra notes!

So, we still need to prove closure. Notice that $\displaystyle det(A) \neq 0 \Longleftrightarrow A \text{ has an inverse}$. So, we can construct an inverse for $\displaystyle AB$. It is just $\displaystyle B^{-1}A^{-1}$...! - Dec 9th 2009, 02:21 AMsaambrethanks
thanks for your reply. was very helpful. cheers