For G=GL(2,R), for f, show that f is not an inner automorphism

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Homework Help Overview

The problem involves the group G=GL(2,R) and a function f defined by f(A)=(A^tr)^-1 for all A in GL(2,R). The objective is to demonstrate that f is not an inner automorphism, which entails showing that there is no matrix B in G such that f(A)=BAB^-1 for all A in G.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the center of GL(2,R) and the implications of the function f on elements of this center. There is an exploration of specific matrices that could serve as B and the reasoning behind their validity in GL(2,R). Questions arise regarding the nature of inner automorphisms and the behavior of f when applied to elements in the center.

Discussion Status

The discussion is ongoing, with participants examining the relationship between the function f and the inner automorphism g defined as g(A) = BAB^-1. Some guidance has been provided regarding the restriction of f to the center of G, and participants are actively questioning the implications of this restriction.

Contextual Notes

There is a focus on the properties of matrices in GL(2,R) and the constraints imposed by the definition of inner automorphisms. The exploration includes the need to consider matrices that are not in GL(2,R) and the implications of such choices on the validity of the argument.

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1. Homework Statement :

Let G=GL(2,R) and define f:G->G by f(A)=(A^tr)^-1, where A^tr represents the transpose of A for all A in GL(2,R). Prove that f is not an inner automorphism, hence prove that there does not exist a fixed matrix B in G such that f(A)=BAB^-1 for all A in G

2. Homework Equations :

Before doing this question, we were asked to find the Z(G) [centre of G=GL(2,R)].
I know that an inner automorphism on G is given by as stated in the question f(A)=BAB^-1

3. The Attempt at a Solution :

So my attempt at the solution was first to find the centre of GL(2,R)
Hence Z(G) is given by the set of matrices of the form aI, where a is a scalar and I is the 2x2 identity.
My teacher had told us a hint and said that for us to solve this question we would have to look at the restriction of f to the centre of G=GL(2,R).
After obtaining Z(G), I can't see the restriction that f places on Z(G).
I thought of the zero matrix for B, then obviously that would not be an inner automorphism because all forms of BAB^-1 for A in GL(2,R) would equal the zero matrix. But then it hit me that the zero matrix is not in GL(2,R) so that solution attempt failed.
Then I thought what about a matrix of the form where the first row was zeros and then the bottom row was scalars, but then those matrices do not belong in GL(2,R) since their determinant is 0.
I also tried multiplying arbitrary matrices, but that also got me no where.

Please help me : )
Thank you so much!
 
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Answer this question: what are you trying to gain by trying out several values of B? Indeed the ones you tried are not in GL(2,R), so won't work. But the point is to show that f(A) is not equal to BAB-1 for all B in GL(2,R). Proving it isn't so for some B won't work.

Your teacher's advice is good. Let's say that g(A) = BAB-1 is an inner automorphism. If M is in the center of GL(2,R), what is g(M)?
 
mmm

if g(M) for M in the centre of Z(G), that implies that M=BMB^-1 or that MB=BM?
 
Ok, so g(M)=M for M in the center where g is an inner automorphism. Does f(M)=M for all M in the center?
 

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