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

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.

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?

Dick