# Problem on inner automorphism

• March 7th 2013, 09:10 AM
Problem on inner automorphism
f: GLn(R)-->GLn(R) is defined by f(A)= (A-1)T . Show that f is not an
inner automorphism.

My Solution:
Assume that there is C in GLn(R) such that CAC-1 = (A-1)T for all A in GLn(R).
Then for all A in GLn(R)
det(CAC-1) = det (A-1)T
=> det(C). det(A). Det(C-1) = det(A-1)
=> det(A) = det(A-1).

But, the above is not true for all A in GLn(R).
Therefore f is not an inner automorphism of GLn(R).

Is this correct?
• March 7th 2013, 03:03 PM
Nehushtan
Re: Problem on inner automorphism
Yes, your proof is correct. (Nod)

You could also provide a counterexample. For instance, let $n=2$ and $A=\begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}$. Then $\left(A^{-1}\right)^{\mathrm T}=\begin{pmatrix} \frac12 & 0 \\ 0 & \frac12 \end{pmatrix}$. Thus $\det CAC^{-1}=\det A=4$ for any $C\in\mathrm{GL}_2(\mathbb R)$ whereas $\det\left(A^{-1}\right)^{\mathrm T}=\frac14\ne\det A$.
• March 7th 2013, 07:16 PM
You could also provide a counterexample. For instance, let $n=2$ and $A=\begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}$. Then $\left(A^{-1}\right)^{\mathrm T}=\begin{pmatrix} \frac12 & 0 \\ 0 & \frac12 \end{pmatrix}$. Thus $\det CAC^{-1}=\det A=4$ for any $C\in\mathrm{GL}_2(\mathbb R)$ whereas $\det\left(A^{-1}\right)^{\mathrm T}=\frac14\ne\det A$.