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?
Please help!
Re: Problem on inner automorphism
Yes, your proof is correct. (Nod)
You could also provide a counterexample. For instance, let $\displaystyle n=2$ and $\displaystyle A=\begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}$. Then $\displaystyle \left(A^{-1}\right)^{\mathrm T}=\begin{pmatrix} \frac12 & 0 \\ 0 & \frac12 \end{pmatrix}$. Thus $\displaystyle \det CAC^{-1}=\det A=4$ for any $\displaystyle C\in\mathrm{GL}_2(\mathbb R)$ whereas $\displaystyle \det\left(A^{-1}\right)^{\mathrm T}=\frac14\ne\det A$.
Re: Problem on inner automorphism
Quote:
Originally Posted by
Nehushtan
Yes, your proof is correct. (Nod)
You could also provide a counterexample. For instance, let $\displaystyle n=2$ and $\displaystyle A=\begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}$. Then $\displaystyle \left(A^{-1}\right)^{\mathrm T}=\begin{pmatrix} \frac12 & 0 \\ 0 & \frac12 \end{pmatrix}$. Thus $\displaystyle \det CAC^{-1}=\det A=4$ for any $\displaystyle C\in\mathrm{GL}_2(\mathbb R)$ whereas $\displaystyle \det\left(A^{-1}\right)^{\mathrm T}=\frac14\ne\det A$.
Thanks :)