Problem on inner automorphism

f: GL_{n}(R)-->GL_{n}(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 GL_{n}(R) such that CAC^{-1} = (A^{-1})^{T} for all A in GL_{n}(R).

Then for all A in GL_{n}(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 GL_{n}(R).

Therefore f is not an inner automorphism of GL_{n}(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 :)