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!

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 $\displaystyle \det\left(A^{-1}\right)^{\mathrm T}=\frac14\ne\det A$.

Quote:

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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 $\displaystyle \det\left(A^{-1}\right)^{\mathrm T}=\frac14\ne\det A$.

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Thanks :)