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I read in my book that the said isomorphism holds, but I am confused as to what exactly GL(Rn) is. Can someone help clarify this and how to show the above isomorphism.

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For a field $\mathbb F$, the group $GL(n,\mathbb F)$ is defined to be the invertible $n\times n$ matrices with entries in $\mathbb F$, with matrix multiplication as the group operation.

For a vector space $V$ over a field $\mathbb F$, the group $GL(V;\mathbb F)$ is defined to be the group of $\mathbb F$-linear transformations from $V$ to itself, with function composition as the group operation. Taking $V$ to be the vector space $\mathbb F^n$ (i.e., the vector space whose elements are column vectors of size $n$ with entries in $\mathbb F$ and the obvious definitions of addition and scalar multiplication), we see that the isomorphism between $GL(n,\mathbb F)$ and $GL(\mathbb F^n;\mathbb F)$ is given by taking a linear transformation and writing it as a matrix in the standard basis of column vectors, namely $$ \begin{pmatrix}1\\0\\\vdots\\0\end{pmatrix},\ldots,\begin{pmatrix}0\\\vdots\\0\\1\end{pmatrix}. $$

Thus $GL(n,\mathbb F)$ and $GL(\mathbb F^n;\mathbb F)$ are isomorphic. Specializing to the case $\mathbb F=\mathbb R$ answers your question.

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    $\begingroup$ The group of automorphisms of the ring $F^n$ is not $\text{GL}(n,F)$. Rather it is the semidirect product of $S_n$ and $\text{Aut}(F)^n$. $\endgroup$ – Lord Shark the Unknown Jun 2 at 4:52
  • $\begingroup$ Good point, I was careless. Rephrasing in terms of vector space automorphisms. $\endgroup$ – pre-kidney Jun 2 at 4:56
  • $\begingroup$ @JyrkiLahtonen incorporated your comment in the answer by specifying the base field in the notation. $\endgroup$ – pre-kidney Jun 2 at 5:31

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