# Show that GL(n,R) is isomorphic to GL($R^n$)

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.

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.

• 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 $\text{Aut}(F)^n$. – Lord Shark the Unknown Jun 2 at 4:52
• Good point, I was careless. Rephrasing in terms of vector space automorphisms. – pre-kidney Jun 2 at 4:56
• @JyrkiLahtonen incorporated your comment in the answer by specifying the base field in the notation. – pre-kidney Jun 2 at 5:31