Matrix group as subgroup of $GL(n,\mathbb{R})$ or $GL(n,\mathbb{C}$)?

As far as I understand, one has (at least?) two choices to introduce infinite matrix groups:

Either, one can say they are all subgroups of the general linear group over the complex numbers numbers $\mathbf{GL}(n,\mathbb{C})$. Then, $\mathbf{GL}(n,\mathbb{R})$ is a subgroup of $\mathbf{GL}(n,\mathbb{C})$.

Or one can say that a infinite matrix group is a subgroup of the general linear group $\mathbf{GL}(n,\mathbb{R})$. In this case we can say that $\mathbf{GL}(m,\mathbb{C})\subset \mathbf{GL}(2m,\mathbb{R})$, while a single complex number is represented using a $2\times 2$ matrix.

Is there any particular (e.g. technical) reason to prefer one over the other?

edit:

These following comments from Gallier ("Geometric Methods and Applications", second edition, 2011, p.475) made me wonder:

"As in the real case, the groups $\mathbf{GL}(n, \mathbb{C})$, $SL(n, \mathbb{C})$, $\mathbf{U}(n)$, and $\mathbf{SU}(n)$ are also topological groups (viewed as subspaces of $\mathbb{R}^{2n^2}$ ), and in fact, smooth real manifolds. Such objects are called (real) Lie groups. [...]

It is also possible to define complex Lie groups, which means that they are topological groups and smooth complex manifolds. It turns out that $\mathbf{GL}(n, \mathbb{C})$ and $\mathbf{SL}(n, \mathbb{C})$ are complex manifolds, but not $\mathbf{U}(n)$ and $\mathbf{SU}(n)$. One should be very careful to observe that even though the Lie algebras ${\frak sl}(n, \mathbb{C})$, ${\frak u}(n)$, and ${\frak su}(n)$ consist of matrices with complex coefficients, we view them as real vector spaces. The Lie algebra ${\frak sl}(n, \mathbb{C})$ is also a complex vector space, but ${\frak u}(n)$ and ${\frak su}(n)$ are not! Indeed, if A is a skew-Hermitian matrix, iA is not skew-Hermitian, but Hermitian!"

So, all infinite matrix groups are real manifolds, but only some of them are complex manifolds. So, isn't it conceptually "nicer" to say that $\mathbf{GL}(m,\mathbb{C})\subset \mathbf{GL}(2m,\mathbb{R})$?

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You showed the definitions are equivalent. What reasons do you imagine would then lead to prefer one over the other? –  Marc van Leeuwen Feb 14 '12 at 13:05
Didactic reasons for instance. But after your comment, I guess it is best to mention that both are equivalent. Thanks! –  Hauke Strasdat Feb 14 '12 at 13:35
Perhaps not technical, but I much prefer $GL(n,\mathbb{R})\leq GL(n,\mathbb{C})$ simply because $\mathbb{C}$ is an extension of $\mathbb{R}$ –  Sid Raval Feb 14 '12 at 14:39
An $n \times n$ matrix over $\mathbb C$ can be represented by $2n$ real numbers, whereas a $2n \times 2n$ matrix over $\mathbb R$ requires $4n$ so, as a computationally inclined person, I prefer ${\rm GL}(n,{\mathbb C})$. –  Derek Holt Feb 14 '12 at 21:48
@fretty: that's included in the present definition: you can embed $\mathbb{R}^n$ as an additive group into $GL(n+1,\mathbb{R})$ see here; and $M_n(\mathbb{R})$ is just $\mathbb{R}^{n^2}$. –  t.b. Aug 26 '12 at 13:11