Can some one give a reference/proof of the following (possibly positive) result?
$GL(r,\mathbb{C})$ is diffeomorphic to $U(n)$ and Euclidean space.
This is a (possible) special case of Iwasawa theorem
If $G$ is a connected Lie group, then $G$ is diffeomorphic to direct product of a maximal compact subgroup $H$ and Eucldiean space.
Here $Gl(r,\mathbb{C})$ is connected Lie group, $U(n)$ is maximal compact subgroup. So, it seems $GL(r,\mathbb{C})$ is diffeomorphic to $U(n)$ and Euclidean space. Can an easy proof/reference be given for this?
The Iwasawa theorem is not terribly easy, to my recollection. If you change your statement as in Rob Arthan's comment, it is correct. But for most of the classical groups you are probably interested in, there are simple proofs, as in the following (your specific case).
Let $\mathcal H_n$ be the space of positive Hermitian forms on $\Bbb C^n$; this is a convex open subset of the space of $n \times n$ Hermitian (that is, complex self-adjoint) matrices. This space has real dimension $n^2$; any convex open subset of a real vector space of dimension $k$ is diffeomorphic to $\Bbb R^k$ (see here for a more complicated proof of a more general fact).
The group $GL_n(\Bbb C)$ acts smoothly on $\mathcal H_n$ by pullback of forms (or if you're thinking of these as matrices, it acts by conjugation). It acts transitively because every $n$-dimensional complex Hilbert space is isomorphic, so given $H, H' \in \mathcal H_n$, there must be some isomorphism $A: (\Bbb C^n, H) \to (\Bbb C^n, H')$; this isomorphism is $A \in GL_n(\Bbb C)$. If we write $g$ for the standard inner product corresponding to the diagonal matrix, the stabilizer of $g$ is $U(n)$ (by definition!). So $\mathcal H_n$ is a homogeneous space, and hence sits into a principal bundle $$U(n) \to GL_n(\Bbb C) \to GL_n(\Bbb C)/U(n) \cong \mathcal H_n \cong \Bbb R^{n^2}.$$ Because every bundle over a contractible space is smoothly trivial, we can find a ($U(n)$-equivariant!) diffeomorphism $GL_n(\Bbb C) \cong U(n) \times \Bbb R^{n^2}$.