Are $\mathbb R^{n \times m}$ and $\mathbb R^{nm}$ diffeomorphic? Are $\mathbb R^{n \times m}$ and $\mathbb R^{nm}$ diffeomorphic?
I am improving my insight about matrix spaces. Is $\mathbb R^{n \times m}$ (the space of real valued matrices with n rows and m columns) diffeomorphic to $\mathbb R^{nm}$, the space of real valued vectors of $n*m$ components?
 A: Of course (modulo one subtlety about smooth structures). If the function was any simpler, it'd be the identity function. We just take the coordinates and write them in a straight line instead of in a box. No, I'm not kidding.
The subtlety is that a priori there could be more smooth structures for $n,m>>0$, so the identity map might not be a diffeomorphism. I assume your problem gives these spaces their standard smooth structures. But even if it didn't, $\mathbb{R}^N$ has a unique smooth structure for $N \neq 4$, so the only possible issues could arise with some non-standard smooth structure when $n=m=2$ or one $n,m=4$ and the other $=1$.
A: Each finite-dimensional real vector space $V$ of dimension $r$ has a canonical smooth structure. This is obtained by choosing any vector space isomorphism $\phi : V \to \mathbb R^r$ and taking it as the single chart of an atlas $A_\phi =\{\phi\}$. This atlas generates a smooth structure on $V$. This smooth structure does not depend on the choice of $\phi$: If $\psi$ is another vector space isomorphism , then the only transition maps between  $A_\phi$ and  $A_\psi$ are $\psi \circ \phi^{-1}$ and $\phi \circ \psi^{-1}$ which are smooth (recall that all linear maps on $\mathbb R^r$ are smooth).
This answers your question in the affirmative: All real vector spaces of the same dimension (endowed with the canonical smooth structure) are diffeomorphic.
However, $\mathbb R^k$ also has non-canonical smooth structures. See for example Clarification between charts being incompatible and smooth structures being "exotic". But unless $r = 4$ all smooth structures on $\mathbb R^4$ produce diffeomorphic smooth manifolds. These are highly non-trivial results.
Anyway, there is no reason to endow $\mathbb R^k$ with a non-canonical smooth structure.
