TITLE: Show that every k-dimensional vector subspace V of $R^N$ is a manifold diffeomorphic to $R^k$.
QUESTION [7 upvotes]: I'm actually in a exercise of the book " Differential Topology " of Guillemin and Pollack.
Show that every k-dimensional vector subspace $V$ of $R^N$ is a manifold diffeomorphic to $R^k$, and that all linear maps on $V$ are smooth.
Definition :  k-dimensional manifold meaning that each point x possesses a neighborhood $V$ in $X$ which is diffeomorphic to an open set $U$ of $R^k$.
I have trouble well operate on this problem.
I looked for a solution on the internet. Here's what I found:
1. Show that every k-dimensional vector subspace V of $R^N$ is a manifold diffeomorphic to $R^k$, and that all linear maps on V are smooth. 
Choose any basis {$v_1, . . . , v_k$} of V, where $v_i  ∈ R^N$. Since V is k-dimensional, there are exactly k vectors in this basis, and the vi  are linearly independent.  Choose a basis {$e_1, . . . , e_k$} of $R^k$. Let $φ(v_i) := e_i$ for all i. Since {$v_i$} are linearly independent, $φ$ extends linearly to a map on all of V, and since the set {$e_i$} is linearly independent, $φ : V  → R^k$ is surjective.  It is clearly invertible for the same reasoning, where $φ^{−1}(e_i) → v_i$, and $$φ^{-1}: R^k → V$$ is the linear extension of the map on the basis element. Now $φ$ is one-to-one; the fact that it is smooth follows from the next exercise. Similarly, $φ^{−1}$ is a linear map that is smooth by the next exercise. Thus $φ$ is a diffeomorphism. The fact that all linear maps on V are smooth also follows from the next exercise.
2. Suppose that U is an open set in $R^n$. Prove that if $L: U → R^m$ is a linear map, then $dL_x = L$ for all $x ∈ U$. Hint. Suppose $0 ∈ U$ and try to prove the statement first at $x = 0$.
Let L be a linear map. Then for any $h∈R^n$,
$$dL_x(h) = \lim\limits_{\begin{array}{l} t \to 0\\\end{array}} \frac{L(x + th) − L(x))}{t} =\lim\limits_{\begin{array}{1} t \to 0\\\end{array}}\frac {L(x) + tL(h) − L(x)}{t} = \lim\limits_{\begin{array}{1} t \to 0\\\end{array}} \frac {tL(h)}{t} = L(h)$$
I'm not sure what happen in the bold section, and also it is not clear to me that this proof respects definiton at the beginning. Does someone could explain to me?

REPLY [8 votes]: $\newcommand{\Reals}{\mathbf{R}}\newcommand{\Basis}{\mathbf{e}}$Let $V$ be a $k$-dimensional subspace of $\Reals^{N}$, let $(v_{i})_{i=1}^{k}$ be an ordered basis of $V$, and let $(\Basis_{i})_{i=1}^{k}$ denote the standard basis of $\Reals^{k}$.
The mapping $\varphi:V \to \Reals^{k}$ defined by
$$
\varphi(x_{1} v_{1} + \dots + x_{k} v_{k}) = (x_{1}, \dots, x_{k})
$$
is linear, with (linear) inverse
$$
\varphi^{-1}(x_{1}, \dots, x_{k}) = x_{1} v_{1} + \dots + x_{k} v_{k},
$$
hence is bijective, and consequently an isomorphism of vector spaces.
Since the component functions in each direction are smooth (linear polynomials in Cartesian coordinates), $\varphi$ is a diffeomorphism, hence is a manifold chart for $V$.
That is, if $x = x_{1} v_{1} + \dots + x_{k} v_{k}$ is an arbitrary element of $V$, then viewing $V$ as an open subset of itself, $\varphi$ is a diffeomorphism to $U = \Reals^{k}$, an open subset of $\Reals^{k}$.