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?
$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Basis}{\mathbf{e}}$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 $\Reals^{k}$.