I can't understand why a finite dimensional vector space is a smooth manifold. I think it's a trivial thing, but I can't deal with it.
Let $V$ a finite-dimensional real vector space. $V$ is a topological manifold of dimension $n$. This because $V\simeq\mathbb{R}^n$.
Now, each basis $(E_1,\dots, E_n)$ for $V$ defines a basis isomorphism $E\colon\mathbb{R}^n\to V$ by $$E(x)=\sum_{i=1}^nx_iE_i.$$
This map is a homeomorphism, then $\big(V,E^{-1}\big)$ is a chart. if $\big(F_1,\dots,F_m\big)$ is any other basis and $F(x)=\sum_jx_jF_j$ is the correspondig isomorphism, then there is some invertible matrix $\big(A_j^i \big)$ such that $E_i=\sum_j A_j^{i} F_j$ for each $i$. Up to this point it seems to have been applied, without further explanation.
The transition map between the two charts $\big(V, E^{-1}\big)$, $\big(V, F^{-1}\big)$ is given by $F^{-1}\circ E(x)=\tilde{x}$, where $\tilde{x}=(\tilde{x_1},\dots, \tilde{x_n})$ is determined by $$\sum_{j=1}^n \tilde{x_j} F_j=\sum_{i=1}^{n} x_iE_i=\sum_{i,j}^{n}x_iA_j^{i} F_j$$
Question. Why this thing?
It follows that $\tilde{x}=\sum_i A_i^{j}x_i$.
Thus, the map sending $x$ to $\tilde{x}$ is an invertible linear map.
Question Why the map sending $x$ to $\tilde{x}$ is an invertible linear map?
Thanks!
$\tilde{x}=F^{-1}\circ E(x)$, so $F(\tilde{x})=E(x)$, i.e. by definition of $F$ and $E$ $$ \sum_j\tilde{x}_j F_j=\sum_jx_j E_j$$
and since $E_j=\sum_i A^i_j F_j$
$$ \sum_j\tilde{x}_j F_j=\sum_jx_j E_j=\sum_{j,i}x_j A^i_j F_j$$
you might recognize that $\tilde{x}_j=\sum_i A^i_j x_i$ is just the matrix vector multiplication $\tilde{x}=Ax$. Since $A$ maps a basis to a basis, $A$ is an invertible matrix.