Action of the linear group over a vector space What is a simple proof that the action of $GL_n(\mathbb{K})$ is transitive on 
$K-{0_\mathbb{K}}$ ?
I don't understand the one in my book...
I have an idea but it does not work the thing out completely.
Let $E$ be a finite-dimensional vector space, therefore isomorphic to $\mathbb{R}^n$. We want to prove that for any $x,y\in E-{0_E}$, there exists $f\in GL_n(\mathbb{K})$ such that $f(x)=y$.
If all coordinates of $x$ are different from $0$ in a certain base $(e_1,...,e_n)$, then $x=\sum_{i=1}^{n}x_ie_i$, $y=\sum_{i=1}^{n}y_ie_i$, and $f$ such that$$\forall i\in [1,n], f(e_i)=\frac{y_i}{x_i}e_i$$ does the job : $$f(x)=f(\sum_{i=1}^{n}x_ie_i)=\sum_{i=1}^{n}x_if(e_i)=y$$
The matrix of $f$ in the said base is diagonal, with diagonal elements equal to $\frac {y_i}{x_i}$ therefore $f\in GL_n(\mathbb{K})$.
Now what if some of the coordinates of $x$ are equal to zero ? I mean, is it obvious that there is a base in which the coordinates of $x$ are all different from $0$ provided $x$ is different from $\vec0$ ?
Thanks in advance,
 A: The simplest proof is that $GL_n(K)$ acts by changing bases and any vector can be taken as the first vector in a basis.
For something a little more elementary, let $A_x$ be a matrix whose first column is $x$ and whose remaining columns are whatever you want so long as $A_x$ ends up invertible (it should be clear to you that you can always choose the remaining columns to make this so).
Then $A_xe_1 = x$ and $A_ye_1 = y$ so $A_yA_x^{-1}x = y$.
As to your more specific question: Yes it's true that for any vector there is a basis such that it's coordinates in that basis are all non-zero.  This can be seen using a change of basis argument, but I don't think it's the clearest way to think about this problem.
A: Fix a basis $\mathcal{B} = \{v_1, \ldots, v_n\}$ for the vector space $E$.  Let
$$
[x]_{\mathcal{B}} = \begin{bmatrix} x_1 \\ \vdots \\ x_1 \end{bmatrix} \in \Bbb{K}^n
$$
be the coordinate representation.  In other words,
$$
x = \sum_{i = 1}^{n} x_i v_i \in E.
$$
Now choose a basis $\{[w_2], \ldots, [w_n]\}$ for $E / \Bbb{K} x$.  Lift each $[w_i] = w_i + \Bbb{K}x$ to a representative $w_i \in E$.
Now construct the matrix
$$
A_{\mathcal{B}}(x) = \begin{bmatrix}
\begin{bmatrix} x \end{bmatrix}_{\mathcal{B}}, \begin{bmatrix} w_2 \end{bmatrix}_{\mathcal{B}}, \cdots, \begin{bmatrix} w_i \end{bmatrix}_{\mathcal{B}}
\end{bmatrix} \in M_n(\Bbb{K}). 
$$
By construction, $A_{\mathcal{B}}(x) \in GL_n(\Bbb{K})$ and $A_{\mathcal{B}}(x) v_1 = x$.
Now, for any $y \in E$, you can repeat the construction to form $A_{\mathcal{B}}(y) \in GL_n(\Bbb{K})$, such that $A_{\mathcal{B}}(y) v_1 = y$.  Now,
$$
A = A_{\mathcal{B}}(y) A_{\mathcal{B}}(x)^{-1}
$$
does the trick:
$$
Ax = y.
$$
