# 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,

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.

• "it should be clear to you that you can always choose the remaining columns to make this so" is exactly the statement "any vector can be taken as the first vector in a basis". Nov 22, 2013 at 18:15
• Well, it's exactly that statement if you're comfortable with $GL_n$ representing base changes. If you're not then one of those statements is pretty abstract and potentially confusing, whereas the other is not abstract and potentially clearer.
– Jim
Nov 22, 2013 at 18:17
• "...any vector different from zero can be taken as the first vector in a basis." Nov 22, 2013 at 20:11

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: $$Ax = y.$$