I am working on a problem to find the orbits of the general linear group $\mathrm{GL}_n(\mathbb{R})$, acting on $\mathbb{R}^n$, with the invertible matrix $A$ acting on a column vector $x \in \mathbb{R}^n$ by taking it to the vector $Ax$.

I have already verified that this is a group action, but I'm not sure how to show that there are only two orbits.

Thanks for your help.

  • $\begingroup$ The choice of the field in GL does not matter for the result. $\endgroup$
    – Aelx
    Jan 14 at 23:38

2 Answers 2


The two orbits are $\{0\}$ and $\mathbb R^n \setminus \{0\}$. It's pretty clear the first is an orbit, so for the second you have to prove that if $v, w \neq 0$ then there exists an $A$ such that $Av = w$. To do this I would suggest thinking of $A$ as a linear transformation of the vector space $\mathbb R^n$.


This is equivalent to Jim's perfectly good answer, but it helps to think in terms of linear algebra, and things work just as well over any field. Let $V$ be an $n$-dimensional vector space over a field $K$. The group ${\rm GL}(V)$ of invertible linear transformations from $V$ to $V$ is defined independently of choice of basis of $V$ but is isomorphic to ${\rm GL}(n,K)$ (the group of invertible $n \times n$ matrices with entries from $K$) (in many ways-just choose any basis for $V$ and replace an ivertible linear transformation $V \to V$ by its matrix with respect to the chosen basis). By linear algebra, a linear transformation from $V \to V$ is invertible if and only if it sends a basis for $V$ to another basis. On the other hand, given two bases for $V$, there is a unique invertible linear transformation from $V$ to $V$ which (for each $i$) sends the $i$-th vector in the first basis to the $i$-th vector in the second. Now take two non-zero vectors $v, w \in V$.Each extends to a basis of $V$, so there is an invertible linear transformation which sends the first basis to the second (in order), and in particular sends $v$ to $w$.


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