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I'm wondering whether this question is wrong...

It says show that $\operatorname{orb}(p)$ where $p=[1,1]$ is the set $\{[x,y]: xy\neq0\}$ but if we apply the matrix $\begin{bmatrix}1&0\\1&-1\end{bmatrix}$

we get the vector $[1,0]$ which has $xy=0$.

Is it me or the book who is wrong or am I missing the point completely?

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    The orbit of a nonzero vector under GL is the set of nonzero vectors. Maybe they meant $x^2+y^2 \neq 0$ or $x,y \neq 0$.2012-07-05

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You are right, the book is wrong. It should be : $\mathrm{orb}(p) = \mathbb{R}^2\backslash \{(0,0)\}$

Indeed, if $x$ is any non zero vector, you can find $A \in GL(\mathbb{R}^2)$ that sends $p$ to $x$ (this property is known as transitivity), and for any invertible $A$, $Ap \neq (0,0)$ since $p \neq (0,0)$.