Orbits of symplectic group $Sp(2n,K)$ acting on $K^{2n}$ Let $J_n$ be a matrix of standard non-degenerate symplectic form
$$J_n=\begin{bmatrix}
0 & -1 & 0 & 0 & \cdots & 0 & 0 \\
1 & 0 & 0 & 0 & \cdots & 0 & 0 \\
0 & 0 & 0 & -1 & \cdots & 0 & 0 \\
0 & 0 & 1 & 0 & \cdots & 0 & 0 \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & 0 & 0 & \cdots & 0 & -1 \\
0 & 0 & 0 & 0 & \cdots & 1 & 0 \\
\end{bmatrix}$$
With coeffitients in the field $K$. I define the symplectic group to be
$$Sp(2n,K)=\{A\in GL_{2n}(K) \ | \ A^TJ_nA=J_n\}$$
This group acts on vector space $K^{2n}$ by standard multiplication of matrix by vector $A \cdot v$
Question is to show that $Sp(2n,K)$ has 2 orbits: $K^{2n}\setminus\{0\}$ and $\{0\}$.
I tried to construct a matrix $A\in Sp(2n,K)$ that transforms vector $e_1=(1, 0, \cdots , 0)$ to nonzero vector $v=(x_1, x_2, \cdots, x_{2n})$ but managed to do it only for the case $n=1$. How do I prove this for any $n$?
 A: You've not really explained your context, but as a raw linear algebra question this is not immediate or trivial. For broader context, this is a very special case of Witt's theorem about extension of isometries from subspaces to the whole space, in non-degenerate alternating(symplectic), quadratic, and unitary spaces. The alternating/symplectic case is somewhat easier.
In any case, a too matrix-bound approach may have more baggage than we'd desire...
In your somewhat-more-limited case, an induction on the dimension can succeed. Given a vector $f_{2n}$, the non-degeneracy promises a vector $f_{2n-1}$ so that $\langle f_{2n-1},f_{2n}\rangle=1$. The "orthogonal complement" $W=\{v:\langle v,f_{2n-1}\rangle=\langle v,f_{2n}\rangle=0\}$ is a non-degenerate alternating space. By induction on $n$ (and, as you've done, the $n=1$ case can be done very explicitly), the automorphism group of that smaller orthogonal group is transitive on non-zero vectors...
There are various ways of composing the remainder of the argument, but the existence of the complementary vector $f_{2n-1}$ is a key...
