We know that corresponding to any skew-symmetric bilinear form $f$, there exists a basis with respect to which the matrix of $f$ will look like $$\begin{pmatrix}J& &\\ &J& \\ & & \ddots\\ & & &O\end{pmatrix}, \qquad J=\begin{pmatrix}0 &1\\-1& 0\end{pmatrix}. $$ This is called symplectic basis. Now my question is: given any skew-symmetric bilinear form, how can I find a symplectic basis?
Say for $f(x,y)=2(x_1y_2-x_2y_1)+3(x_1y_3-x_3y_1)+x_2y_3-x_3y_2$, how can I get hold of a symplectic basis? Can someone help me?
Addendum Will the basis $\{(1,0,0),(0,1/2,0),(1/2,-3/2,1)\}$ work?
There is a general algorithm for this. Take $f \colon V \times V \to \mathbb R$ a skew-symmetric form. Define $$\ker f := \{v \in V: f(v,{}\cdot{}) = 0\}.$$ This is a subspace of $V$, so we can now work on a complement of it. Say $V = \ker f \oplus U$. It might well be that $U=\{0\}$, but then $f$ is just the zero map, so we assume $U \neq \{0\}$.
Take $e_1 \neq 0$ to be an element in $U$. Since $f(e_1,{}\cdot{}) \neq 0$, there is $e_2$ in $U$ such that $f(e_1,e_2) = 1$ (up to rescaling). By skew-symmetry we have $f(e_2,e_1) = -1$ and $f(e_i,e_i) = 0$ for $i=1,2$. Call $U_{12}$ the subspace spanned by $e_1,e_2$. Now we have $V = \ker f \oplus U_1 \oplus U'$, with $U'$ complement of $U_1$ in $U$ defined as $U' = \{v \in U: f(v,u) = 0 \text{ for all } u \in U_1\}$.
Now take $0 \neq e_3 \in U'$ with $f(e_1,e_3)=f(e_2,e_3)=0$. Again, $f(e_3,{}\cdot{}) \neq 0$, so there is $\tilde{e}_4 \in U'$ such that $f(e_3,\tilde{e}_4) \neq 0$. Say $f(\tilde{e}_4,e_1) = a$ and $f(\tilde{e}_4,e_2) = b$, and set $$e_4 := \tilde{e}_4 -be_1+ae_2.$$ Then $f(e_4,e_1) = f(\tilde{e}_4-be_1+ae_2,e_1) = a-a=0$ and $f(e_4,e_2) = f(\tilde{e}_4-be_1+ae_2,e_2) = b-b=0$. Rescale $e_4$ so that $f(e_3,e_4)=1$. Now you have a two-dimensional subspace $U_{34}$ of $U'$ generated by $e_3,e_4$ with $f(e_3,e_4) = -f(e_4,e_3) = 1$ and $f(e_i,e_i) = 0$ for $i=3,4$, and $V = \ker f \oplus U_{12} \oplus U_{34} \oplus U''$.
Since $V$ is finite-dimensional you can go on like this and conclude in a finite number of steps.
Notice that $f$ restricted to $U_{12}$, etc., is of the form of $J$, and on $\ker f$ it is identically zero. In this way you get the basis wanted.
In your particular case, $f \colon \mathbb R^3 \times \mathbb R^3 \to \mathbb R$. Notice that $f$ is degenerate, because $f((1,-3,2),{}\cdot{}) = 0$. So $\ker f = \mathbb R(1,-3,2)$. Now take $e_1 \neq 0$ in a complement of $\ker f$, e.g. $e_1 = (3,1,0)$. Then $f(e_1,y) = 6y_2-2y_1+10y_3$. Now choose $e_2 = (0,1/6,0)$, so that $f(e_1,e_2) = 1$. Hence $(1,-3,2), (3,1,0), (0,1/6,0)$ is your basis.