Action of automorphism group of $\mathbb{P}^3(k)$ Suppose you have 3 skew lines in $\mathbb{P}^3(k)$. Then it is known that you can find a projective transformation that puts them in basically any form you like, for example:
$L_1: x_0=x_1=0\quad \quad L_2:x_2=x_3=0\quad\quad  L_3: x_0=x_2 ,\:x_1=x_3$
I am trying to understand why. I can surely consider the two planes $U_1, U_2$ in $k^4$ that correspond to these lines. Once I put the first two planes in the form of $L_1$, $L_2$, I can consider the matrices that would send the subspace corresponding to $L_1$ to itself, the subspace corresponding to $L_2$ to itself and put the third plane in the $L_3$ form above. They would be $4 \times 4$ matrices of this form 
\begin{bmatrix}A&0\\0&B\end{bmatrix}
Where $A$ and $B$ are $2 \times 2$. Can anyone help? 
This is basically a follow-up question to a comment on this question
There is a unique quadric through three disjoint lines

Essentially this is just choosing the "right" basis.  I am going to abuse notation and write $L_i$ for the 2-planes in $k^4$.
Choose any basis $e_2,e_3$ of $L_1$.  Since $L_i\cap L_j=0$ we have $k^4=L_i\oplus L_j$ and hence $e_0=L_2\cap (e_2+L_3)$ and $e_1=L_2\cap(e_3+L_3)$ exist and satisfy $e_0,e_1\in L_2$ and $e_0-e_2,e_1-e_3\in L_3$ (alternatively, $e_0$ is the $L_2$ component of $e_2$ in the decomposition $k^4=L_2\oplus L_3$, similarly $e_1$).  Moreover, if $e_0,e_1$ are linearly dependent then some nontrivial linear combination of $e_2,e_3$ lies on $L_3$, contradicting $L_1\cap L_3=0$.
Now with respect to $e_0,e_1,e_2,e_3$ of $k^4$, the equation of $L_i$ takes the desired form.  Hence also on $\Bbb P^3$.
