A group $G$ acts $k$-transitive on some set $X$ if for every two $k$-tupels $(x_1, \ldots, x_k)$ and $(y_1, \ldots, y_k)$ there exists some $g \in G$ such that $$ g\cdot x_1 = y_1, \ldots, g\cdot x_k = y_k. $$ (if $k = 1$ we simple say $G$ acts transitively).
Let $K$ some field, and $V$ be a finite dimensinoal vector space over $K$. Then denote by $SL(V)$ the space of all linear transformations with determinant $1$, the so called special linear group, and further denote by $PSL(V)$ the projective special linear group
Now I am asked to prove that $SL(V)$ and $PSL(V)$ act $2$-transitively on every linear subspace of dimension $1$, but I think this could be sharpened, but I nowhere find this result so I am asking if I overlooked something. My claim is:
Proposition: Let $n = \dim(V)$, then if $n > 1$ the group $SL(V)$ as well as $PSL(V)$ act $n$-transitively on $\{ U \le V : \dim U = 1 \}$, and if $n = 1$ these groups acts $k$-transitively for every $k$.
Proof: i) Let $n > 1$ and let $U_1, \ldots, U_n$ be distinct one-dimensional subspaces of $V$, then they are generated by vectors $u_1, \ldots, u_n$ which are linear independent because these subspaces are distinct, further let $W_1, \ldots, W_n$ be another set of distinct lineary independent one-dimensional subspaces with linear independent generating vectors $w_1, \ldots, w_n$. Then define the linear map $A : V \to V$ by $$ A(u_i) = w_i $$ and set $\alpha := \det(A)$. By the linear independence it is $\alpha \ne 0$. Now define $$ A'(u_1) = \frac{1}{\alpha} w_1 \quad \mbox{and} \quad A'(u_i) = w_i, ~ i=2,\ldots, n. $$ Then if fix $u_1, \ldots, u_n$ as a basis for $V$, and consider the matrix $[A] = (a_{ij})$ with respect to that basis, by definition we have for $$ [A'] = \left\{ \begin{array}{ll} \frac{1}{\alpha} a_{ij} & i = 1 \\ a_{ij} & i \ne 1 \end{array}\right. $$ so that by the linearity of the determinant function in its rows we have $$ \det(A') = \frac{1}{\alpha} \det(A) = \frac{\det A}{\det A} = 1 $$ therefore $A' \in SL(V)$.
Further in the projective special linear space, two $n$ distinct vectors correspond two $n$ distinct linear $1$-dimensional subspaces, and by the above $SL(V)$ act $n$-transitively on them, so $PSL(V)$ too.
For the case $n = 1$, all $1$-dimensional subspaces equal $V$, and so the identity map maps all $k$-tupel of linear subspace onto each other. $\square$
So is my proof correct, and why everywhere I find just the result that they act $2$-transitively?
The action will not be $n$ transitive, for instance, if $n\geq 3$ and if you take a $k$-uple, $k\geq 3$ of lines that are all contained in a $2$ dimensional subspace of $V$, then any translates by $SL(V)$ will have that same property.
However, if the ground field $\Bbbk$ is algebraically closed, $PSL(V)$ will act simply transitively on all circuits of lines, that is $(n+1)$-uples of lines such that any $n$-uple is linearly independent. This is because if $(\ell_1,\dots,\ell_{n},\ell_{n+1})$ is such a circuit, then taking a nonzero vector $x$ in $\ell_{n+1}$, it decomposes as $$x=x_1+\cdots+x_n$$ for some nonzero $x_i\in \ell_i$. If $(\ell_1^0,\dots,\ell_{n}^0,\ell_{n+1}^0)$ is a reference circuit (and $x^0=x_1^0+\cdots+x_n^0$ are defined similarly) then the linear automorphism $g$ defined by sending the $x_i^0$ to the $x_i$ (for $i=1,\dots,n$) will also send $x^0$ to $x$, and thus will have $g\cdot(\ell_1^0,\dots,\ell_{n}^0,\ell_{n+1}^0)=(\ell_1,\dots,\ell_{n},\ell_{n+1})$. To convert this $g$ to an element $g'$ of $SL(V)$ you consider $g'=\lambda g$ where $\lambda^n\det(g)=1$. Such a $\lambda$ always exists because $\Bbbk$ is assumed to be algebraically closed.
Simple transitivity is easy enough to prove, I'll just leave this hanging for now unless you're interested.