The group action of special linear group of a vector space $V$ on $V/\{0\}$ is transitive

Question

I learn from wiki that the group action of special linear group of a vector space $V$ on $V/\{0\}$ is transitive. I don't know how to prove this.

Answer 1

Take $v,w \in V$. Sends $v$ to $w$, complete $v$ in a basis $B$ and $w$ in a basis $B'$. Send the remaining element of $B$ to $B'$ and you get your $A \in GL(V)$ such that $Av = w$.

In fact you have better : for any basis $B$,$B'$ there is a unique element $A \in GL(V)$ which sends $B$ to $B'$.

Edit : I did realize with the answer of Marc Bogaert that I didn't answer the question ! In fact, consider the same transformation as before, and let $\lambda$ be its determinant. Take the same transformation, and pick one $u \in B$, and send it on $1/\lambda f(u)$ (where $f$ was the old transformation) and don't change the other image. You can easily check that the linear transformation has determinant $1$. This shows you that $SL(V)$ acts transitively on any $n-1$-uple of linearly independant vectors, and this is the best you can have.

Answer 2

Let $e_1, e_2, \ldots, e_n$ be a basis for $V$. if $u$ and $v$ are two given nonzero vectors. Then let $i$ be such that the component $v_i \neq 0$ (this is possible since $v \neq 0$) and let $j$ be such that $u_j \neq 0$. Let $E_{i,j}$ be a linear transformation that maps $e_i$ to $e_j$ . Such a transformation exists and $\in SL(V)$ indeed its matrix is the identity matrix where the columns $i$ and $j$ have been switched (if necessary multiply with $-1$ to make the determinant $1$ ). Let $T_v \in SL(V)$ be a transformation that maps $e_i$ into $v$, you can take a transformation with matrix the identity matrix were the $i$-th column is replaced with the components of $v$, if we replace an element on the diagonal and in another column with $1/v_i$ then $T_v$ has determinant $1$. If we do the same for $u$ giving $T_u$ then the composition $T^{-1}E_{i,j}T_u$ maps $v$ in $u$ and has determinant $1$ .