If $x,y \in V$ are linearely independent, then there exists a transvection $\tau$ with $\tau(x)=y$ Let $V$ be a $n$-dimensional $K$ vector space. A $\tau \in \operatorname{GL}(V)$ is called a transvection, if there exists a $(n-1)$-dimensional $\tau$-invariant subvector space $W$ of $V$ with 
$$\tau_{W} = id_W \text{ and } \tau(v)-v \in W \quad \forall v \in W$$
Task: 
Prove: If $x,y \in V$ are linearly independent, then there exists a transvection $\tau$ with $\tau(x)=y$.
All the mappings of $\tau$ that I can find either do not satisfy $\tau(x)=y$ or $\tau(v)-v \in W $.
What I have so far: 
$x,y$ are linearly independent 
$\Rightarrow \operatorname{V} \geq 2$
$\Rightarrow \text{because} W=<y> \Rightarrow \operatorname{dim}(V) > dim(W) > 0$
$\Rightarrow \text{so } W \text{exists and thus } \tau$.
I have found a similiar post on that task.
But it seems not to be the answer to my task. If yes, 
please help me clarifying that answer. 

Since $\{ x, y \}$ are linearly independent, we also have that $\{ x - y, x \}$ are linearly independent. Complete this set to a basis
$$ w_1 = x - y, w_2, \dots, w_{n-1}, w_n = x $$
for $V$ and set $W = \operatorname{span} \{ w_1, \dots, w_{n-1} \}$. Define a linear map $\tau \colon V \rightarrow V$ by requiring that
$$ \tau(w_i) = w_i, \,\,\, 1 \leq i \leq n - 1, \\
\tau(w_n) = y. $$
Then clearly $\tau|_{W} = \operatorname{id}|_{W}$ and $\tau(w_n) = \tau(x) = y$. Finally, any $v \in V$ can be written as $v = w + ax$ for some $w \in W$ and $a \in K$ and then
$$ \tau(v) - v = \tau(w + ax) - (w + ax) = \tau(w) + a\tau(x) - w - ax = w + ay - w - ax = (-a)(x - y) \in W. $$
