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. 

Choose a basis where the first to vectors are $x$ and $y$ then the transformation $\tau$ corresponding to the cyclic permutation matrix of order $2$: 
$$ 
\begin{pmatrix}
0 & 1 & 0 & \ldots & 0 \\
1 & 0 & 0 & \ldots & 0 \\
0 & 0 & 1 & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & 1 \\
\end{pmatrix}
$$
will do the trick. Note that $W$ is the kernel of $\tau - \tau^0$ which has rank $1$ since $
\begin{pmatrix}
-1 & 1 \\
1 & -1 \\
\end{pmatrix}
$ has rank $1$, so dim($W$) = $n-1$.
