Following screenshot is of a proof from the book Classical groups and geometric algebra by Larry C. Groove. I have some doubt in the proof. My question:
How $\tau$ becomes a transvection on $V$?
To show $\tau$ is a transvection with fixed hyperplane $W$ we must show $\tau(z)-z \in W$ for all $z\in V$. Now an arbitrary element $z\in V$ can be expressed as $z=w+ax$ for some $w\in W$ and $a\in F$. But then $\tau(z)-z=w+ay-w-ax=a(y-x)$. Now if $(y-x)\in W$, then $y\in W$, which is a contradiction. Am I doing any mistake? Help me. Thanks.
I am writing a tentative solution of my own question. To begin with we note that $v=x+y$ for some $x\in W_1$ and $y\in W_2$. Since $v \notin W_1 \cup W_2$, it must be the case that $x\notin W_2$ and $y\notin W_1$. Thus $\{x,y\}$ is linearly independent set in $V$, so is $\{x,-y\}$. We already have $W_1=W_1\cap W_2 +\langle x \rangle$ and $W_2=W_1 \cap W_2 + \langle y \rangle$ and we define $W=W_1\cap W_2 +\langle v \rangle$. Thus $W$ is a hyperplane in $V$, moreover both $x,y$ lie outside $W$. Then we define a linear operator $\tau$ on $V$ such that $\tau(w)=w$, if $w\in W$ and $\tau(x)=-y$. First note that $\tau\in GL(V)$. Also any element $z\in V$ is of the form $z=w+ax$ for some $w\in W$ and $a\in F$, where $F$ is the ambient field. Thus $\tau(z)-z=w-ay-w-ax=-av\in W$, hence $\tau$ is a transvection on $V$. We also note that $\tau(W_1)=\tau(W_1\cap W_2 + \langle x \rangle)\subseteq W_2$, and hence by the dimension consideration we have $\tau(W_1)=W_2$. Thus we finally obtain a transvection $\tau$ on $V$ which sends $W_1$ to $W_2$ and fixes $v$.