Isometry of bilinear spaces This is Corollary 1.3.10 from Scharlau's book Quadratic and Hermitian Forms.

3.10 Corollary Two bilinear spaces are isometric if and only if they have the same dimension and the same regular components.
Proof. If $(V,b)$ and $(V,b_1)$ satisfy the hypotheses and if $$V=V^\bot\oplus W, \qquad V_1=V_1^\bot\oplus W_1$$
  then $\alpha\oplus\beta\colon V\to V_1$ is an isometry where $\alpha\colon V^\bot\to V_1^\bot$ is an arbitrary isomorphism and $\beta\colon W\to W_1$ is an isometry.

I don't know how to show explicitly that direct sum satisfy isometry of bilinear space please help it out.
 A: Recall that $V^\bot=\{x\in V; (\forall v\in V) b(x,v)=0\}$.
Any vector $v\in V$ can be uniquely written as $v=v_0+w_0$ where $v_0\in V^\bot$ and $w_0\in W$. Let us denote $\gamma=\alpha\oplus\beta$. We have
\begin{align*}
b_1(\gamma(v), \gamma(v')) 
&\overset{(1)}= b_1(\alpha(v_0)+\beta(w_0),\alpha(v'_0)+\beta(w'_0)) \\
&\overset{(2)}= b_1(\alpha(v_0),\alpha(v'_0))+b_1(\alpha(v_0),\beta(w'_0))+b_1(\beta(w_0),\alpha(v'_0))+b_1(\beta(w_0),\beta(w'_0)) \\
&\overset{(3)}= b_1(\beta(w_0),\beta(w'_0)) \\
&\overset{(4)}= b(w_0,w'_0) \\
&\overset{(5)}= b(v_0,v'_0)+b(v_0,w'_0)+b(w_0,v'_0)+b(w_0,w'_0) \\
&\overset{(6)}= b(v_0+w_0,v'_0+w'_0) \\
&\overset{(7)}= b(v,v')
\end{align*}
In steps $(2)$, $(6)$ we have simply used bilinearity.
Step $(4)$ is the fact that $\beta$ is isometry.
In step $(3)$ we used the fact that $\alpha(v_0),\alpha(v'_0)\in V_1^\bot$ which means that $b_1$ vanishes whenever these vectors are plugged into $b_1$.
Step $(5)$ is similar, just that we are using $b$ rather than $b_1$ and the fact that $v_0,v'_0\in V^\bot$.
