Prove the uniqueness of an inner product Question:
Let V be an inner product space s.t. $V=W_1 \oplus  W_2$. $W_1, W_2$ with the inner products $\langle ,\rangle_{W_i}$. Prove there exists a single inner space product $\langle ,\rangle_V$ s.t :
a. $W_2=W_1^\top$
b. if $u_1,u_2 \in W_i$ then $\langle u_1,u_2\rangle_V=\langle u_1,u_2\rangle_{W_i}$
Thoughts
Only thing I thought about is saying that 2 of these exist and showing they are the same one... don't really know how to do that though..
Thanx

Assume the inner product on $V$ exists. Then $W_2$ is the orthogonal complement of $W_1$; if $P_1$ and $P_2$ denote the orthogonal projections of $V$ onto $W_1$ and $W_2$ respectively, we have, for all $v,v'\in V$,
\begin{gather}
v=P_1(v)+P_2(v)\\
\langle v,v'\rangle_V=
\langle P_1(v),P_1(v')\rangle_V+\langle P_2(v),P_2(v')\rangle_V
\end{gather}
Property (b) then implies
$$
\langle v,v'\rangle_V=
\langle P_1(v),P_1(v')\rangle_{W_1}+\langle P_2(v),P_2(v')\rangle_{W_2}
$$
Now, what are $P_1(v)$ and $P_2(v)$? They're the unique vectors $w_1\in W_1$ and $w_2\in W_2$ such that $v=w_1+w_2$, because these vectors satisfy the property of being the orthogonal projections of $v$. Thus we can write
$$
\langle v,v'\rangle_V=
\langle w_1,w_1'\rangle_{W_1}+\langle w_2,w_2'\rangle_{W_2}
$$
where $v=w_1+w_2$, $v'=w_1'+w_2'$, $w_1,w_1'\in W_1$, $w_2,w_2'\in W_2$.
The verification that the last formula defines an inner product on $V$ is easy: just check the required properties.
