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

I will show that norm on $V$ is uniquely determined. And since this norm satisfy parallelogram law then it defines inner product.
Let $v\in V$ be written as $v = w_1 + w_2$, where $w_i \in W_i$
Then $$\|v\|^2 = \langle v,v\rangle_V = \langle w_1,w_1\rangle_{W_1} + \langle w_2,w_2\rangle_{W_2} + \langle w_1,w_2\rangle_V +  \langle w_2,w_1\rangle_V$$
Because we require $\langle w_1,w_2\rangle_V = 0$, then 
$$\|v\|^2 = \langle v,v\rangle_V = \langle w_1,w_1\rangle_{W_1} + \langle w_2,w_2\rangle_{W_2} = \|w_1\|^2_{W_1} + \|w_2\|^2_{W_2}$$
So $\|\cdot\|_V$ is determined uniquely by $\|\cdot\|_{W_1}$ and $\|\cdot\|_{W_2}$.
