Let $V$, a vector space over $\mathbb{F}$ and $W_1,W_2 \subseteq V$ such that $W_1 \oplus W_2 = V$. For $i=1,2$, let $\langle , \rangle $ on $W_i$. Prove that there is a unique inner product on $V$ such that:
- $W_2 = W^\perp_1$
- For $i=1,2$ and for all $v,u\in W_i$: $\langle v, u \rangle = \langle v, u \rangle _i$
Uniqueness:
Let $v = w_1 + w_2$ and $u = w_1' + w_2'$
$$\langle u, v \rangle_V = \langle w_1,w_1' \rangle + \langle w_1,w_2' \rangle + \langle w_2,w_1' \rangle + \langle w_2,w_2' \rangle = \langle w_1,w_1' \rangle + \langle w_2,w_2' \rangle $$
So the inner product is determined uniquely by $\langle, \rangle_i $ for $i=1,2$.
How do I show the existence of such inner product?
You can define $$ \langle p, q \rangle = \langle p_1, q_1 \rangle_1 + \langle p_2, q_2\rangle_2 $$ where $p = p_1 + p_2$ is the decomposition of $p$ into a sum of vectors in $W_1$ and $W_2$, and similarly for $q = q_1 + q_2$. Because it's a direct sum, this decomposition is unique, so this expression is well-defined.
And if $p \in W_1$ and $q \in W_2$, we get $\langle p, q \rangle = 0$, as needed for the orthogonality condition.