Inner product in a direct sum of a dimensional space Supposer that $V = W_{1} \oplus W_{2}$, $f_{1}$ and $f_{2}$ are inner product at $W_{1}$ and $W_{2}$, respectively. Show that there is only one inner product $f$ in $V$ such that
i) $W_{2} = W_{1}^{\bot}$;
ii)$f(\alpha,\beta) = f_{k}(\alpha,\beta)$ when $\alpha,\beta \in W_{k}$, $k = 1,2.$
my first impression is that this $f$ is some kind of a sum of $f_{1}$ and $f_{2}$ but how can i prove this?
 A: An inner product $f$ on $V$ is a map $f: V \times V \to \mathbb C$ that is linear in the first argument, conjugate-linear in the second argument, and positive-definite. (If you're working with inner products over $\mathbb R$, just replace "conjugate-linear" with "linear".)
So if $f$ is any inner product on $V$ satisfying i) and ii), then for each $v = (w_1',w_2') \in V$ the equation $T_v(w_1,w_2) = f((w_1,w_2),v)$ defines a linear map $T_v: W_1\oplus W_2 \to \mathbb C$ such that by ii) $T_v(w_1,0) = f_1(w_1,w_1')$ and $T_v(0,w_2) = f_2(w_2,w_2')$. Therefore $T_v(w_1,w_2) = T_v[(w_1,0) + (0,w_2)] = T_v(w_1,0) + T_v(0,w_2) = f_1(w_1,w_1') + f_2(w_2,w_2')$. In other words, you can show in this way that $f((w_1,w_2),(w_1',w_2')) = f_1(w_1,w_1') + f_2(w_2,w_2')$ for all $(w_1,w_2) \in V$.
Think about the above derivation. It more-or-less already shows you that the function $f((w_1,w_2),(w_1',w_2')) = f_1(w_1,w_1') + f_2(w_2,w_2')$ is the only candidate for an inner product on $V$ satisfying ii). You should verify that (a) $f$ is an inner product and (b) both i) and ii) are satisfied by $f$.
