Suppose \(V = W_1 \oplus W_2\) and that \(f_1\) and \(f_2\) are inner products on \(W_1\) and \(W_2\), respectively. Show that there is a unique inner product \(f\) on \(V\) such that
\(W_2 = W_1^{\perp}\);
\(f(\alpha, \beta) = f_k(\alpha, \beta)\), when \(\alpha, \beta\) are in \(W_k\), \(k = 1, 2\).
Solution
Goal. For \(V = W_1 \oplus W_2\) with inner products \(f_1, f_2\) on \(W_1, W_2\), construct the unique inner product \(f\) on \(V\) with \(W_2 = W_1^\perp\) and \(f|_{W_k} = f_k\).
1. Define \(f\) on \(V\).
1.1. Every \(v \in V\) writes uniquely as \(v = w_1 + w_2\) with \(w_k \in W_k\).
\(V = W_1 \oplus W_2\) is a direct sum.
1.2. Define \(f(w_1 + w_2, w_1' + w_2') \definedas f_1(w_1, w_1') + f_2(w_2, w_2')\).
this is the orthogonal direct sum of the two inner products.
2. \(f\) is an inner product.
2.1. \(f\) is bilinear and symmetric.
each \(f_k\) is bilinear and symmetric, and the sum is componentwise.
2.2. \(f\) is positive definite.
\(f(v, v) = f_1(w_1, w_1) + f_2(w_2, w_2) \ge 0\), and equality forces \(f_1(w_1, w_1) = f_2(w_2, w_2) = 0\), hence \(w_1 = w_2 = 0\) by positive definiteness of each \(f_k\).
3. \(W_2 = W_1^\perp\).
3.1. \(W_2 \subseteq W_1^\perp\).
for \(w_1 \in W_1\) and \(w_2 \in W_2\), \(f(w_1, w_2) = f_1(w_1, 0) + f_2(0, w_2) = 0\).
3.2. \(W_1^\perp \subseteq W_2\).
if \(v = w_1 + w_2 \in W_1^\perp\), then for all \(w_1' \in W_1\), \(0 = f(v, w_1') = f_1(w_1, w_1')\), so \(w_1 = 0\) (since \(f_1\) is nondegenerate), hence \(v = w_2 \in W_2\).
4. \(f|_{W_k} = f_k\).
for \(\alpha, \beta \in W_1\), \(f(\alpha, \beta) = f_1(\alpha, \beta) + f_2(0, 0) = f_1(\alpha, \beta)\); similarly for \(W_2\).
5. Uniqueness.
5.1. Any inner product satisfying (a) and (b) must equal \(f\).
- fixes \(f\) on each \(W_k\); (a) forces \(f(w_1, w_2) = 0\) for \(w_1 \in W_1, w_2 \in W_2\); bilinearity then determines \(f\) on all of \(V = W_1 \oplus W_2\).
6. Q.E.D.
2, 3, 4 give existence; 5 gives uniqueness.