Let \(V\) be a vector space, let \(W_1, \ldots, W_k\) be subspaces of \(V\), and let \[ V_j = W_1 + \cdots + W_{j-1} + W_{j+1} + \cdots + W_k. \] Suppose that \(V = W_1 \oplus \cdots \oplus W_k\). Prove that the dual space \(V^*\) has the direct-sum decomposition \(V^* = V_1^0 \oplus \cdots \oplus V_k^0\).
Solution
For each \(j\), let \[ E_j:V\to W_j \] be the projection associated to the direct-sum decomposition \(V=W_1\oplus\cdots\oplus W_k\); thus if \(v=w_1+\cdots+w_k\) with \(w_i\in W_i\), then \(E_jv=w_j\) [@HK71].
1. Every functional on \(V\) is a sum of elements of the annihilators \(V_j^0\).
Let \(f\in V^*\) and define \[ f_j=f\circ E_j. \] If \(v\in V_j=\sum_{i\ne j}W_i\), then its \(W_j\)-component is zero, so \(E_jv=0\). Hence \(f_j(v)=0\), and therefore \[ f_j\in V_j^0. \] Also \[ E_1+\cdots+E_k=I, \] so for every \(v\in V\), \[ (f_1+\cdots+f_k)(v) =f((E_1+\cdots+E_k)v)=f(v). \] Thus \[ V^*=V_1^0+\cdots+V_k^0. \]
2. The sum is direct.
Suppose \[ g_1+\cdots+g_k=0, \qquad g_j\in V_j^0. \] Fix \(i\) and take \(w\in W_i\). For \(j\ne i\), one has \(W_i\subseteq V_j\), so \(g_j(w)=0\). Evaluating the displayed zero sum at \(w\) therefore gives \[ g_i(w)=0. \] Hence \(g_i\) vanishes on \(W_i\). By definition it also vanishes on \(V_i\), and \[ V=W_i\oplus V_i. \] Therefore \(g_i=0\) on all of \(V\). Since \(i\) was arbitrary, every \(g_i=0\). Consequently the sum is direct, and \[ \boxed{V^*=V_1^0\oplus\cdots\oplus V_k^0}. \]