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 $V^* = V_1^0 \oplus \cdots \oplus V_k^0$.
My attempt: By theorem 16 section 3.5, $\dim(V_i^0)=\dim(V)-\dim(V_i)$. Since $V=V_i+W_i$, we have $\dim(V)=\dim(V_i)+\dim (W_i)-\dim (V_i\cap W_i$), by theorem 6 section 2.3. Since $W_1,…,W_k$ are independent, we have $V_i\cap W_i=\{0\}$ and $\dim(V_i\cap W_i)=0$. So $\dim(V)-\dim (V_i)=\dim(W_i)$. Since $V=W_1\oplus …\oplus W_k$, we have $\dim (V)=\sum_{i=1}^k\dim(W_i)$. Thus $$\sum_{i=1}^k\dim(V_i^0)=\sum_{i=1}^k\dim (V)-\dim (V_i)=\sum_{i=1}^k\dim(W_i)=\dim(V)=\dim(V^*).$$ Hence $\dim(V^*)=\sum_{i=1}^k\dim(V_i^0)$. We show $V^*=V_1^0+…+V_k^0$. Let $f\in V^*$. Let $B_i=\{\alpha_{i1},…,\alpha_{in_i}\}$ be basis of $W_i$. Since $V=W_1\oplus …\oplus W_k$, we have $B=\bigcup_{i=1}^kB_i$ is basis of $V$. $\forall i\in J_k$, define $f_i\in L(V,F)$ such that $f_i(\alpha_{pq})=f(\alpha_{pq})$, if $p=i$ and $f_i(\alpha_{pq})=0$, if $p\neq i$. It’s easy to check $f_i(V_i)=\{0\}$, i.e. $f_i\in V_i^0$. Let $\alpha \in V$. Then $\exists \alpha_i\in W_i$ such that $\alpha =\alpha_1+…+\alpha_k$. Since $\alpha_j\in W_j=\text{span}(B_j)$, we have $\alpha_j=\sum_{i=1}^{n_j}c_{ji}\cdot \alpha_{ji}$. So $f(\alpha_j)=\sum_{i=1}^{n_j}c_{ji}\cdot f(\alpha_{ji})= \sum_{i=1}^{n_j}c_{ji}\cdot f_j(\alpha_{ji})=f_j(\alpha_j)$. Thus $$f(\alpha)=f(\alpha_1)+…+f(\alpha_k)=f_1(\alpha_1)+…+f_k(\alpha_k)=f_1(\alpha)+…+f_k(\alpha)=f_1+…+f_k(\alpha).$$ So $f=f_1+…+f_k$. Hence $V^*=V_1^0+…+V_k^0$. By exercise 2 section 6.6, $V^*= V_1^0\oplus …\oplus V_k^0$. Is my proof correct?
Here and here is extremely clever way to show $V_1^0,…,V_k^0$ are independent and $V^*= V_1^0+…+V_k^0$, respectively.
