Let $V$ be a vector space. For any set $S\subset V$, we define $S^0=\{f\in V^\ast\,:\, f(v)=0,\forall v\in S\}$. If $\dim V<\infty$ and $W_1,W_2\subset V$ are vector subspaces, then show that $W_1^0+W_2^0=(W_1\cap W_2)^0$.
I was able to show that $W_1^0+W_2^0\subset (W_1\cap W_2)^0$: Let $f\in W_1^0+W_2^0$. Then there are $f_i\in W_i^0$, $i=1,2$ so that $f=f_1+f_2$. If $v\in W_1\cap W_2$, then $f(v)=f_1(v)+f_2(v)=0$, since $v\in W_1$, $f_1\in W_1^0$ and $v\in W_2$, $f_2\in W_2^0$. Therefore $f\in (W_1\cap W_2)^0$.
I don't have any idea of how to do the converse inclusion!
Let $H = W_1 \cap W_2$. There exist $U_1$ subspace of $W_1$, and $U_2$ subspace of $W_2$ such that \begin{align} W_1 &= H \oplus U_1 \\ W_2 &= H \oplus U_2 \end{align} Note that $U_1 \cap U_2 = \{ 0 \}$, so that $U_1 + U_2 = U_1 \oplus U_2$. Let $a$ (resp. $b$) be a basis of $U_1$ (resp. of $U_2$), and $a^*$ (resp. $b^*$) be the corresponding dual basis in $U_1^*$ (resp. $U_2^*$). For all $\alpha \in a$ and all $\beta \in b$, we have \begin{equation} \alpha^* (\beta) = 0 = \beta^* (\alpha) \end{equation}
Let $h \in H^0$. Writing $0_a$ (resp. $0_b$) for the null vector of $U_1$ (resp. of $U_2$), we have \begin{equation} h = (h + {0_b}^*) + {0_a}^* \in {W_1}^0 + {W_2}^0 \end{equation}