Let \(W_1\) and \(W_2\) be subspaces of a finite-dimensional vector space \(V\).
Prove that \((W_1 + W_2)^0 = W_1^0 \cap W_2^0\).
Prove that \((W_1 \cap W_2)^0 = W_1^0 + W_2^0\).
Solution
For (a), a functional \(f\) annihilates \(W_1+W_2\) exactly when it annihilates every vector \(w_1+w_2\) with \(w_i\in W_i\). This is equivalent to annihilating both \(W_1\) and \(W_2\). Hence \[ (W_1+W_2)^0=W_1^0\cap W_2^0. \]
For (b), every \(f_1\in W_1^0\) and \(f_2\in W_2^0\) satisfies \[ (f_1+f_2)(w)=0 \] for every \(w\in W_1\cap W_2\), so \[ W_1^0+W_2^0\subseteq (W_1\cap W_2)^0. \] Let \(n=\dim V\) and \(d_i=\dim W_i\), \(d=\dim(W_1\cap W_2)\). By part (a), \[ \dim(W_1^0\cap W_2^0) =\dim (W_1+W_2)^0 =n-(d_1+d_2-d). \] Therefore \[ \dim(W_1^0+W_2^0) =(n-d_1)+(n-d_2)-[n-d_1-d_2+d] =n-d. \] But \(\dim (W_1\cap W_2)^0=n-d\). Since the former is contained in the latter and they have equal dimension, \[ (W_1\cap W_2)^0=W_1^0+W_2^0. \]