Use Theorem 20 to prove the following. If \(W\) is a subspace of a finite-dimensional vector space \(V\) and if \(\{g_1, \ldots, g_r\}\) is any basis for \(W^0\), then \[ W = \bigcap_{i=1}^{r} N_{g_i}. \]
Solution
Let \[ K=\bigcap_{i=1}^r N_{g_i}. \] Every \(w\in W\) is annihilated by every \(g_i\in W^0\), so \(W\subseteq K\).
Conversely, if \(v\in K\), then \(g_i(v)=0\) for every basis vector \(g_i\) of \(W^0\). Hence every \(g\in W^0\) satisfies \(g(v)=0\), because \(g\) is a linear combination of the \(g_i\). Thus \[ v\in (W^0)^0. \] By the finite-dimensional double-annihilator theorem, \[ (W^0)^0=W. \] Therefore \(K\subseteq W\), and hence \[ W=\bigcap_{i=1}^r N_{g_i}. \]