Let \(V\) be a finite-dimensional vector space over the field \(F\) and let \(W\) be a subspace of \(V\). If \(f\) is a linear functional on \(W\), prove that there is a linear functional \(g\) on \(V\) such that \(g(\alpha) = f(\alpha)\) for each \(\alpha\) in the subspace \(W\).
Solution
Choose a basis \(\{\alpha_1,\ldots,\alpha_r\}\) of \(W\) and extend it to a basis \[ \{\alpha_1,\ldots,\alpha_r,\alpha_{r+1},\ldots,\alpha_n\} \] of \(V\). Define \(g\) on this basis by \[ g(\alpha_i)=f(\alpha_i)\quad(1\le i\le r), \qquad g(\alpha_i)=0\quad(r<i\le n), \] and extend linearly to \(V\).
If \(w\in W\), then \(w=\sum_{i=1}^r c_i\alpha_i\), and hence \[ g(w)=\sum_{i=1}^r c_i g(\alpha_i) =\sum_{i=1}^r c_i f(\alpha_i)=f(w). \] Thus \(g\) is a linear functional on \(V\) extending \(f\).