Let \(W\) be a finite-dimensional real subspace of a complex vector space \(V\). Show that \(W\) satisfies condition (b) of Exercise 14 if and only if every basis of \(W\) is also a basis of \(V\).
Solution
Condition (b) of Exercise 14 says that every \(\alpha\in V\) has a unique expression \[ \alpha=\beta+i\gamma, \qquad \beta,\gamma\in W. \]
1. If condition (b) holds, every real basis of \(W\) is a complex basis of \(V\).
Let \[ \mathcal B=(w_1,\ldots,w_m) \] be any basis of the finite-dimensional real vector space \(W\). Given \(\alpha\in V\), write uniquely \[ \alpha=\beta+i\gamma, \qquad \beta,\gamma\in W. \] Expanding \[ \beta=\sum_j a_jw_j, \qquad \gamma=\sum_j b_jw_j, \qquad a_j,b_j\in\mathbb R, \] gives \[ \alpha=\sum_j(a_j+ib_j)w_j. \] Thus \(\mathcal B\) spans \(V\) over \(\mathbb C\).
For complex linear independence, suppose \[ \sum_j(a_j+ib_j)w_j=0. \] Then \[ \sum_j a_jw_j+i\sum_jb_jw_j=0. \] By uniqueness in condition (b), both real components vanish. Since \(\mathcal B\) is real-linearly independent, all \(a_j\) and \(b_j\) are zero. Hence \(\mathcal B\) is a complex basis of \(V\).
2. If every real basis of \(W\) is a complex basis of \(V\), then condition (b) holds.
Choose a real basis \[ \mathcal B=(w_1,\ldots,w_m) \] of \(W\). By hypothesis it is also a complex basis of \(V\). Therefore every \(\alpha\in V\) has unique complex coordinates \[ \alpha=\sum_j z_jw_j, \qquad z_j=a_j+ib_j,\quad a_j,b_j\in\mathbb R. \] Set \[ \beta=\sum_j a_jw_j, \qquad \gamma=\sum_j b_jw_j. \] Then \(\beta,\gamma\in W\) and \[ \alpha=\beta+i\gamma. \]
If also \(\alpha=\beta'+i\gamma'\) with \(\beta',\gamma'\in W\), expand \(\beta',\gamma'\) in the same real basis. Equality of the resulting complex coordinates forces equality of all real and imaginary coefficients, so \(\beta'=\beta\) and \(\gamma'=\gamma\). Thus the decomposition is unique.
Hence \(W\) satisfies condition (b) if and only if every real basis of \(W\) is a complex basis of \(V\) [@HK71].