Let \(V\) be a complex vector space. A function \(J\) from \(V\) into \(V\) is called a conjugation if \(J(\alpha + \beta) = J(\alpha) + J(\beta)\), \(J(c\alpha) = \overline{c}J(\alpha)\), and \(J(J(\alpha)) = \alpha\), for all scalars \(c\) and all \(\alpha, \beta\) in \(V\). If \(J\) is a conjugation show that:
The set \(W\) of all \(\alpha\) in \(V\) such that \(J\alpha = \alpha\) is a vector space over \(\RR\) with respect to the operations defined in \(V\).
For each \(\alpha\) in \(V\) there exist unique vectors \(\beta, \gamma\) in \(W\) such that \(\alpha = \beta + i\gamma\).
Solution
Let \[ W=\{\alpha\in V:J\alpha=\alpha\}. \]
1. The fixed-point set \(W\) is a real vector subspace of \(V\).
Since \(J(0)=0\), one has \(0\in W\). If \(\alpha,\beta\in W\), then \[ J(\alpha+\beta)=J\alpha+J\beta=\alpha+\beta, \] so \(\alpha+\beta\in W\).
If \(r\in\mathbb R\) and \(\alpha\in W\), then \(\overline r=r\), hence \[ J(r\alpha)=\overline r\,J\alpha=r\alpha. \] Thus \(r\alpha\in W\). Therefore \(W\) is a vector space over \(\mathbb R\).
2. Every vector has a decomposition into fixed real and imaginary parts.
For \(\alpha\in V\), define \[ \beta=\frac{\alpha+J\alpha}{2}, \qquad \gamma=\frac{\alpha-J\alpha}{2i}. \] Using conjugate linearity and \(J^2=I\), \[ J\beta =\frac{J\alpha+\alpha}{2} =\beta. \] Also, because \(\overline{1/(2i)}=-1/(2i)\), \[ J\gamma =-\frac{J\alpha-\alpha}{2i} =\frac{\alpha-J\alpha}{2i} =\gamma. \] Thus \(\beta,\gamma\in W\). Directly, \[ \beta+i\gamma =\frac{\alpha+J\alpha}{2} +\frac{\alpha-J\alpha}{2} =\alpha. \]
3. The decomposition is unique.
Suppose \[ \alpha=\beta+i\gamma =\beta'+i\gamma', \qquad \beta,\gamma,\beta',\gamma'\in W. \] Applying \(J\) gives \[ J\alpha=\beta-i\gamma=\beta'-i\gamma'. \] Adding this equality to the original one gives \(2\beta=2\beta'\), hence \(\beta=\beta'\). Subtracting gives \(2i\gamma=2i\gamma'\), hence \(\gamma=\gamma'\). Therefore the representation \[ \boxed{\alpha=\beta+i\gamma,\qquad \beta,\gamma\in W} \] is unique [@HK71].