Let \(V\) be \(C^2\), with the standard inner product. For which vectors \(\alpha\) in \(V\) is there a positive linear operator \(T\) such that \(\alpha = T\epsilon_1\)?
Solution
Write \[ \alpha=\binom ab\in\mathbb C^2. \] We claim that such a positive operator exists exactly when \[ \boxed{a\in\mathbb R_{>0},\qquad b\in\mathbb C\text{ arbitrary}.} \]
1. Positivity forces the first coordinate to be positive real.
Suppose \(T\) is positive and \(T\epsilon_1=\alpha\). Then \[ a=(T\epsilon_1\mid\epsilon_1). \] Since \(T\) is positive and \(\epsilon_1\ne0\), this scalar is strictly positive. In particular \(a\) is real and \(a>0\).
2. Every vector with positive real first coordinate occurs.
Assume \(a>0\) is real and \(b\in\mathbb C\). Choose a real number \(d\) satisfying \[ d>\frac{|b|^2}{a}, \] and define \[ A=\begin{pmatrix} a&\overline b\\ b&d \end{pmatrix}. \] This matrix is Hermitian, its first leading principal minor is \(a>0\), and \[ \det A=ad-|b|^2>0. \] Hence \(A\) is positive definite. Let \(T\) be the operator with standard matrix \(A\). Then \(T\) is positive and its first column is \[ A\epsilon_1=\binom ab=\alpha. \] Thus every such \(\alpha\) is realized.