How to extract the vector from a rank-$1$ matrix?

Question

Given Hermitian and positive semidefinite rank-$1$ matrix ${\bf Z} \in \mathbb C^{N \times N}$, how to find vector $\mathbf z \in \mathbb C^{N \times 1}$ such that $\mathbf Z = \mathbf z \mathbf z^{H}$?

Answer 1

$z$ is an eigenvector of $Z$, i.e. $Z z = (z^H z) z = |z|^2 z$. However you can only find $z$ up to a complex multiplier of modulus $1$. So, $z e^{i \theta}$ also satisfies eigenvalue / eigenvector pair.

Answer 2

Such $z$ does not always exist. For example, if

$$ Z= \begin{bmatrix} 1 & 1 \\ 2 & 2\end{bmatrix} $$

and there exists $z=(z_1,z_2)^T$ such that $Z=zz^T$, then $1=z_1z_2=2$.

If $Z$ is an Hermitian semidefinite matrix, then such $z$ always exists. In fact, there exists an orthonormalized basis in which the matrix $Z$ has a real-diagonal form. Since the rank of $Z$ is $1$, then $$ S^{-1}ZS=\operatorname{diag}(t^2,0,\ldots,0),\ t\in\mathbb{R}. $$ Let $z'=(t,0,\ldots,0)^T$. Then $z=S^{-1}z'S$.