How should I build a SU(4) matirx with a C4 vector?

Question

I have a complex vector $S=[S_1,S_2,S_3,S_4]$ with $|S_1|^2+|S_2|^2+|S_3|^2+|S_4|^2=1$. My question is how to bulid a matix $C\in SU(4)$ while

\begin{equation}C= \left( \begin{array}{cccccc} S_1 & S_2 & S_3 & S_4 \\ *&*&*&*\\ *&*&*&*\\ *&*&*&*\\ \end{array} \right) \end{equation}

Many thanks in advance.

Answer 1

For convenience, I'll call your vector $z$ instead. So, we want to construct a unitary matrix whose determinant is $1$ and whose first row is $z^T$. If $z=(c,0,0,0)^T$ for some unit complex number $c$, then clearly the diagonal matrix $C=\operatorname{diag}(c,\bar{c},1,1)$ will do. Suppose $z$ is not a multiple of $(1,0,0,0)^T$. Then $C$ can be constructed easily in a few steps:

1. Let $u$ and $w$ be the two vectors such that $u_j=|z_j|$ and $w_j=\exp(i\arg z_j)$ for each $j$. Since $z$ is a unit vector, so is $u$.
2. Let $e_1=(1,0,0,0)^T$ and $H$ be the Householder reflection matrix that maps $e_1$ to $u$, i.e. let $H=I-2vv^T$, where $v=\frac{u-e_1}{\|u-e_1\|}$. It is well known that such a matrix $H$ is a real orthogonal matrix with determinant $-1$ and whose first row is $u^T$.
3. Let $c=-\prod_{j=1}^n\bar{w}_j$ (note the minus sign and the complex conjugate). Then the matrix $C=\operatorname{diag}(1,\ldots,1,c)\,H\,\operatorname{diag}(w)$ would satisfy the given requirements.