Parametrization of real-valued SU(N) I want to construct a $SU(N)$ matrix $V$, with the following property:


*

*All the elements of the first row are given, i.e. $V_{1,j}=a_i$ (with $\sum_i a_i^2=1$)

*All matrix elements are real, i.e. $V_{i,j} \in \mathbb{R}$

How can I find a matrix $V$ that satifies the criteria? Specifically, how can I find the matrix elements as a function of $a_i$, i.e. $V_{i,j}(a_i)$?


Special case: SU(2)
$$
   V=
  \left[ {\begin{array}{cc}
   a_1 & a_2 \\
   V_{2,1} & V_{2,2} \\
  \end{array} } \right]
$$
We easily find $V_{2,1}=-a_2$ and $V_{2,2}=a_1$.

Special case: SU(3)
$$
   V=
  \left[ {\begin{array}{cc}
   a_1 & a_2 & a_3 \\
   V_{2,1} & V_{2,2} & V_{2,3} \\
   V_{3,1} & V_{3,2} & V_{3,3} \\
  \end{array} } \right]
$$
Here already I cannot find any feasible way to represent $V_{i,j}$ as a function of $a_1, a_2, a_3$. I have tried to use the generators of SU(3), the Gell-Mann matrices $\lambda_i$. In particular, $\lambda_2$, $\lambda_5$, $\lambda_7$ are the generators for real-valued SU(3) matrices. However, the resulting equation system involves multiple trigonometric functions for which I cannot solve $V_{i,j}(a_1, a_2, a_3)$.
The matrix $V$ is not unique, I just want any solution.
 A: Since you state that already the $3\times 3$ case would be very useful to you:
$$
  V=\left( \begin{array}{ccc}
  a_1 & a_2 & a_3 \newline
  -\frac{a_2 }{\sqrt{a_1^2 +a_2^2 } } & \frac{a_1 }{\sqrt{a_1^2 +a_2^2 } } & 0 \newline
   -\frac{a_1 a_3 }{\sqrt{a_1^2 +a_2^2 } } & -\frac{a_2 a_3 }{\sqrt{a_1^2 +a_2^2 } } & \sqrt{a_1^2 +a_2^2 }
   \end{array} \right)
$$
A: In addition to the comments I made above about continuous solutions, I thought I'd point out a solution that works for all $n$ with only one point of discontinuity, namely
$$
(a_1\ a_2\ \ldots\ a_n) = (1\ 0\ \ldots\ 0).
$$
Away from this point, one can start with the following formulae:
$$
V_{i,1}= V_{1,i} = a_i\qquad\text{and}\qquad
V_{i,j} = \delta_{ij} - \frac{a_ia_j}{(1-a_1)}\quad\text{when}\ 1<i,j\le n
$$
Note that the resulting matrix is both orthgonal and symmetric.  However, it has determinant $-1$, so reversing a single row, say, the last one, will give a matrix in $\mathrm{SO}(n)$.
A: One way to formulate your problem is via Stiefel bundles. 
In your case there is the bundle $SU(n) \to S^{2n-1}$ given by taking the first row vector of a special unitary matrix.   You are asking for a way to reverse the process, i.e. if you have a unit row vector you want to complete it to not just a Hermitian basis but one that has determinant one. 
The idea is to consider complex mirror reflections $M_p$ in the complex hyperplane orthogonal to vectors $p \in S^{2n-1}$.  Let $q \in S^{2n-1}$ and let $p \in H_q S^{2n-1}$, this  is meant to indicate the hemi-sphere of $S^{2n-1}$ centred at the point $q$.  Then the composite
$$M_q \circ M_p$$
should be essentially the hermitian matrix you are looking for.  Likely it will only be giving you the first column vector the one you want (depending on how you think about matrices, i.e. perhaps you will need to take the transpose).  That vector will be the vector twice as far from p as q is, i.e. take the great circle from p to q, and go twice as far.  That's why there is the discontinuity at the point antipodal to p.  
Does this sound reasonable?  Hmm, on second thought, I'm running a bit on autopilot here.  I am not certain if you get every vector in $S^{2n-1}$ via this construction.   In the orthogonal group $O_n \to S^{n-1}$ this construction works fine.  I'll see if this construction can be fixed for $SU(n)$.  But I might need to sleep on it. 
