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I am working with a matrix of the following form:

$$ A =\begin{pmatrix} a_{11} & Q & \ldots & Q\\ a_{21} & Q & \ldots & Q\\ \vdots & \vdots & \ddots & \vdots\\ a_{n1} & Q &\ldots & Q \end{pmatrix} $$

where the $a_{ij}$ elements are real and predetermined, the $Q$'s are placeholders and not necessarily equal to one another, and $A$ is square of size $n$x$n$. I am looking to find values for all $Q$'s such that $A$ is unitary. To do this I have attempted to set up a system of nonlinear equations of the form $AA^\dagger=I$ which yields a system of $n^2-n$ unknown $Q$'s, but only $n^2/2 +n/2$ equations after removing any duplicate equations. Therefore, for $n=2$ the system is over-determined and for $n>3$ the system is under-determined.

My question is, is there a method in which I can solve for $Q$'s to force $A$ to be unitary given these constraints for any size n?

The issue that I found with this method is that once I apply the Gram-Schmidt process my column of $a_{11} ... a_{n1}$ is no longer the same. How do I apply this procedure while keeping the first column constant? – thespaceman – 2020-10-16T01:42:54.137

1Ah.. so you mean the vector $u_1 = \begin{pmatrix} a_{11} \ a_{21} \ \vdots \ a_{n1} \end{pmatrix}$ has a normalization factor in front of it now, right? – KAJ226 – 2020-10-16T02:35:10.357

1It is anyway a necessary condition that your first column is normalised. Otherwise $A$ can never be unitary. Picking random linearly independent vectors for the rest and performing Gram-Schmidt seems the optimal strategy to achieve your goal. Alternatively, you can directly pick random vectors from the orthocomplement of the preceding columns. This will

notchange the first column. – Markus Heinrich – 2020-10-16T10:33:01.123The Gram-Schmidt method is exactly what I need when the first column is already normalized which is the case for me. – thespaceman – 2020-10-16T16:29:39.407