Construct $n \times n$ special orthogonal matrices, all the entries of the last row equalling $\frac{1}{\sqrt{n}}$

Question

For $n=l+1$, pursuant to the instruction immediately following eq. (23) in https://arxiv.org/abs/1607.03364 I desire "$Q\in \mathrm{SO}(l+1)$ with elements in the last row being $Q_{(l+1)j} = \sqrt{p_j}$; $p_j\geq 0$, and $\sum_{j=1}^{l+1} p_j = 1$."

Is there a general construction available? It would seem that such $Q$'s would certainly not be unique.

Given--as angryavian emphasizes in his answer--the nonuniqueness of the candidates, can one reasonably aim to construct matrices with "simple" characteristics, that might facilitate subsequent analyses? The simplest in nature would seem to be a matrix in which all entries have the same absolute value (Hadamard matrices being an example).

Answer 1

If the last row is given to you, then you can apply the Gram-Schmidt process to extend it to an orthonormal basis over $\mathbb{R}^{l+1}$, and let these be the remaining rows of your matrix. This gives you an orthogonal matrix $Q$. If its determinant is $-1$, then you can negate one of the rows to change its determinant to $1$. There is definitely not a unique $Q$.

Answer 2

Let $e_n=(0,\ldots,0,1)^T$. When $v$ is a unit vector such that $v$ and $e_n$ are linearly independent, we can construct the Householder reflection matrix $H=I-2uu^T$ where $u=\frac{v-e_n}{\|v-e_n\|}$. It is well-known that $H$ is a real orthogonal matrix whose determinant is $-1$ and whose last row is $v^T$.

In your case, since $v=(\sqrt{p_1},\ldots,\sqrt{p_n})^T$ is an entrywise nonnegative unit vector, $v$ and $e_n$ are linearly independent if and only if $p_n\ne1$. Therefore, you may take $$ Q= \begin{cases} I&\text{if } p_n=1,\\ \operatorname{diag}(-1,1,\ldots,1)H&\text{if } p_n\ne1. \end{cases} $$