Creating a tight frame of $\mathbb{R}^{n}$ when already knowing some of its vectors. I'm wondering whether or not there's an optimal way for adding rows to a given matrix $S\in\mathbb{R}^{m\times mn}$, $m\leq n$, so that the columns of the resulting matrix form an orthogonal system of vectors of equal norm. 
I've been struggling with this problem for quite some time now and the reason is that I don't want to change my starting matrix $S$. 
This is equivalent to creating a tight frame of $\mathbb{R}^{mn+v}$, $v\geq 0$  when you already know some of its vectors. Up to this point, I have succeeded in creating a tight frame given any number of its vectors but I don't know how to minimize $v$.
 A: If the number $v$ of additional rows is not prespecified but chosen at will, it is always possible to append a square matrix to $S$ to form an $m(n+1)\times mn$ matrix $A$ that has mutually orthogonal columns of equal norms, although I am not sure if this is qualified as "optimal". Anyway, the construction is conceptually easy. Let $S=U(\Sigma,0)V^\top$ be a singular value decomposition. Take
$$
T = \pmatrix{\sqrt{\sigma_1^2I_m-\Sigma^2}\\ &\sigma_1I_{m(n-1)}}V^\top,
\quad A=\pmatrix{S\\ T}.
$$
Note that $\sigma_1I_m-\Sigma\succeq0$ has a real square root because it is positive semidefinite. Now
$$
A^\top A=S^\top S+T^\top T=V\pmatrix{\Sigma^2\\ &0}V^\top
+V\pmatrix{\sigma_1^2I_m-\Sigma^2\\ &\sigma_1^2I_{m(n-1)}}V^\top=\sigma_1^2I_{mn}.
$$
Hence the columns of $A$ are mutually orthogonal and each column has norm $\sigma_1$.
If $m+v$ is required to be equal to $mn$, i.e. if the completed matrix $A$ is required to be square, clearly the completion is possible if and only if $SS^T=\alpha I_m$ for some $\alpha>0$. If this condition is satisfied, you may simply fill in the other rows by Gram-Schmidt process.
