A $n\times n$ matrix $A=[a\_{ij}]$ is called "good", if there exists some $k$ and a set of $k\times k$ complex unitaries $U\_i$, $1\leq i\leq n$ , such that $tr(U\_i^{+}U\_j)=ka\_{ij}$, where $U^{+}$ denotes the conjugate transposed matrix of $U$.


Let $S=${$A|A$ is good}, Is there any useful method to check if a given matrix is "good"? Or how to describe $S$ completely? 


In quantum information, maximally entangled states are widely concerned, which can be consider as a vector, and really connected to unitaries, the mother problem is: 


Describe the set of matrice $A$, where $A=P^{+}P$, with P be a $n\times k$ matrix whose columns are all maximally entangled states.


Is $S$ a closed set? Is it a convex set?



Matrices whose entries are essentially traces of products of unitary matrices.
I think I can give a partial answer. 


*(I will use the OP's notation of + for the adjoint)*


1) It is clear that any "good" matrix has every diagonal entry equal to 1. 


2) If A is "good", then it is positive. Indeed, being good means that there exist $k\in\mathbb{N}$ and $U\_1,\ldots,U\_n$ unitaries in $M\_k(\mathbb{C})$ with 
$$
A=\frac1k\,\mbox{tr}^{(n)}\left(\begin{bmatrix}U\_1\\ U\_2\\
\vdots \\ U\_n\end{bmatrix} \,
\begin{bmatrix}U\_1\\ U\_2\\
\vdots \\ U\_n\end{bmatrix}^+\right),
$$
where $\mbox{tr}^{(n)}$ is the map that replaces each $k\times k$ block by its trace. Since
the trace has abelian range it is completely positive, so $\mbox{tr}^{(n)}$ is positive and thus
$A$ is positive.


3) For $n=2$, the converse holds (i.e. any positive matrix with 1 in the diagonal is "good"). Indeed, let $A=\begin{bmatrix}1&a\\ a&1\end{bmatrix}$, with $|a|\leq1$. Let $k=2$, $U\_1=I\_2$,
$U\_2=\begin{bmatrix}a&\sqrt{1-|a|^2}\\ -\sqrt{1-|a|^2}&a\end{bmatrix}$. Then
$$
A=\frac12\,\mbox{tr}^{(n)}\left(\begin{bmatrix}U\_1\\ U\_2\end{bmatrix} \,
\begin{bmatrix}U\_1\\ U\_2\end{bmatrix}^+\right).
$$


4) So the question remains, is every $n\times n$ positive matrix with diagonal 1, "good"? 
For $n\geq3$, I still cannot see too much in this direction. On the one hand, being free to choose $k$ gives an incredible amount of unitaries to play with. On the other, positivity is hard to make precise as a relation among the coefficients, and so I cannot see an obvious way to choose the unitaries in a satisfactory way (somehow the restriction given by positivity should appear in the choice of the unitaries, and I couldn't make sense of it).


5) If the conjecture were true (and it is for $n=2$) this makes the set $S$ closed (a norm-limit of positive matrices with diagonal one is going to be again positive with diagonal 1). Otherwise, from the definition alone, I don't really see it. 


