Matrices whose entries are essentially traces of products of unitary matrices. 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?
 A: I'm reading this question almost a year after it was asked, but my answer might be useful.
The set that you call good matrices is studied in the paper by Dykema and Juschenko Matrices of unitary moments, under the notation $\mathcal F_n$. The motivation for the paper is the Connes embedding problem. And as far as I understand it, its conclusion is probably that a very precise description of $\mathcal F_n$ for all $n$ is very difficult (because such a description would perhaps lead to an answer to Connes's question).
But the paper already answers some of your and Martin's questions. Among other, they prove that $\mathcal F_n$ is closed (hence compact), convex, stable under Schur (or Hadamard) product (Proposition 1.4).
They compare $\mathcal F_n$ with the set $\Theta_n$ of positive matrices with $1$ on the diagonal. As you notice, $\mathcal F_n$ and $\Theta_n$ coincide if $n=1,2,3$. But for $n \geq 4$ they differ (Corollary 2.11).
However, interestingly, $\mathcal F_n \cap M_n(\mathbb R) = \Theta_n \cap M_n(\mathbb R)$ (Theorem 3.1). 
A: 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. 
