1$. So, $VV^{\dagger}$ satisfies the conditions of a density matrix (with rank-$1$).

Now, let $V_1, \ldots, V_d$ be $d$ normalized vectors in $\mathbb{C}^{d\times 1}\setminus \{\mathbf{0}\}$ and $\gamma_1,\ldots
,\gamma_d$ be non-negative weights with $ \sum_{j=1}^d\gamma_j =1$. Put $$\label{formula}
\nu = \sum_{i=1}^d \gamma_i V_i V_i^{\dagger} .$$ Then $ \nu $ is clearly a density matrix: it is Hermitian (as a sum of Hermitian matrices), it is semi-definite positive (same reason) and $${\rm Tr}(\nu) = \sum_{i=1}^d \gamma_i {\rm Tr}
(V_i V_i^{\dagger}) =1 .$$

Moreover, note that any density matrix can be written in such way, as we know that for any density matrix $\rho$, $$\label{formula-diag}
\rho= U \Lambda U^{\dagger}$$ and just write $U=(U_1|\dots|U_d)$ with the $U_i$’s being [*orthogonal*]{}, where $ \Lambda = {\rm diag}
(\Lambda_1, \ldots, \Lambda_n): \Lambda_1
\geq \ldots \geq \Lambda_n \geq 0,
\sum_{i = 1}^d  \Lambda_i = 1$.

The only difference in (\[formula\]) is that we do not require that the $V_i$’s are orthogonal. Thus, it is easier to simulate a matrix $\rho$ by simulating the $V_i$’s and $\gamma_i's$ in  than by simulating $U$ and $\Lambda$ in . Also, note that the $\gamma_i$’s are not necessarily the eigenvalues of $\rho$.

We define the prior definition on $\rho$, $\pi({\rm d}\rho)$, by $$\begin{aligned}
 