Random variate of a singular Wishart distribution with non-integral degrees of freedom Let's say that I have a random variable $X$ that follows a singular Wishart distribution with $\nu$ degrees of freedom and a shape matrix $\Sigma$ that is $p\times p$. We can write "symbolically":
$$X\sim W_p(\nu,\Sigma)$$
where $\nu\le p-1$ for the singular case.
If $\nu$ is integral then I can get a random variate by simply taking $\nu$ independent random normal variables $Y_i=\mathcal N(0,\Sigma)$, where $\mathcal N(0,\Sigma)$ is the normal distribution with no mean and a covariance of $\Sigma$. Then my random variate will be:
$X=\sum_{i=1}^\nu Y_iY_i'$. If $\nu$ is less or equal to $p-1$, this random variate will be non-invertible with probability of one.
Let's say that I would like to get a random variate for a non-integral value of $\nu$, for example $\nu=\frac12$. In this situation, I obviously can't use the summation of normal variables. What would be the appropriate way to get a random variate for such value of $\nu$?
Also, I am not sure it makes perfect sense to have a random variate for $\nu=\frac12$ as if I sum two independent variates of this kind then I can't ensure that the rank of my random variate will be $1$ as what it should be normally ($1=\nu+\nu$).

The answer to this question is negative, we must have an integral number of degrees of freedom $\nu$ if $\nu\le p-1$. To prove it, we can look at the singular Wishart, defined as:
$$W_p(\nu,\Sigma)=\sum_{i=1}^nY_iY_i'$$
with $Y_i\sim\mathcal N(0,\Sigma)$, and $\Sigma$ a positive definite matrix.
The rank of the random variate will be $\nu$ almost surely if $\nu\le p-1$ and $\nu$ is integral. In order to keep the property that $W_p(\nu_1,\Sigma)+W_p(\nu_2,\Sigma)=W_p(\nu_1+\nu_2,\Sigma)$ for all $\nu_1$ and $\nu_2$, we must ensure that $\nu_1$ and $\nu_2$ are integral numbers. This can be seen from the definition of the singular Wishart above.
We can try to prove that there exists a way to define $W_p(\frac12,\Sigma)$ so adding two of these independent random variates will create a matrix of rank $1$ but that requires the two variates to share the same (degenerate) eigenspace. Unless we constraint it explicitly then we can't have this property. So we need to have $\nu$ integral until we reach full rank.
