 Solution

Since $M⪰0$ but $M\notin S$, its eigenvalues satisfyThe continuous decreasing functionfalls from $q(0)>s$ to zero, so there is a unique $\rho >0$ for which $q(\rho )=s$.

PutThen $\Pi ⪰0$ and $\mathrm{Tr}\Pi ={\sum }_i{p}_i=s$, so $\Pi \in S$. Moreoverand thereforeFor any $Z\in S$,Whenever $p_i>0$, the corresponding eigenvalue of $A$ is exactly $\rho$, while terms with $p_i=0$ contribute nothing. ConsequentlyIt follows thatPart (a), equivalently the formula for projection onto a positive semidefinite trace ball, now gives