Are orthogonal positive semidefinite matrices each with trace equal to one necessarily rank-one projections?

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Let $n$ be a positive integer and let $P_1,\dots,P_n\in M_n(\mathbb{C})$ be a collection of positive semidefinite $n\times n$ matrices satisfying the following properties:

  • $\operatorname{Tr}P_i=1$ for each $i\in\{1,\dots,n\}$
  • $P_iP_j = 0$ whenever $i\neq j$.

It is necessarily the case that there is an orthonormal basis $u_1,\dots,u_n$ such that $P_i = u_i u_i^*$ for each $i\in\{1,\dots,n\}$?


My thoughts:

  • The matrices $P_1,\dots,P_n$ are normal and commuting, so they are simultaneously diagonalizable. In particular, we might as well assume without loss of generality that the are diagonal.
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Well, you've gotten to the point where they are diagonal. Note that a product of diagonal matrices is $0$ iff they do not share non-zero entries, i.e. for any $i \neq j$ and $1 \leq k \leq n$ we get $(P_i)_{k,k} \neq 0 \Rightarrow (P_j)_{k,k} = 0$. Use the pigeon hole principle to conclude.