Let $A,B$ be positive matrices on a finite-dimensional space, and suppose that $A+B=I$.
In the special case of $A,B$ being projectors, we know that this implies that they must be orthogonal, as shown for example here and links therein.
Can something be said about the more general case of $A,B\ge0$?
If $A,B$ have orthogonal support, it is not hard to see that they must each equal the identity on their supports. We can therefore, I think, restrict ourselves to consider only the cases in which $\operatorname{im}(A)=\operatorname{im}(B)$, as we know that the restriction of the operators on every subspace in which only one of the two acts equals the identity (more precisely, one of the two operators will act like the identity and the other like $0$).
Best Answer
On second thought, this is actually trivial. If $A+B=I$ then $B=I-A$, thus $[A,B]=0$. This means $A$ and $B$ are always mutually diagonalisable and such that each pair of eigenvalues corresponding to the same eigenvector sums to $1$.
Indeed, this is not even about positive matrices. The same kind of argument works for arbitrary diagonalisable matrices summing to the identity.
More generally, if $A$ is diagonalisable, and $A+B=I$, then $B$ is also diagonalisable, and in some basis we have $A=\operatorname{diag}(\lambda_1,...,\lambda_n)$ and $B=\operatorname{diag}(1-\lambda_1,...,1-\lambda_n)$. If furthermore $A\ge0$ is (Hermitian and) positive semidefinite, then there is an orthonormal basis with respect to which $A=\operatorname{diag}(\lambda_1,...,\lambda_n)$ with $0\le \lambda_i\le 1$, and again $B=\operatorname{diag}(1-\lambda_1,...,1-\lambda_n)$.