Probability of making an error in identifying quantum state

Question

In my notes I have the following:

In a measurement described by POVM $\{\pî_j\}$, we associate $\pî_j$ with state $\rhô_j$, that is if outcome $j$ is obtained, we take this to indicate that the state sent was $\rhô_j$.

The probability of making an error is: $$ P_{\text{err}} = 1 - > P_{\text{corr}} = 1 - \sum_j P(\rhô_j) P(j | \rhô_j) = 1 - \sum_j > p_j \operatorname{Tr}(\rhô_j \pî_j) $$

Any POVM minimizing error satisfies: $$1. \ \ \sum_i p_i \rhô_i \pî_i -p_j \rhô_j \geq 0 \quad \forall j$$ $$2. \ \ \pî_i (p_i \rhô_i - p_j > \rhô_j) \pî_j = 0 \quad \forall i,j $$

The first is necessary and sufficient, while the second is necessary but not sufficient.

Consider the case of just two states $\rhô_0$ and $\rhô_1$. The conditions reduce to, according to my notes: $$\\pî_0 (p_0 \rhô_0 - p_1 \rhô_1) \pî_1 = 0 $$

However I cannot get this from the first equations:

$$\sum_i p_i \rhô_i \pî_i - p_j \rhô_j \geq 0 \quad \rightarrow \quad (p_1 \rhô_1 \pî_1 + p_2 \rhô_2 \pî_2) - p_2 \rhô_2 \geq 0 \quad \text{and} \quad (p_1 \rhô_1 \pî_1 + p_2 \rhô_2 \pî_2) - p_1 \rhô_1 \geq 0$$

Also it says "We find that $\pî_0$, $\pî_1$ are projectors onto the eigenstates of the operator $p_0 \rhô_0 - p_1 \rhô_1$ with positive ($\pî_0$) and negative ($\pî_1$) eigenvalues". How do we get that from this?

Probability of making an error in identifying quantum state

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