I've come across this problem in Nielsen & Chuang's Quantum Information book (problem 2.64)
Suppose Bob is given a quantum state chosen from a set $|ψ_1 \rangle, . . . , |ψ_m\rangle$ of linearly independent states. Construct a POVM $\{E_1, E_2, . . ., E_{m+1}\}$ such that if outcome $E_i$ occurs, $1 ≤ i ≤ m$, then Bob knows with certainty that he was given the state $|ψ_i\rangle$. (The POVM must be such that $\langle ψ_i|E_i|ψ_i \rangle > 0$ for each $i$.)
This is my proposed solution:
Denote by $|\phi_i\rangle$ the (unique? I guess it doesn't matter) vector orthogonal to the subspace spanned by $\{ | \psi_j \rangle \}_{j \neq i}$ and define
$$E_i = \sum_{i\neq j} | \phi_j \rangle \langle\phi_j |$$
Then $\langle \psi_j | E_i | \psi_j\rangle = 0$ by construction, and $\langle \psi_i | E_i | \psi_i\rangle > 0$. The last operator is defined to satisfy completeness:
$$E_{m+1} = \mathbb{I} - \sum_{j=1}^m E_j.$$
So, when get gets outcome $i$, he knows it can't have been any of the other $\psi_j$'s, so it must have been $\psi_i$ for sure. If he gets outcome $m+1$, he doesn't know anything. Is this correct?
What happens if now we introduce another vector to the set: $\psi \rangle = a |ψ_1 \rangle + b |ψ_2 \rangle$, i.e. drop the linear independence condition (just on a simple example here). How would that affect the $E_j$'s, is it still possible to construct a POVM like that?