I've been trying to solve exercise 2.73 (p.g 105) in Nielsen Chuang, and I'm not sure if i'v been overthinking it and the answer is as simple as i've described below or if I am missing something, or i'm just wrong!
Ex 2.73:
Let $\rho$ be a density operator. A minimal ensemble for $\rho$ is an ensemble $\{p_i,|\psi_i\rangle\}$ containing a number of elements equal to the rank of $\rho$. Let $|\psi\rangle$ be any state in the support of $\rho$. Show that there is a minimal ensemble for $\rho$ that contains $|\psi\rangle$, and moreover that in any such ensemble $|\psi\rangle$ must appear with probability
$p_i=\frac{1}{\langle\psi_i|\rho^{-1}|\psi_i\rangle}$
where $p^{-1}$ is defined to be the inverse of $\rho$, when $\rho$ is considered as an operator acting only on the support of $\rho$
My answer so far is:
$\rho$ is positive the therefore has a spectral decomposition $\rho=\sum_k\lambda_k|k\rangle\langle k|$.
The density operator cann be defined as $\rho=\sum_kp_k|k\rangle\langle k| = \sum_k|\hat{k}\rangle\langle \hat{k}|$, where $|\hat{k}\rangle=\sqrt{\lambda_k}|k\rangle$, and therefore $|k\rangle = \frac{|\hat{k}\rangle}{\sqrt{\lambda_k}} $.
For any $|\psi_i\rangle = \sum_k c_{ik}|k\rangle$, using the above definition of $|k\rangle$:
$|\psi_i\rangle = \sum_k \frac{c_{ik}}{\sqrt{\lambda_k}}|\hat{k}\rangle$
The density operator is given by $\rho=\sum_i|\psi_i\rangle\langle\psi_i|$, therefore
$\rho = \sum_{i}\sum_{k}\frac{c_{ik}^2}{\lambda_k}|\hat{k}\rangle \langle\hat{k}|$.
By the definition of $\rho$ is can be seen that $p_i = \sum_{k}\frac{c_{ik}^2}{\lambda_k}$.
--- reading this back i'm not sure this is correct at all :(
For the second part working backwards a bit:
$\langle \psi_i|\rho^{-1}|\psi_i\rangle = \langle \psi_i|\sum_k \left( \frac{1}{\lambda_k}|k\rangle\langle k| \right) |\psi_i\rangle = \sum_k \frac{1}{\lambda_k}\langle \psi_i|k\rangle\langle k |\psi_i\rangle = \sum_{i,k} \frac{1}{\lambda_k}c_{i,k}^2\langle i|k\rangle \langle k |i\rangle $
Given that $|i\rangle$ is of basis $|k \rangle$, $\langle k |i\rangle = \langle i |k\rangle = 1 $ if $i=k$, therefore
$\langle \psi_i|\rho^{-1}|\psi_i\rangle = \sum_{k} \frac{c_{i,k}^2}{\lambda_k}$, so it then follows that
$p_i = \frac{1}{\sum_{k} \frac{c_{i,k}^2}{\lambda_k}}$
However the above result does not match with the result I got for $p_i$ in the first part, so one of them is wrong...