Sketch of proof: We need the following fact:
Fact: Any density operator has a minimal ensemble.
This can be proven with the spectral theorem. However, this is the only thing that we need the spectral theorem for.
Recall that an operator $\rho$ is positive semidefinite iff for all vectors $|\phi \rangle$, we have $\langle \phi |\rho | \phi \rangle > 0$. For part one, use the fact that $\langle \psi|\rho|\psi \rangle$ to show that there exists an $\alpha$ with $0<\alpha<1$ for which $\sigma = \rho - \alpha |\psi \rangle \langle \psi |$ is positive semdefinite. Now, take any ensemble $\{p_i,|\psi_i\rangle \}$ for $\hat \sigma = \frac{\sigma}{1-\alpha}$, and show that $\{(1-\alpha)p_i,|\psi_i\rangle\} \cup \{\alpha, |\psi\rangle\}$ is an ensemble for $\rho$.
I'm not quite sure what they're asking for part $2$, but here are my thoughts. Let $r$ denote the rank of $\rho$, and let $\alpha_* = \frac{1}{\langle \psi |\rho^{-1}|\psi\rangle}$. It suffices to note/show that $\sigma = \rho - \alpha |\psi \rangle \langle \psi |$ will fail to be positive semidefinite for $\alpha > \alpha_*$, and that $\sigma$ will have rank $r$ (instead of $r-1$) when $\alpha < \alpha_*$.
Proof that this is the correct value for $\alpha_*$: With the Schur complement, we see that $\rho - \alpha xx^\dagger$ is positive semidefinite iff the matrix
$$
M = \pmatrix{\rho & x\\x^\dagger & \alpha^{-1}}
$$
is positive semidefinite. By taking the Schur complement relative to $\rho$, we find that $M$ is positive semidefinite iff $\alpha^{-1} - x^\dagger\rho^{-1}x \geq 0$, which is to say that $\alpha \leq \alpha_* = \frac{1}{x^\dagger\rho^{-1}x}$, as was desired.
A matrix version of the proof given on the QIT SE site:
Let $D = \operatorname{diag}(p_1,\dots,p_r)$, and let $a_1,\dots,a_r$ be the linearly independent vectors (corresponding to $\hat \psi_i = \sqrt{p_i}\psi_i$. Let $A$ be the matrix with columns $A$; we have $\rho = AA^\dagger$. Note that
$$
A = \rho \rho^{-1} A = AA^\dagger \rho^{-1} A= A[A^\dagger\rho^{-1}A].
$$
$A$ has linearly independent columns and is therefore left-cancellable. Conclude that $A^\dagger \rho^{-1} A = I_{r}$. If we consider the $j,j$ entry, we have
$$
1 = a_j^\dagger\rho^{-1}a_j \leadsto 1 = \langle \hat \psi_j | \rho^{-1} | \hat \psi_j \rangle = p_j \cdot \langle \psi_j | \rho^{-1} | \psi_j \rangle.
$$
Second version of the matrix proof: $\rho - \alpha xx^\dagger$ is positive semidefinite iff $\rho^{-1/2}[\rho - \alpha xx^\dagger] \rho^{-1/2} = I - \alpha (\rho^{-1/2}x)(\rho^{-1/2}x)^\dagger$ is postive semidefinite. It's easy to see that the lowest eigenvalue of this matrix is $1-\lambda$, where $\lambda$ is the largest eigenvalue of $\alpha (\rho^{-1/2}x)(\rho^{-1/2}x)^\dagger$. Because this matrix has rank $1$, we see that
$$
\lambda = \operatorname{Tr}(\alpha (\rho^{-1/2}x)(\rho^{-1/2}x)^\dagger) =
\alpha (\rho^{-1/2}x)^\dagger) (\rho^{-1/2}x) = \alpha x^\dagger \rho^{-1} x.
$$
We reach a threshold at $\lambda = 1$, i.e. $\alpha = \frac{1}{x^\dagger \rho^{-1} x}$.
I figured it out i think. The statement is simply untrue.
Consider the example
$$\rho_1 =\frac{1}{2} ( | 1 \rangle \langle1| +|2 \rangle \langle 2 | ),$$
$$\rho_2 = |2 \rangle \langle 2 |. $$
Where $| i \rangle \in \mathcal{H} , i=1, 2$ are an orthonormal system.
Then
$$2 \rho_1 + (1-2)\rho_2 = | 1 \rangle \langle1|.$$
A pure state. So convexity of the lin. combination is necessary on the RHS of the problem.
Best Answer
Since $\langle \psi|\psi\rangle=1$, we have $$ (|\psi\rangle\langle\psi|)^2=|\psi\rangle\langle\psi|\psi\rangle\langle\psi|=|\psi\rangle\langle\psi| $$ Then $|\psi\rangle\langle\psi|$ is a projection, and its eigenvalues are $0$ and $1$. Since we know that the trace of $|\psi\rangle\langle\psi|$ is one we conclude that, counting multiplicities, the eigenvalues of $|\psi\rangle\langle\psi|$ are $1,0,\ldots,0$. So its diagonal form is \begin{equation} \hat{\rho}= {\begin{bmatrix} 1 & 0 & \cdots & 0 \\ 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 0 \\ \end{bmatrix}} \end{equation}