Problem statement
Given is the density matrix of a spin-1/2 system which was set up in a state of superposition $$ \varrho = \begin{pmatrix} \frac{3}{4} & \frac{1}{4} \\ \frac{1}{4} & \frac{1}{4} \\ \end{pmatrix} $$ Now I'd like to find convex combinations, such that $$ \varrho = \sum\limits_{i=1}^{N} p_i \; \chi_i \chi_i^\dagger $$ where $p_i \in [0,1]$ and $\chi_i \in \mathbb{C}^2$. Furthermore the $\chi_i$ need to be normalized, i.e. $$ \| \chi_i \| = 1 $$ and also $$ \sum_i p_i = 1 $$ The $\chi_i$ don't need to be orthogonal.
I don't even know where to start, because this case is not as easy as finding linear combinations, where there is no such condition as $p_i \in [0,1]$. How do I satisfy this condition?
Remarks
$(\cdot)^\dagger$ is the hermite conjugate.
$\chi_i \chi_i^\dagger$ denotes a matrix product. Example: $$ \chi_i = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \\ \implies \chi_i \chi_i^\dagger = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} $$
I believe you can just use unitary diagonalization to find such a decomposition, and the fact that $\varrho$ is a density matrix should guarantee that it is in fact a convex combination. In particular we need the fact that $\varrho$ is positive-semidefinite Hermitian with trace $1$ - according to Wikipedia this is true for anything one would call a density matrix, and it is certainly true for your example.
Apply the spectral theorem to $\varrho$: since it is Hermitian, it has an orthonormal basis of eigenvectors $v_1,v_2$ with corresponding real eigenvalues $\lambda_1, \lambda_2$. We can thus write $\varrho = \sum_{i=1}^2 \lambda_i v_i v_i^\dagger$.
Now, why does this decomposition satisfy your convexity requirements?