How do I show the separability of this density matrix? I am stuck since a longer time regarding this exercise where I need to work with a density matrix of the given form
$$\displaystyle \rho_{AB}(X)= \frac{1}{N+\text{tr}X^2}\left( {\begin{array}{cc}
    I & X \\
   X & X^2 \\
  \end{array} } \right)$$
where $X$ is an $N\times N$ hermitian matrix in the quantum system $B$ with the Hilbert space $\mathbb{C}^N$, $I$ is the identity matrix and $\rho_{AB}$ a state on the quantum system $AB$ with Hilbert space $\mathbb{C}^2 \otimes \mathbb{C}^N$.
Question: How do I show now, that $\rho_{AB}(X)$ is separable by using the spectral theorem on $X$?
 A: Let $|i\rangle$ be the eigenbasis of the Hilbert space of system B which diagonalizes $X$
$$
X = \sum_{i=1}^N x_i |i\rangle\langle i|
$$
where $x_i \in \mathbb{R}$ and define
$$
|\psi_k\rangle = \frac{1}{\sqrt{1 + x_k^2}} (|0\rangle + x_k |1\rangle)
$$
$$
\rho_k = |\psi_k\rangle\langle\psi_k| = \frac{1}{1 + x_k^2}\begin{pmatrix}
1 & x_k \\
x_k & x_k^2
\end{pmatrix}.
$$
Now, note that
$$
\rho_{AB}(X) = \sum_{i=1}^N \lambda_i \rho_i \otimes |i\rangle\langle i|\tag1
$$
where
$$
\lambda_i = \frac{1 + x_i^2}{N + \mathrm{tr}X^2}
$$
and the right hand side of $(1)$ is separable.

In order to see the equality $(1)$, consider each of the four blocks of the matrix $\rho_{AB}(X)$ in turn. The top left block is
$$
\langle 0|\rho_{AB}(X)|0\rangle = \frac{1}{N + \mathrm{tr}X^2} I.
$$
The bottom right block is
$$
\langle 1|\rho_{AB}(X)|1\rangle = \frac{x_i^2}{N + \mathrm{tr}X^2} \sum_{i=1}^N |i\rangle\langle i| = \frac{1}{N + \mathrm{tr}X^2} \sum_{i=1}^N x_i|i\rangle\langle i| x_i|i\rangle\langle i| = \frac{1}{N + \mathrm{tr}X^2} X^2.
$$
The top right block is
$$
\langle 0|\rho_{AB}(X)|1\rangle = \frac{x_i}{N + \mathrm{tr}X^2} \sum_{i=1}^N |i\rangle\langle i| = \frac{1}{N + \mathrm{tr}X^2} \sum_{i=1}^N x_i|i\rangle\langle i| = \frac{1}{N + \mathrm{tr}X^2} X
$$
and similarly for the bottom left block.
