I need to construct a non-negative matrix with desired eigenvalues. To that end, I came up with a block matrix of the following form:

$$ \mathbf{M} = \begin{vmatrix} \mathbf{A} & \mathbf{b} \\\ \mathbf{c^T} & \it{d} \end{vmatrix} $$

Where matrix $\mathbf{A} \in \mathbb{R}^{n \times n}$ is free parameter, and vectors $\mathbf{b}$, $\mathbf{c} \in \mathbb{R}^{n}$ and scalar $\it{d}$ depend on $\mathbf{A}$ and a set of desired eigenvalues (given as diagonal elements of matrix $\mathbf{\Lambda} \in \mathbb{R}^{n \times n}$)

Equation that relates $\mathbf{c}$ to $\mathbf{A}$ and $\mathbf{\Lambda}$ is quite complicated and involves an inverse of the solution to a Sylvester equation. Namely

$$ \mathbf{c^T} = \left[\text{diag}(\mathbf{\Lambda}) - \mathbf{1^T}(\text{tr}(\mathbf{\Lambda}-\mathbf{A}) + \sigma)\right]\mathbf{K^{-1}} $$

where $\mathbf{K}$ is a solution to a following Sylvester equation:

$$ \mathbf{A}\mathbf{K}-\mathbf{K}\mathbf{\Lambda} = (\mathbf{A}-\sigma\mathbf{I}_n)\mathbf{1} \cdot\mathbf{1^T} $$

Now, I want to find constraints on parameter $\mathbf{A}$ which will render $\mathbf{M}$ non-negative. $\mathbf{A}$ itself must be non-negative of course, and $\mathbf{b}$ and $d$ have a straightforward dependence on $\mathbf{A}$, so it's easy. I have no idea, however, what to do with $\mathbf{c}$, because there's this $\mathbf{K^{-1}}$ which I don't have a solution for (only for $\text{vec}(\mathbf{K})$). How could I approach the question?

If needed, here's the dependence of $\mathbf{b}$ and $d$ on $\mathbf{A}$ and $\mathbf{\Lambda}$:

$$ \mathbf{b} = (\sigma \mathbf{I}_n - \mathbf{A})\mathbf{1} $$

$$ d = \text{tr}(\mathbf{\Lambda} - \mathbf{A}) + \sigma $$

and a solution for $\mathbf{K}$

$$ \text{vec}(\mathbf{K}) = (\mathbf{I}_n \otimes\mathbf{A} - \mathbf{\Lambda} \otimes \mathbf{I}_n)^{-1}(\mathbf{1} \otimes \mathbf{b}) $$

Given these relations, the resulting matrix $\mathbf{M}$ is equivalent to $$ \begin{vmatrix} \mathbf{A} & \mathbf{b} \\\ \mathbf{c^T} & \it{d} \end{vmatrix} = \begin{vmatrix} \mathbf{K} & \mathbf{1} \\\ \mathbf{1^T} & 1 \end{vmatrix} \begin{vmatrix} \mathbf{\Lambda} & \mathbf{0} \\\ \mathbf{0} & \sigma \end{vmatrix} \begin{vmatrix} \mathbf{K} & \mathbf{1} \\\ \mathbf{1^T} & 1 \end{vmatrix}^{-1} $$

As a last remark, $\mathbf{\Lambda}$ need not be diagonal, but its eigenvalues must be chosen as an input argument.