nonegative inverse eigenvalue problem 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:
$$
\mathbf{A}\mathbf{K}-\mathbf{K}\mathbf{\Lambda} = (\mathbf{A}-\sigma\mathbf{I}_n)\mathbf{1} \cdot\mathbf{1^T}
$$
Questions:
How 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 analytically? Or is there an efficient numerical way to solve the problem (e.g. given $\mathbf{\Lambda}$, find $A$ that maximizes the smallest element of $M$)?

Details:
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 vectorized 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
$$
\mathbf{M} = 
\begin{vmatrix}
\mathbf{A} & \mathbf{b} \\
\mathbf{c^T} & \it{d} \\
\end{vmatrix} = 
\begin{vmatrix}
\mathbf{K} & \mathbf{1} \\
\mathbf{1^T} & 1 \\
\end{vmatrix}
\begin{vmatrix}
\mathbf{\Lambda} & \mathbf{0} \\
\mathbf{0} & \sigma \\
\end{vmatrix}
\begin{vmatrix}
\mathbf{K} & \mathbf{1} \\
\mathbf{1^T} & 1 \\
\end{vmatrix}^{-1}
$$
$\mathbf{\Lambda}$ need not be diagonal, but its eigenvalues must be chosen as an input argument.

UPD: ok, sorry for a confusing question. I'll try to clear things up with a 2 $\times$ 2 special case. It can be shown that a 2 $\times$ 2 matrix of the form
$$
\mathbf{M} = 
\begin{vmatrix}
a & \sigma - a \\
a - \lambda & \sigma - (a - \lambda) \\
\end{vmatrix}
$$
has equal row-sum ($\sigma$) and a set of eigenvalues $\{\lambda, \sigma\}$ for any arbitrary choice of $a$. Now, because all equations are scalar, it is easy to find constraints on $a$ and $\lambda$ that will result in a non-negative $\mathbf{M}$ by solving a simple system of inequalities. The result is, for $\mathbf{M}$ to be non-negative
$$
\begin{cases}
\begin{align}
0 \le & a \le \sigma\\
\lambda \le & a \le \lambda + \sigma \\
-\sigma < &\lambda < \sigma \\
\end{align}
\end{cases}
$$
The original question is an attempt to generalize this problem to $(n+1)\times(n+1)$, where $a$ becomes $\mathbf{A} \in \mathbb{R}^{n \times n}$ and so forth. You can see that equations for $\mathbf{b}$, $\mathbf{c}$ and $d$ in the first section of the post collapse to the scalar equations for entries of $\mathbf{M}$ here, when $n$ is set to 1.
 A: Forgive me if I have completely misunderstood you, but it seems in the comments you are after a matrix (call it $\mathbf{M}$)  that has:


*

*some desired eigenvalues

*equal row sum.


If they are the only constraints, consider a square arbitrary diagonalizable matrix, $\mathbf{M}$:
$$\mathbf{M}=\mathbf{S}\mathbf{D}\mathbf{S}^{-1}.$$
The column vectors of $\mathbf{S}$ are the right eigenvectors of matrix $\mathbf{M}$ and $\mathbf{D}$ is the diagonal matrix containing the eigenvalues of $\mathbf{M}$. To construct your $\mathbf{M}$ you would then simply fill $\mathbf{D}$ with the desired eigenvalues and then, to insure that the row sums of $\mathbf{M}$ are equal, as you noted, you make every element of one of the column vectors of $\mathbf{S}$ (call it $E_1$) to be equal to $1$, then fill up the rest of $\mathbf{S}$ however you please so long as $\text{det}[\mathbf{S}]\ne0$, then finally calculate the inverse $\mathbf{S}^{-1}$ and you have your $\mathbf{M}$. The row sum would then be equal to the eigenvalue (call it $e_1$) corresponding to $E_1$ as can be seen by noting that the right hand side of: $$e_1E_1=\mathbf{M}E_1$$ gives a column vector containing the row sums and the left hand side is a column vector full of eigenvalue $e_1$.
Example:
Take $\mathbf{D}$ as the matrix containing your given set of eigenvalues on the diagonal:
$$\mathbf{D}= \left[ \begin {array}{ccc} \sigma&0&0\\ 0&1&0
\\ 0&0&2\end {array} \right], $$
then construct $\mathbf{S}$:
$$\mathbf{S}= \left[ \begin {array}{ccc} 1&2&3\\ 1&3&4
\\ 1&4&6\end {array} \right],$$ 
then find the inverse:
$$ \mathbf{S}^{-1}=\left[ \begin {array}{ccc} 2&0&-1\\ -2&3&-1
\\ 1&-2&1\end {array} \right], 
$$
then build $\mathbf{M}$:
$$ \mathbf{M}=\mathbf{S}\mathbf{D}\mathbf{S}^{-1}=\left[ \begin {array}{ccc} 2\,\sigma+2&-6&-\sigma+4
\\ 2\,\sigma+2&-7&-\sigma+5\\ 2\
\sigma+4&-12&-\sigma+8\end {array} \right], 
$$
$$\text{row sum}=\sigma.$$
I'm sure this is not what you're asking for as I ignored the Sylvestre equation e.t.c, but I'll post it just in case. I'll delete it when you highlight my ignorance...
A: This is not an answer, but rather a simplification of the problem that can't be fit into the comments. 
Without loss of generality, one can assume that $A$ is diagonal. Why? Well suppose $A = U \Sigma U^T$ is the (full) SVD of $A$. Then we can convert the problem to a diagonal one via the similarity transformation,
$$\begin{bmatrix}
U^T \\ & 1
\end{bmatrix}
\begin{bmatrix}
A & b \\ c^T & d
\end{bmatrix}
\begin{bmatrix}
U \\ & 1
\end{bmatrix} = 
\begin{bmatrix}
\Sigma & U^T b \\ c^T U & d
\end{bmatrix}.$$
Since similarity transforms preserve eigenvalues and since $b$ and $c$ could be anything, we might as well instead compute the necessary $f,g$ in
$$\begin{bmatrix}
\Sigma & f \\ g^T & d
\end{bmatrix}.$$
